hypothesis h0 and h1

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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Module 9: Hypothesis Testing With One Sample

Null and alternative hypotheses, learning outcomes.

• Describe hypothesis testing in general and in practice

The actual test begins by considering two  hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 : The null hypothesis: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt.

H a : The alternative hypothesis : It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make adecision. There are two options for a  decision . They are “reject H 0 ” if the sample information favors the alternative hypothesis or “do not reject H 0 ” or “decline to reject H 0 ” if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in  H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30

H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

H 0 : The drug reduces cholesterol by 25%. p = 0.25

H a : The drug does not reduce cholesterol by 25%. p ≠ 0.25

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:

H 0 : μ = 2.0

H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 66 H a : μ __ 66

• H 0 : μ = 66
• H a : μ ≠ 66

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

H 0 : μ ≥ 5

H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 45 H a : μ __ 45

• H 0 : μ ≥ 45
• H a : μ < 45

In an issue of U.S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

H 0 : p ≤ 0.066

H a : p > 0.066

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : p __ 0.40 H a : p __ 0.40

• H 0 : p = 0.40
• H a : p > 0.40

Concept Review

In a  hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using less than, greater than, or not equals symbols, i.e., (≠, >, or <). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.

Formula Review

H 0 and H a are contradictory.

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Hypothesis Testing – A Deep Dive into Hypothesis Testing, The Backbone of Statistical Inference

• September 21, 2023

Explore the intricacies of hypothesis testing, a cornerstone of statistical analysis. Dive into methods, interpretations, and applications for making data-driven decisions.

In this Blog post we will learn:

• What is Hypothesis Testing?
• Steps in Hypothesis Testing 2.1. Set up Hypotheses: Null and Alternative 2.2. Choose a Significance Level (α) 2.3. Calculate a test statistic and P-Value 2.4. Make a Decision
• Example : Testing a new drug.
• Example in python

1. What is Hypothesis Testing?

In simple terms, hypothesis testing is a method used to make decisions or inferences about population parameters based on sample data. Imagine being handed a dice and asked if it’s biased. By rolling it a few times and analyzing the outcomes, you’d be engaging in the essence of hypothesis testing.

Think of hypothesis testing as the scientific method of the statistics world. Suppose you hear claims like “This new drug works wonders!” or “Our new website design boosts sales.” How do you know if these statements hold water? Enter hypothesis testing.

2. Steps in Hypothesis Testing

• Set up Hypotheses : Begin with a null hypothesis (H0) and an alternative hypothesis (Ha).
• Choose a Significance Level (α) : Typically 0.05, this is the probability of rejecting the null hypothesis when it’s actually true. Think of it as the chance of accusing an innocent person.
• Calculate Test statistic and P-Value : Gather evidence (data) and calculate a test statistic.
• p-value : This is the probability of observing the data, given that the null hypothesis is true. A small p-value (typically ≤ 0.05) suggests the data is inconsistent with the null hypothesis.
• Decision Rule : If the p-value is less than or equal to α, you reject the null hypothesis in favor of the alternative.

2.1. Set up Hypotheses: Null and Alternative

Before diving into testing, we must formulate hypotheses. The null hypothesis (H0) represents the default assumption, while the alternative hypothesis (H1) challenges it.

For instance, in drug testing, H0 : “The new drug is no better than the existing one,” H1 : “The new drug is superior .”

2.2. Choose a Significance Level (α)

When You collect and analyze data to test H0 and H1 hypotheses. Based on your analysis, you decide whether to reject the null hypothesis in favor of the alternative, or fail to reject / Accept the null hypothesis.

The significance level, often denoted by $α$, represents the probability of rejecting the null hypothesis when it is actually true.

In other words, it’s the risk you’re willing to take of making a Type I error (false positive).

Type I Error (False Positive) :

• Symbolized by the Greek letter alpha (α).
• Occurs when you incorrectly reject a true null hypothesis . In other words, you conclude that there is an effect or difference when, in reality, there isn’t.
• The probability of making a Type I error is denoted by the significance level of a test. Commonly, tests are conducted at the 0.05 significance level , which means there’s a 5% chance of making a Type I error .
• Commonly used significance levels are 0.01, 0.05, and 0.10, but the choice depends on the context of the study and the level of risk one is willing to accept.

Example : If a drug is not effective (truth), but a clinical trial incorrectly concludes that it is effective (based on the sample data), then a Type I error has occurred.

Type II Error (False Negative) :

• Symbolized by the Greek letter beta (β).
• Occurs when you accept a false null hypothesis . This means you conclude there is no effect or difference when, in reality, there is.
• The probability of making a Type II error is denoted by β. The power of a test (1 – β) represents the probability of correctly rejecting a false null hypothesis.

Example : If a drug is effective (truth), but a clinical trial incorrectly concludes that it is not effective (based on the sample data), then a Type II error has occurred.

Balancing the Errors :

In practice, there’s a trade-off between Type I and Type II errors. Reducing the risk of one typically increases the risk of the other. For example, if you want to decrease the probability of a Type I error (by setting a lower significance level), you might increase the probability of a Type II error unless you compensate by collecting more data or making other adjustments.

It’s essential to understand the consequences of both types of errors in any given context. In some situations, a Type I error might be more severe, while in others, a Type II error might be of greater concern. This understanding guides researchers in designing their experiments and choosing appropriate significance levels.

2.3. Calculate a test statistic and P-Value

Test statistic : A test statistic is a single number that helps us understand how far our sample data is from what we’d expect under a null hypothesis (a basic assumption we’re trying to test against). Generally, the larger the test statistic, the more evidence we have against our null hypothesis. It helps us decide whether the differences we observe in our data are due to random chance or if there’s an actual effect.

P-value : The P-value tells us how likely we would get our observed results (or something more extreme) if the null hypothesis were true. It’s a value between 0 and 1. – A smaller P-value (typically below 0.05) means that the observation is rare under the null hypothesis, so we might reject the null hypothesis. – A larger P-value suggests that what we observed could easily happen by random chance, so we might not reject the null hypothesis.

2.4. Make a Decision

Relationship between $α$ and P-Value

When conducting a hypothesis test:

We then calculate the p-value from our sample data and the test statistic.

Finally, we compare the p-value to our chosen $α$:

• If $p−value≤α$: We reject the null hypothesis in favor of the alternative hypothesis. The result is said to be statistically significant.
• If $p−value>α$: We fail to reject the null hypothesis. There isn’t enough statistical evidence to support the alternative hypothesis.

3. Example : Testing a new drug.

Imagine we are investigating whether a new drug is effective at treating headaches faster than drug B.

Setting Up the Experiment : You gather 100 people who suffer from headaches. Half of them (50 people) are given the new drug (let’s call this the ‘Drug Group’), and the other half are given a sugar pill, which doesn’t contain any medication.

• Set up Hypotheses : Before starting, you make a prediction:
• Null Hypothesis (H0): The new drug has no effect. Any difference in healing time between the two groups is just due to random chance.
• Alternative Hypothesis (H1): The new drug does have an effect. The difference in healing time between the two groups is significant and not just by chance.

Calculate Test statistic and P-Value : After the experiment, you analyze the data. The “test statistic” is a number that helps you understand the difference between the two groups in terms of standard units.

For instance, let’s say:

• The average healing time in the Drug Group is 2 hours.
• The average healing time in the Placebo Group is 3 hours.

The test statistic helps you understand how significant this 1-hour difference is. If the groups are large and the spread of healing times in each group is small, then this difference might be significant. But if there’s a huge variation in healing times, the 1-hour difference might not be so special.

Imagine the P-value as answering this question: “If the new drug had NO real effect, what’s the probability that I’d see a difference as extreme (or more extreme) as the one I found, just by random chance?”

For instance:

• P-value of 0.01 means there’s a 1% chance that the observed difference (or a more extreme difference) would occur if the drug had no effect. That’s pretty rare, so we might consider the drug effective.
• P-value of 0.5 means there’s a 50% chance you’d see this difference just by chance. That’s pretty high, so we might not be convinced the drug is doing much.
• If the P-value is less than ($α$) 0.05: the results are “statistically significant,” and they might reject the null hypothesis , believing the new drug has an effect.
• If the P-value is greater than ($α$) 0.05: the results are not statistically significant, and they don’t reject the null hypothesis , remaining unsure if the drug has a genuine effect.

4. Example in python

For simplicity, let’s say we’re using a t-test (common for comparing means). Let’s dive into Python:

Making a Decision : “The results are statistically significant! p-value < 0.05 , The drug seems to have an effect!” If not, we’d say, “Looks like the drug isn’t as miraculous as we thought.”

5. Conclusion

Hypothesis testing is an indispensable tool in data science, allowing us to make data-driven decisions with confidence. By understanding its principles, conducting tests properly, and considering real-world applications, you can harness the power of hypothesis testing to unlock valuable insights from your data.

More Articles

Correlation – connecting the dots, the role of correlation in data analysis, sampling and sampling distributions – a comprehensive guide on sampling and sampling distributions, law of large numbers – a deep dive into the world of statistics, central limit theorem – a deep dive into central limit theorem and its significance in statistics, skewness and kurtosis – peaks and tails, understanding data through skewness and kurtosis”, similar articles, complete introduction to linear regression in r, how to implement common statistical significance tests and find the p value, logistic regression – a complete tutorial with examples in r.

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing

Examples of null and alternative hypotheses

• Writing null and alternative hypotheses
• P-values and significance tests
• Comparing P-values to different significance levels
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• Estimating P-values from simulations
• Using P-values to make conclusions

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Null hypothesis significance testing: a short tutorial

Cyril pernet.

1 Centre for Clinical Brain Sciences (CCBS), Neuroimaging Sciences, The University of Edinburgh, Edinburgh, UK

Version Changes

Revised. amendments from version 2.

This v3 includes minor changes that reflect the 3rd reviewers' comments - in particular the theoretical vs. practical difference between Fisher and Neyman-Pearson. Additional information and reference is also included regarding the interpretation of p-value for low powered studies.

Peer Review Summary

Although thoroughly criticized, null hypothesis significance testing (NHST) remains the statistical method of choice used to provide evidence for an effect, in biological, biomedical and social sciences. In this short tutorial, I first summarize the concepts behind the method, distinguishing test of significance (Fisher) and test of acceptance (Newman-Pearson) and point to common interpretation errors regarding the p-value. I then present the related concepts of confidence intervals and again point to common interpretation errors. Finally, I discuss what should be reported in which context. The goal is to clarify concepts to avoid interpretation errors and propose reporting practices.

The Null Hypothesis Significance Testing framework

NHST is a method of statistical inference by which an experimental factor is tested against a hypothesis of no effect or no relationship based on a given observation. The method is a combination of the concepts of significance testing developed by Fisher in 1925 and of acceptance based on critical rejection regions developed by Neyman & Pearson in 1928 . In the following I am first presenting each approach, highlighting the key differences and common misconceptions that result from their combination into the NHST framework (for a more mathematical comparison, along with the Bayesian method, see Christensen, 2005 ). I next present the related concept of confidence intervals. I finish by discussing practical aspects in using NHST and reporting practice.

Fisher, significance testing, and the p-value

The method developed by ( Fisher, 1934 ; Fisher, 1955 ; Fisher, 1959 ) allows to compute the probability of observing a result at least as extreme as a test statistic (e.g. t value), assuming the null hypothesis of no effect is true. This probability or p-value reflects (1) the conditional probability of achieving the observed outcome or larger: p(Obs≥t|H0), and (2) is therefore a cumulative probability rather than a point estimate. It is equal to the area under the null probability distribution curve from the observed test statistic to the tail of the null distribution ( Turkheimer et al. , 2004 ). The approach proposed is of ‘proof by contradiction’ ( Christensen, 2005 ), we pose the null model and test if data conform to it.

In practice, it is recommended to set a level of significance (a theoretical p-value) that acts as a reference point to identify significant results, that is to identify results that differ from the null-hypothesis of no effect. Fisher recommended using p=0.05 to judge whether an effect is significant or not as it is roughly two standard deviations away from the mean for the normal distribution ( Fisher, 1934 page 45: ‘The value for which p=.05, or 1 in 20, is 1.96 or nearly 2; it is convenient to take this point as a limit in judging whether a deviation is to be considered significant or not’). A key aspect of Fishers’ theory is that only the null-hypothesis is tested, and therefore p-values are meant to be used in a graded manner to decide whether the evidence is worth additional investigation and/or replication ( Fisher, 1971 page 13: ‘it is open to the experimenter to be more or less exacting in respect of the smallness of the probability he would require […]’ and ‘no isolated experiment, however significant in itself, can suffice for the experimental demonstration of any natural phenomenon’). How small the level of significance is, is thus left to researchers.

What is not a p-value? Common mistakes

The p-value is not an indication of the strength or magnitude of an effect . Any interpretation of the p-value in relation to the effect under study (strength, reliability, probability) is wrong, since p-values are conditioned on H0. In addition, while p-values are randomly distributed (if all the assumptions of the test are met) when there is no effect, their distribution depends of both the population effect size and the number of participants, making impossible to infer strength of effect from them.

Similarly, 1-p is not the probability to replicate an effect . Often, a small value of p is considered to mean a strong likelihood of getting the same results on another try, but again this cannot be obtained because the p-value is not informative on the effect itself ( Miller, 2009 ). Because the p-value depends on the number of subjects, it can only be used in high powered studies to interpret results. In low powered studies (typically small number of subjects), the p-value has a large variance across repeated samples, making it unreliable to estimate replication ( Halsey et al. , 2015 ).

A (small) p-value is not an indication favouring a given hypothesis . Because a low p-value only indicates a misfit of the null hypothesis to the data, it cannot be taken as evidence in favour of a specific alternative hypothesis more than any other possible alternatives such as measurement error and selection bias ( Gelman, 2013 ). Some authors have even argued that the more (a priori) implausible the alternative hypothesis, the greater the chance that a finding is a false alarm ( Krzywinski & Altman, 2013 ; Nuzzo, 2014 ).

The p-value is not the probability of the null hypothesis p(H0), of being true, ( Krzywinski & Altman, 2013 ). This common misconception arises from a confusion between the probability of an observation given the null p(Obs≥t|H0) and the probability of the null given an observation p(H0|Obs≥t) that is then taken as an indication for p(H0) (see Nickerson, 2000 ).

Neyman-Pearson, hypothesis testing, and the α-value

Neyman & Pearson (1933) proposed a framework of statistical inference for applied decision making and quality control. In such framework, two hypotheses are proposed: the null hypothesis of no effect and the alternative hypothesis of an effect, along with a control of the long run probabilities of making errors. The first key concept in this approach, is the establishment of an alternative hypothesis along with an a priori effect size. This differs markedly from Fisher who proposed a general approach for scientific inference conditioned on the null hypothesis only. The second key concept is the control of error rates . Neyman & Pearson (1928) introduced the notion of critical intervals, therefore dichotomizing the space of possible observations into correct vs. incorrect zones. This dichotomization allows distinguishing correct results (rejecting H0 when there is an effect and not rejecting H0 when there is no effect) from errors (rejecting H0 when there is no effect, the type I error, and not rejecting H0 when there is an effect, the type II error). In this context, alpha is the probability of committing a Type I error in the long run. Alternatively, Beta is the probability of committing a Type II error in the long run.

The (theoretical) difference in terms of hypothesis testing between Fisher and Neyman-Pearson is illustrated on Figure 1 . In the 1 st case, we choose a level of significance for observed data of 5%, and compute the p-value. If the p-value is below the level of significance, it is used to reject H0. In the 2 nd case, we set a critical interval based on the a priori effect size and error rates. If an observed statistic value is below and above the critical values (the bounds of the confidence region), it is deemed significantly different from H0. In the NHST framework, the level of significance is (in practice) assimilated to the alpha level, which appears as a simple decision rule: if the p-value is less or equal to alpha, the null is rejected. It is however a common mistake to assimilate these two concepts. The level of significance set for a given sample is not the same as the frequency of acceptance alpha found on repeated sampling because alpha (a point estimate) is meant to reflect the long run probability whilst the p-value (a cumulative estimate) reflects the current probability ( Fisher, 1955 ; Hubbard & Bayarri, 2003 ).

The figure was prepared with G-power for a one-sided one-sample t-test, with a sample size of 32 subjects, an effect size of 0.45, and error rates alpha=0.049 and beta=0.80. In Fisher’s procedure, only the nil-hypothesis is posed, and the observed p-value is compared to an a priori level of significance. If the observed p-value is below this level (here p=0.05), one rejects H0. In Neyman-Pearson’s procedure, the null and alternative hypotheses are specified along with an a priori level of acceptance. If the observed statistical value is outside the critical region (here [-∞ +1.69]), one rejects H0.

Acceptance or rejection of H0?

The acceptance level α can also be viewed as the maximum probability that a test statistic falls into the rejection region when the null hypothesis is true ( Johnson, 2013 ). Therefore, one can only reject the null hypothesis if the test statistics falls into the critical region(s), or fail to reject this hypothesis. In the latter case, all we can say is that no significant effect was observed, but one cannot conclude that the null hypothesis is true. This is another common mistake in using NHST: there is a profound difference between accepting the null hypothesis and simply failing to reject it ( Killeen, 2005 ). By failing to reject, we simply continue to assume that H0 is true, which implies that one cannot argue against a theory from a non-significant result (absence of evidence is not evidence of absence). To accept the null hypothesis, tests of equivalence ( Walker & Nowacki, 2011 ) or Bayesian approaches ( Dienes, 2014 ; Kruschke, 2011 ) must be used.

Confidence intervals

Confidence intervals (CI) are builds that fail to cover the true value at a rate of alpha, the Type I error rate ( Morey & Rouder, 2011 ) and therefore indicate if observed values can be rejected by a (two tailed) test with a given alpha. CI have been advocated as alternatives to p-values because (i) they allow judging the statistical significance and (ii) provide estimates of effect size. Assuming the CI (a)symmetry and width are correct (but see Wilcox, 2012 ), they also give some indication about the likelihood that a similar value can be observed in future studies. For future studies of the same sample size, 95% CI give about 83% chance of replication success ( Cumming & Maillardet, 2006 ). If sample sizes however differ between studies, CI do not however warranty any a priori coverage.

Although CI provide more information, they are not less subject to interpretation errors (see Savalei & Dunn, 2015 for a review). The most common mistake is to interpret CI as the probability that a parameter (e.g. the population mean) will fall in that interval X% of the time. The correct interpretation is that, for repeated measurements with the same sample sizes, taken from the same population, X% of times the CI obtained will contain the true parameter value ( Tan & Tan, 2010 ). The alpha value has the same interpretation as testing against H0, i.e. we accept that 1-alpha CI are wrong in alpha percent of the times in the long run. This implies that CI do not allow to make strong statements about the parameter of interest (e.g. the mean difference) or about H1 ( Hoekstra et al. , 2014 ). To make a statement about the probability of a parameter of interest (e.g. the probability of the mean), Bayesian intervals must be used.

The (correct) use of NHST

NHST has always been criticized, and yet is still used every day in scientific reports ( Nickerson, 2000 ). One question to ask oneself is what is the goal of a scientific experiment at hand? If the goal is to establish a discrepancy with the null hypothesis and/or establish a pattern of order, because both requires ruling out equivalence, then NHST is a good tool ( Frick, 1996 ; Walker & Nowacki, 2011 ). If the goal is to test the presence of an effect and/or establish some quantitative values related to an effect, then NHST is not the method of choice since testing is conditioned on H0.

While a Bayesian analysis is suited to estimate that the probability that a hypothesis is correct, like NHST, it does not prove a theory on itself, but adds its plausibility ( Lindley, 2000 ). No matter what testing procedure is used and how strong results are, ( Fisher, 1959 p13) reminds us that ‘ […] no isolated experiment, however significant in itself, can suffice for the experimental demonstration of any natural phenomenon'. Similarly, the recent statement of the American Statistical Association ( Wasserstein & Lazar, 2016 ) makes it clear that conclusions should be based on the researchers understanding of the problem in context, along with all summary data and tests, and that no single value (being p-values, Bayesian factor or else) can be used support or invalidate a theory.

What to report and how?

Considering that quantitative reports will always have more information content than binary (significant or not) reports, we can always argue that raw and/or normalized effect size, confidence intervals, or Bayes factor must be reported. Reporting everything can however hinder the communication of the main result(s), and we should aim at giving only the information needed, at least in the core of a manuscript. Here I propose to adopt optimal reporting in the result section to keep the message clear, but have detailed supplementary material. When the hypothesis is about the presence/absence or order of an effect, and providing that a study has sufficient power, NHST is appropriate and it is sufficient to report in the text the actual p-value since it conveys the information needed to rule out equivalence. When the hypothesis and/or the discussion involve some quantitative value, and because p-values do not inform on the effect, it is essential to report on effect sizes ( Lakens, 2013 ), preferably accompanied with confidence or credible intervals. The reasoning is simply that one cannot predict and/or discuss quantities without accounting for variability. For the reader to understand and fully appreciate the results, nothing else is needed.

Because science progress is obtained by cumulating evidence ( Rosenthal, 1991 ), scientists should also consider the secondary use of the data. With today’s electronic articles, there are no reasons for not including all of derived data: mean, standard deviations, effect size, CI, Bayes factor should always be included as supplementary tables (or even better also share raw data). It is also essential to report the context in which tests were performed – that is to report all of the tests performed (all t, F, p values) because of the increase type one error rate due to selective reporting (multiple comparisons and p-hacking problems - Ioannidis, 2005 ). Providing all of this information allows (i) other researchers to directly and effectively compare their results in quantitative terms (replication of effects beyond significance, Open Science Collaboration, 2015 ), (ii) to compute power to future studies ( Lakens & Evers, 2014 ), and (iii) to aggregate results for meta-analyses whilst minimizing publication bias ( van Assen et al. , 2014 ).

[version 3; referees: 1 approved

Funding Statement

The author(s) declared that no grants were involved in supporting this work.

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Referee response for version 3

Dorothy vera margaret bishop.

1 Department of Experimental Psychology, University of Oxford, Oxford, UK

I can see from the history of this paper that the author has already been very responsive to reviewer comments, and that the process of revising has now been quite protracted.

That makes me reluctant to suggest much more, but I do see potential here for making the paper more impactful. So my overall view is that, once a few typos are fixed (see below), this could be published as is, but I think there is an issue with the potential readership and that further revision could overcome this.

I suspect my take on this is rather different from other reviewers, as I do not regard myself as a statistics expert, though I am on the more quantitative end of the continuum of psychologists and I try to keep up to date. I think I am quite close to the target readership , insofar as I am someone who was taught about statistics ages ago and uses stats a lot, but never got adequate training in the kinds of topic covered by this paper. The fact that I am aware of controversies around the interpretation of confidence intervals etc is simply because I follow some discussions of this on social media. I am therefore very interested to have a clear account of these issues.

This paper contains helpful information for someone in this position, but it is not always clear, and I felt the relevance of some of the content was uncertain. So here are some recommendations:

• As one previous reviewer noted, it’s questionable that there is a need for a tutorial introduction, and the limited length of this article does not lend itself to a full explanation. So it might be better to just focus on explaining as clearly as possible the problems people have had in interpreting key concepts. I think a title that made it clear this was the content would be more appealing than the current one.
• P 3, col 1, para 3, last sentence. Although statisticians always emphasise the arbitrary nature of p < .05, we all know that in practice authors who use other values are likely to have their analyses queried. I wondered whether it would be useful here to note that in some disciplines different cutoffs are traditional, e.g. particle physics. Or you could cite David Colquhoun’s paper in which he recommends using p < .001 ( http://rsos.royalsocietypublishing.org/content/1/3/140216) - just to be clear that the traditional p < .05 has been challenged.

What I can’t work out is how you would explain the alpha from Neyman-Pearson in the same way (though I can see from Figure 1 that with N-P you could test an alternative hypothesis, such as the idea that the coin would be heads 75% of the time).

‘By failing to reject, we simply continue to assume that H0 is true, which implies that one cannot….’ have ‘In failing to reject, we do not assume that H0 is true; one cannot argue against a theory from a non-significant result.’

I felt most readers would be interested to read about tests of equivalence and Bayesian approaches, but many would be unfamiliar with these and might like to see an example of how they work in practice – if space permitted.

• Confidence intervals: I simply could not understand the first sentence – I wondered what was meant by ‘builds’ here. I understand about difficulties in comparing CI across studies when sample sizes differ, but I did not find the last sentence on p 4 easy to understand.
• P 5: The sentence starting: ‘The alpha value has the same interpretation’ was also hard to understand, especially the term ‘1-alpha CI’. Here too I felt some concrete illustration might be helpful to the reader. And again, I also found the reference to Bayesian intervals tantalising – I think many readers won’t know how to compute these and something like a figure comparing a traditional CI with a Bayesian interval and giving a source for those who want to read on would be very helpful. The reference to ‘credible intervals’ in the penultimate paragraph is very unclear and needs a supporting reference – most readers will not be familiar with this concept.

P 3, col 1, para 2, line 2; “allows us to compute”

P 3, col 2, para 2, ‘probability of replicating’

P 3, col 2, para 2, line 4 ‘informative about’

P 3, col 2, para 4, line 2 delete ‘of’

P 3, col 2, para 5, line 9 – ‘conditioned’ is either wrong or too technical here: would ‘based’ be acceptable as alternative wording

P 3, col 2, para 5, line 13 ‘This dichotomisation allows one to distinguish’

P 3, col 2, para 5, last sentence, delete ‘Alternatively’.

P 3, col 2, last para line 2 ‘first’

P 4, col 2, para 2, last sentence is hard to understand; not sure if this is better: ‘If sample sizes differ between studies, the distribution of CIs cannot be specified a priori’

P 5, col 1, para 2, ‘a pattern of order’ – I did not understand what was meant by this

P 5, col 1, para 2, last sentence unclear: possible rewording: “If the goal is to test the size of an effect then NHST is not the method of choice, since testing can only reject the null hypothesis.’ (??)

P 5, col 1, para 3, line 1 delete ‘that’

P 5, col 1, para 3, line 3 ‘on’ -> ‘by’

P 5, col 2, para 1, line 4 , rather than ‘Here I propose to adopt’ I suggest ‘I recommend adopting’

P 5, col 2, para 1, line 13 ‘with’ -> ‘by’

P 5, col 2, para 1 – recommend deleting last sentence

P 5, col 2, para 2, line 2 ‘consider’ -> ‘anticipate’

P 5, col 2, para 2, delete ‘should always be included’

P 5, col 2, para 2, ‘type one’ -> ‘Type I’

I have read this submission. I believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard, however I have significant reservations, as outlined above.

The University of Edinburgh, UK

I wondered about changing the focus slightly and modifying the title to reflect this to say something like: Null hypothesis significance testing: a guide to commonly misunderstood concepts and recommendations for good practice

Thank you for the suggestion – you indeed saw the intention behind the ‘tutorial’ style of the paper.

• P 3, col 1, para 3, last sentence. Although statisticians always emphasise the arbitrary nature of p < .05, we all know that in practice authors who use other values are likely to have their analyses queried. I wondered whether it would be useful here to note that in some disciplines different cutoffs are traditional, e.g. particle physics. Or you could cite David Colquhoun’s paper in which he recommends using p < .001 ( http://rsos.royalsocietypublishing.org/content/1/3/140216)  - just to be clear that the traditional p < .05 has been challenged.

I have added a sentence on this citing Colquhoun 2014 and the new Benjamin 2017 on using .005.

I agree that this point is always hard to appreciate, especially because it seems like in practice it makes little difference. I added a paragraph but using reaction times rather than a coin toss – thanks for the suggestion.

Added an example based on new table 1, following figure 1 – giving CI, equivalence tests and Bayes Factor (with refs to easy to use tools)

Changed builds to constructs (this simply means they are something we build) and added that the implication that probability coverage is not warranty when sample size change, is that we cannot compare CI.

I changed ‘ i.e. we accept that 1-alpha CI are wrong in alpha percent of the times in the long run’ to ‘, ‘e.g. a 95% CI is wrong in 5% of the times in the long run (i.e. if we repeat the experiment many times).’ – for Bayesian intervals I simply re-cited Morey & Rouder, 2011.

It is not the CI cannot be specified, it’s that the interval is not predictive of anything anymore! I changed it to ‘If sample sizes, however, differ between studies, there is no warranty that a CI from one study will be true at the rate alpha in a different study, which implies that CI cannot be compared across studies at this is rarely the same sample sizes’

I added (i.e. establish that A > B) – we test that conditions are ordered, but without further specification of the probability of that effect nor its size

Yes it works – thx

P 5, col 2, para 2, ‘type one’ -> ‘Type I’ 

Typos fixed, and suggestions accepted – thanks for that.

Stephen J. Senn

1 Luxembourg Institute of Health, Strassen, L-1445, Luxembourg

The revisions are OK for me, and I have changed my status to Approved.

I have read this submission. I believe that I have an appropriate level of expertise to confirm that it is of an acceptable scientific standard.

Referee response for version 2

On the whole I think that this article is reasonable, my main reservation being that I have my doubts on whether the literature needs yet another tutorial on this subject.

A further reservation I have is that the author, following others, stresses what in my mind is a relatively unimportant distinction between the Fisherian and Neyman-Pearson (NP) approaches. The distinction stressed by many is that the NP approach leads to a dichotomy accept/reject based on probabilities established in advance, whereas the Fisherian approach uses tail area probabilities calculated from the observed statistic. I see this as being unimportant and not even true. Unless one considers that the person carrying out a hypothesis test (original tester) is mandated to come to a conclusion on behalf of all scientific posterity, then one must accept that any remote scientist can come to his or her conclusion depending on the personal type I error favoured. To operate the results of an NP test carried out by the original tester, the remote scientist then needs to know the p-value. The type I error rate is then compared to this to come to a personal accept or reject decision (1). In fact Lehmann (2), who was an important developer of and proponent of the NP system, describes exactly this approach as being good practice. (See Testing Statistical Hypotheses, 2nd edition P70). Thus using tail-area probabilities calculated from the observed statistics does not constitute an operational difference between the two systems.

A more important distinction between the Fisherian and NP systems is that the former does not use alternative hypotheses(3). Fisher's opinion was that the null hypothesis was more primitive than the test statistic but that the test statistic was more primitive than the alternative hypothesis. Thus, alternative hypotheses could not be used to justify choice of test statistic. Only experience could do that.

Further distinctions between the NP and Fisherian approach are to do with conditioning and whether a null hypothesis can ever be accepted.

I have one minor quibble about terminology. As far as I can see, the author uses the usual term 'null hypothesis' and the eccentric term 'nil hypothesis' interchangeably. It would be simpler if the latter were abandoned.

Referee response for version 1

Marcel alm van assen.

1 Department of Methodology and Statistics, Tilburgh University, Tilburg, Netherlands

Null hypothesis significance testing (NHST) is a difficult topic, with misunderstandings arising easily. Many texts, including basic statistics books, deal with the topic, and attempt to explain it to students and anyone else interested. I would refer to a good basic text book, for a detailed explanation of NHST, or to a specialized article when wishing an explaining the background of NHST. So, what is the added value of a new text on NHST? In any case, the added value should be described at the start of this text. Moreover, the topic is so delicate and difficult that errors, misinterpretations, and disagreements are easy. I attempted to show this by giving comments to many sentences in the text.

Abstract: “null hypothesis significance testing is the statistical method of choice in biological, biomedical and social sciences to investigate if an effect is likely”. No, NHST is the method to test the hypothesis of no effect.

Intro: “Null hypothesis significance testing (NHST) is a method of statistical inference by which an observation is tested against a hypothesis of no effect or no relationship.” What is an ‘observation’? NHST is difficult to describe in one sentence, particularly here. I would skip this sentence entirely, here.

Section on Fisher; also explain the one-tailed test.

Section on Fisher; p(Obs|H0) does not reflect the verbal definition (the ‘or more extreme’ part).

Section on Fisher; use a reference and citation to Fisher’s interpretation of the p-value

Section on Fisher; “This was however only intended to be used as an indication that there is something in the data that deserves further investigation. The reason for this is that only H0 is tested whilst the effect under study is not itself being investigated.” First sentence, can you give a reference? Many people say a lot about Fisher’s intentions, but the good man is dead and cannot reply… Second sentence is a bit awkward, because the effect is investigated in a way, by testing the H0.

Section on p-value; Layout and structure can be improved greatly, by first again stating what the p-value is, and then statement by statement, what it is not, using separate lines for each statement. Consider adding that the p-value is randomly distributed under H0 (if all the assumptions of the test are met), and that under H1 the p-value is a function of population effect size and N; the larger each is, the smaller the p-value generally is.

Skip the sentence “If there is no effect, we should replicate the absence of effect with a probability equal to 1-p”. Not insightful, and you did not discuss the concept ‘replicate’ (and do not need to).

Skip the sentence “The total probability of false positives can also be obtained by aggregating results ( Ioannidis, 2005 ).” Not strongly related to p-values, and introduces unnecessary concepts ‘false positives’ (perhaps later useful) and ‘aggregation’.

Consider deleting; “If there is an effect however, the probability to replicate is a function of the (unknown) population effect size with no good way to know this from a single experiment ( Killeen, 2005 ).”

The following sentence; “ Finally, a (small) p-value  is not an indication favouring a hypothesis . A low p-value indicates a misfit of the null hypothesis to the data and cannot be taken as evidence in favour of a specific alternative hypothesis more than any other possible alternatives such as measurement error and selection bias ( Gelman, 2013 ).” is surely not mainstream thinking about NHST; I would surely delete that sentence. In NHST, a p-value is used for testing the H0. Why did you not yet discuss significance level? Yes, before discussing what is not a p-value, I would explain NHST (i.e., what it is and how it is used). 

Also the next sentence “The more (a priori) implausible the alternative hypothesis, the greater the chance that a finding is a false alarm ( Krzywinski & Altman, 2013 ;  Nuzzo, 2014 ).“ is not fully clear to me. This is a Bayesian statement. In NHST, no likelihoods are attributed to hypotheses; the reasoning is “IF H0 is true, then…”.

Last sentence: “As  Nickerson (2000)  puts it ‘theory corroboration requires the testing of multiple predictions because the chance of getting statistically significant results for the wrong reasons in any given case is high’.” What is relation of this sentence to the contents of this section, precisely?

Next section: “For instance, we can estimate that the probability of a given F value to be in the critical interval [+2 +∞] is less than 5%” This depends on the degrees of freedom.

“When there is no effect (H0 is true), the erroneous rejection of H0 is known as type I error and is equal to the p-value.” Strange sentence. The Type I error is the probability of erroneously rejecting the H0 (so, when it is true). The p-value is … well, you explained it before; it surely does not equal the Type I error.

Consider adding a figure explaining the distinction between Fisher’s logic and that of Neyman and Pearson.

“When the test statistics falls outside the critical region(s)” What is outside?

“There is a profound difference between accepting the null hypothesis and simply failing to reject it ( Killeen, 2005 )” I agree with you, but perhaps you may add that some statisticians simply define “accept H0’” as obtaining a p-value larger than the significance level. Did you already discuss the significance level, and it’s mostly used values?

“To accept or reject equally the null hypothesis, Bayesian approaches ( Dienes, 2014 ;  Kruschke, 2011 ) or confidence intervals must be used.” Is ‘reject equally’ appropriate English? Also using Cis, one cannot accept the H0.

Do you start discussing alpha only in the context of Cis?

“CI also indicates the precision of the estimate of effect size, but unless using a percentile bootstrap approach, they require assumptions about distributions which can lead to serious biases in particular regarding the symmetry and width of the intervals ( Wilcox, 2012 ).” Too difficult, using new concepts. Consider deleting.

“Assuming the CI (a)symmetry and width are correct, this gives some indication about the likelihood that a similar value can be observed in future studies, with 95% CI giving about 83% chance of replication success ( Lakens & Evers, 2014 ).” This statement is, in general, completely false. It very much depends on the sample sizes of both studies. If the replication study has a much, much, much larger N, then the probability that the original CI will contain the effect size of the replication approaches (1-alpha)*100%. If the original study has a much, much, much larger N, then the probability that the original Ci will contain the effect size of the replication study approaches 0%.

“Finally, contrary to p-values, CI can be used to accept H0. Typically, if a CI includes 0, we cannot reject H0. If a critical null region is specified rather than a single point estimate, for instance [-2 +2] and the CI is included within the critical null region, then H0 can be accepted. Importantly, the critical region must be specified a priori and cannot be determined from the data themselves.” No. H0 cannot be accepted with Cis.

“The (posterior) probability of an effect can however not be obtained using a frequentist framework.” Frequentist framework? You did not discuss that, yet.

“X% of times the CI obtained will contain the same parameter value”. The same? True, you mean?

“e.g. X% of the times the CI contains the same mean” I do not understand; which mean?

“The alpha value has the same interpretation as when using H0, i.e. we accept that 1-alpha CI are wrong in alpha percent of the times. “ What do you mean, CI are wrong? Consider rephrasing.

“To make a statement about the probability of a parameter of interest, likelihood intervals (maximum likelihood) and credibility intervals (Bayes) are better suited.” ML gives the likelihood of the data given the parameter, not the other way around.

“Many of the disagreements are not on the method itself but on its use.” Bayesians may disagree.

“If the goal is to establish the likelihood of an effect and/or establish a pattern of order, because both requires ruling out equivalence, then NHST is a good tool ( Frick, 1996 )” NHST does not provide evidence on the likelihood of an effect.

“If the goal is to establish some quantitative values, then NHST is not the method of choice.” P-values are also quantitative… this is not a precise sentence. And NHST may be used in combination with effect size estimation (this is even recommended by, e.g., the American Psychological Association (APA)).

“Because results are conditioned on H0, NHST cannot be used to establish beliefs.” It can reinforce some beliefs, e.g., if H0 or any other hypothesis, is true.

“To estimate the probability of a hypothesis, a Bayesian analysis is a better alternative.” It is the only alternative?

“Note however that even when a specific quantitative prediction from a hypothesis is shown to be true (typically testing H1 using Bayes), it does not prove the hypothesis itself, it only adds to its plausibility.” How can we show something is true?

I do not agree on the contents of the last section on ‘minimal reporting’. I prefer ‘optimal reporting’ instead, i.e., the reporting the information that is essential to the interpretation of the result, to any ready, which may have other goals than the writer of the article. This reporting includes, for sure, an estimate of effect size, and preferably a confidence interval, which is in line with recommendations of the APA.

I have read this submission. I believe that I have an appropriate level of expertise to state that I do not consider it to be of an acceptable scientific standard, for reasons outlined above.

The idea of this short review was to point to common interpretation errors (stressing again and again that we are under H0) being in using p-values or CI, and also proposing reporting practices to avoid bias. This is now stated at the end of abstract.

Regarding text books, it is clear that many fail to clearly distinguish Fisher/Pearson/NHST, see Glinet et al (2012) J. Exp Education 71, 83-92. If you have 1 or 2 in mind that you know to be good, I’m happy to include them.

I agree – yet people use it to investigate (not test) if an effect is likely. The issue here is wording. What about adding this distinction at the end of the sentence?: ‘null hypothesis significance testing is the statistical method of choice in biological, biomedical and social sciences used to investigate if an effect is likely, even though it actually tests for the hypothesis of no effect’.

I think a definition is needed, as it offers a starting point. What about the following: ‘NHST is a method of statistical inference by which an experimental factor is tested against a hypothesis of no effect or no relationship based on a given observation’

The section on Fisher has been modified (more or less) as suggested: (1) avoiding talking about one or two tailed tests (2) updating for p(Obs≥t|H0) and (3) referring to Fisher more explicitly (ie pages from articles and book) ; I cannot tell his intentions but these quotes leave little space to alternative interpretations.

The reasoning here is as you state yourself, part 1: ‘a p-value is used for testing the H0; and part 2: ‘no likelihoods are attributed to hypotheses’ it follows we cannot favour a hypothesis. It might seems contentious but this is the case that all we can is to reject the null – how could we favour a specific alternative hypothesis from there? This is explored further down the manuscript (and I now point to that) – note that we do not need to be Bayesian to favour a specific H1, all I’m saying is this cannot be attained with a p-value.

The point was to emphasise that a p value is not there to tell us a given H1 is true and can only be achieved through multiple predictions and experiments. I deleted it for clarity.

This sentence has been removed

Indeed, you are right and I have modified the text accordingly. When there is no effect (H0 is true), the erroneous rejection of H0 is known as type 1 error. Importantly, the type 1 error rate, or alpha value is determined a priori. It is a common mistake but the level of significance (for a given sample) is not the same as the frequency of acceptance alpha found on repeated sampling (Fisher, 1955).

A figure is now presented – with levels of acceptance, critical region, level of significance and p-value.

I should have clarified further here – as I was having in mind tests of equivalence. To clarify, I simply states now: ‘To accept the null hypothesis, tests of equivalence or Bayesian approaches must be used.’

It is now presented in the paragraph before.

Yes, you are right, I completely overlooked this problem. The corrected sentence (with more accurate ref) is now “Assuming the CI (a)symmetry and width are correct, this gives some indication about the likelihood that a similar value can be observed in future studies. For future studies of the same sample size, 95% CI giving about 83% chance of replication success (Cumming and Mallardet, 2006). If sample sizes differ between studies, CI do not however warranty any a priori coverage”.

Again, I had in mind equivalence testing, but in both cases you are right we can only reject and I therefore removed that sentence.

Yes, p-values must be interpreted in context with effect size, but this is not what people do. The point here is to be pragmatic, does and don’t. The sentence was changed.

Not for testing, but for probability, I am not aware of anything else.

Cumulative evidence is, in my opinion, the only way to show it. Even in hard science like physics multiple experiments. In the recent CERN study on finding Higgs bosons, 2 different and complementary experiments ran in parallel – and the cumulative evidence was taken as a proof of the true existence of Higgs bosons.

Daniel Lakens

1 School of Innovation Sciences, Eindhoven University of Technology, Eindhoven, Netherlands

I appreciate the author's attempt to write a short tutorial on NHST. Many people don't know how to use it, so attempts to educate people are always worthwhile. However, I don't think the current article reaches it's aim. For one, I think it might be practically impossible to explain a lot in such an ultra short paper - every section would require more than 2 pages to explain, and there are many sections. Furthermore, there are some excellent overviews, which, although more extensive, are also much clearer (e.g., Nickerson, 2000 ). Finally, I found many statements to be unclear, and perhaps even incorrect (noted below). Because there is nothing worse than creating more confusion on such a topic, I have extremely high standards before I think such a short primer should be indexed. I note some examples of unclear or incorrect statements below. I'm sorry I can't make a more positive recommendation.

“investigate if an effect is likely” – ambiguous statement. I think you mean, whether the observed DATA is probable, assuming there is no effect?

The Fisher (1959) reference is not correct – Fischer developed his method much earlier.

“This p-value thus reflects the conditional probability of achieving the observed outcome or larger, p(Obs|H0)” – please add 'assuming the null-hypothesis is true'.

“p(Obs|H0)” – explain this notation for novices.

“Following Fisher, the smaller the p-value, the greater the likelihood that the null hypothesis is false.”  This is wrong, and any statement about this needs to be much more precise. I would suggest direct quotes.

“there is something in the data that deserves further investigation” –unclear sentence.

“The reason for this” – unclear what ‘this’ refers to.

“ not the probability of the null hypothesis of being true, p(H0)” – second of can be removed?

“Any interpretation of the p-value in relation to the effect under study (strength, reliability, probability) is indeed

wrong, since the p-value is conditioned on H0”  - incorrect. A big problem is that it depends on the sample size, and that the probability of a theory depends on the prior.

“If there is no effect, we should replicate the absence of effect with a probability equal to 1-p.” I don’t understand this, but I think it is incorrect.

“The total probability of false positives can also be obtained by aggregating results (Ioannidis, 2005).” Unclear, and probably incorrect.

“By failing to reject, we simply continue to assume that H0 is true, which implies that one cannot, from a nonsignificant result, argue against a theory” – according to which theory? From a NP perspective, you can ACT as if the theory is false.

“(Lakens & Evers, 2014”) – we are not the original source, which should be cited instead.

“ Typically, if a CI includes 0, we cannot reject H0.”  - when would this not be the case? This assumes a CI of 1-alpha.

“If a critical null region is specified rather than a single point estimate, for instance [-2 +2] and the CI is included within the critical null region, then H0 can be accepted.” – you mean practically, or formally? I’m pretty sure only the former.

The section on ‘The (correct) use of NHST’ seems to conclude only Bayesian statistics should be used. I don’t really agree.

“ we can always argue that effect size, power, etc. must be reported.” – which power? Post-hoc power? Surely not? Other types are unknown. So what do you mean?

The recommendation on what to report remains vague, and it is unclear why what should be reported.

This sentence was changed, following as well the other reviewer, to ‘null hypothesis significance testing is the statistical method of choice in biological, biomedical and social sciences to investigate if an effect is likely, even though it actually tests whether the observed data are probable, assuming there is no effect’

Changed, refers to Fisher 1925

I changed a little the sentence structure, which should make explicit that this is the condition probability.

This has been changed to ‘[…] to decide whether the evidence is worth additional investigation and/or replication (Fisher, 1971 p13)’

my mistake – the sentence structure is now ‘ not the probability of the null hypothesis p(H0), of being true,’ ; hope this makes more sense (and this way refers back to p(Obs>t|H0)

Fair enough – my point was to stress the fact that p value and effect size or H1 have very little in common, but yes that the part in common has to do with sample size. I left the conditioning on H0 but also point out the dependency on sample size.

The whole paragraph was changed to reflect a more philosophical take on scientific induction/reasoning. I hope this is clearer.

Changed to refer to equivalence testing

I rewrote this, as to show frequentist analysis can be used  - I’m trying to sell Bayes more than any other approach.

I’m arguing we should report it all, that’s why there is no exhausting list – I can if needed.

Statistics Made Easy

How to Write a Null Hypothesis (5 Examples)

A hypothesis test uses sample data to determine whether or not some claim about a population parameter is true.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis, which take the following forms:

H 0 (Null Hypothesis): Population parameter =,  ≤, ≥ some value

H A  (Alternative Hypothesis): Population parameter <, >, ≠ some value

Note that the null hypothesis always contains the equal sign .

We interpret the hypotheses as follows:

Null hypothesis: The sample data provides no evidence to support some claim being made by an individual.

Alternative hypothesis: The sample data  does provide sufficient evidence to support the claim being made by an individual.

For example, suppose it’s assumed that the average height of a certain species of plant is 20 inches tall. However, one botanist claims the true average height is greater than 20 inches.

To test this claim, she may go out and collect a random sample of plants. She can then use this sample data to perform a hypothesis test using the following two hypotheses:

H 0 : μ ≤ 20 (the true mean height of plants is equal to or even less than 20 inches)

H A : μ > 20 (the true mean height of plants is greater than 20 inches)

If the sample data gathered by the botanist shows that the mean height of this species of plants is significantly greater than 20 inches, she can reject the null hypothesis and conclude that the mean height is greater than 20 inches.

Read through the following examples to gain a better understanding of how to write a null hypothesis in different situations.

Example 1: Weight of Turtles

A biologist wants to test whether or not the true mean weight of a certain species of turtles is 300 pounds. To test this, he goes out and measures the weight of a random sample of 40 turtles.

Here is how to write the null and alternative hypotheses for this scenario:

H 0 : μ = 300 (the true mean weight is equal to 300 pounds)

H A : μ ≠ 300 (the true mean weight is not equal to 300 pounds)

Example 2: Height of Males

It’s assumed that the mean height of males in a certain city is 68 inches. However, an independent researcher believes the true mean height is greater than 68 inches. To test this, he goes out and collects the height of 50 males in the city.

H 0 : μ ≤ 68 (the true mean height is equal to or even less than 68 inches)

H A : μ > 68 (the true mean height is greater than 68 inches)

Example 3: Graduation Rates

A university states that 80% of all students graduate on time. However, an independent researcher believes that less than 80% of all students graduate on time. To test this, she collects data on the proportion of students who graduated on time last year at the university.

H 0 : p ≥ 0.80 (the true proportion of students who graduate on time is 80% or higher)

H A : μ < 0.80 (the true proportion of students who graduate on time is less than 80%)

Example 4: Burger Weights

A food researcher wants to test whether or not the true mean weight of a burger at a certain restaurant is 7 ounces. To test this, he goes out and measures the weight of a random sample of 20 burgers from this restaurant.

H 0 : μ = 7 (the true mean weight is equal to 7 ounces)

H A : μ ≠ 7 (the true mean weight is not equal to 7 ounces)

Example 5: Citizen Support

A politician claims that less than 30% of citizens in a certain town support a certain law. To test this, he goes out and surveys 200 citizens on whether or not they support the law.

H 0 : p ≥ .30 (the true proportion of citizens who support the law is greater than or equal to 30%)

H A : μ < 0.30 (the true proportion of citizens who support the law is less than 30%)

Additional Resources

Introduction to Hypothesis Testing Introduction to Confidence Intervals An Explanation of P-Values and Statistical Significance

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “How to Write a Null Hypothesis (5 Examples)”

you are amazing, thank you so much

Say I am a botanist hypothesizing the average height of daisies is 20 inches, or not? Does T = (ave – 20 inches) / √ variance / (80 / 4)? … This assumes 40 real measures + 40 fake = 80 n, but that seems questionable. Please advise.

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Table of Contents

In today’s data-driven world , decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life - 

• A teacher assumes that 60% of his college's students come from lower-middle-class families.
• A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

• Here, x̅ is the sample mean,
• μ0 is the population mean,
• σ is the standard deviation,
• n is the sample size.

How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

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Null Hypothesis and Alternate Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average. 

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

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Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

 We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

Steps of Hypothesis Testing

Step 1: specify your null and alternate hypotheses.

It is critical to rephrase your original research hypothesis (the prediction that you wish to study) as a null (Ho) and alternative (Ha) hypothesis so that you can test it quantitatively. Your first hypothesis, which predicts a link between variables, is generally your alternate hypothesis. The null hypothesis predicts no link between the variables of interest.

Step 2: Gather Data

For a statistical test to be legitimate, sampling and data collection must be done in a way that is meant to test your hypothesis. You cannot draw statistical conclusions about the population you are interested in if your data is not representative.

Step 3: Conduct a Statistical Test

Other statistical tests are available, but they all compare within-group variance (how to spread out the data inside a category) against between-group variance (how different the categories are from one another). If the between-group variation is big enough that there is little or no overlap between groups, your statistical test will display a low p-value to represent this. This suggests that the disparities between these groups are unlikely to have occurred by accident. Alternatively, if there is a large within-group variance and a low between-group variance, your statistical test will show a high p-value. Any difference you find across groups is most likely attributable to chance. The variety of variables and the level of measurement of your obtained data will influence your statistical test selection.

Step 4: Determine Rejection Of Your Null Hypothesis

Your statistical test results must determine whether your null hypothesis should be rejected or not. In most circumstances, you will base your judgment on the p-value provided by the statistical test. In most circumstances, your preset level of significance for rejecting the null hypothesis will be 0.05 - that is, when there is less than a 5% likelihood that these data would be seen if the null hypothesis were true. In other circumstances, researchers use a lower level of significance, such as 0.01 (1%). This reduces the possibility of wrongly rejecting the null hypothesis.

Step 5: Present Your Results 

The findings of hypothesis testing will be discussed in the results and discussion portions of your research paper, dissertation, or thesis. You should include a concise overview of the data and a summary of the findings of your statistical test in the results section. You can talk about whether your results confirmed your initial hypothesis or not in the conversation. Rejecting or failing to reject the null hypothesis is a formal term used in hypothesis testing. This is likely a must for your statistics assignments.

Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

Chi-Square 

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

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Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

• The null hypothesis is (H0 <= 90) or less change.
• A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true]. 

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

Level of Significance

The alpha value is a criterion for determining whether a test statistic is statistically significant. In a statistical test, Alpha represents an acceptable probability of a Type I error. Because alpha is a probability, it can be anywhere between 0 and 1. In practice, the most commonly used alpha values are 0.01, 0.05, and 0.1, which represent a 1%, 5%, and 10% chance of a Type I error, respectively (i.e. rejecting the null hypothesis when it is in fact correct).

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A p-value is a metric that expresses the likelihood that an observed difference could have occurred by chance. As the p-value decreases the statistical significance of the observed difference increases. If the p-value is too low, you reject the null hypothesis.

Here you have taken an example in which you are trying to test whether the new advertising campaign has increased the product's sales. The p-value is the likelihood that the null hypothesis, which states that there is no change in the sales due to the new advertising campaign, is true. If the p-value is .30, then there is a 30% chance that there is no increase or decrease in the product's sales.  If the p-value is 0.03, then there is a 3% probability that there is no increase or decrease in the sales value due to the new advertising campaign. As you can see, the lower the p-value, the chances of the alternate hypothesis being true increases, which means that the new advertising campaign causes an increase or decrease in sales.

Why is Hypothesis Testing Important in Research Methodology?

Hypothesis testing is crucial in research methodology for several reasons:

• Provides evidence-based conclusions: It allows researchers to make objective conclusions based on empirical data, providing evidence to support or refute their research hypotheses.
• Supports decision-making: It helps make informed decisions, such as accepting or rejecting a new treatment, implementing policy changes, or adopting new practices.
• Adds rigor and validity: It adds scientific rigor to research using statistical methods to analyze data, ensuring that conclusions are based on sound statistical evidence.
• Contributes to the advancement of knowledge: By testing hypotheses, researchers contribute to the growth of knowledge in their respective fields by confirming existing theories or discovering new patterns and relationships.

Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

• It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
• Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
• Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
• Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

If you are interested in statistics of data science and skills needed for such a career, you ought to explore Simplilearn’s Post Graduate Program in Data Science.

If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

2. What is hypothesis testing and its types?

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves formulating two hypotheses: the null hypothesis (H0), which represents the default assumption, and the alternative hypothesis (Ha), which contradicts H0. The goal is to assess the evidence and determine whether there is enough statistical significance to reject the null hypothesis in favor of the alternative hypothesis.

Types of hypothesis testing:

• One-sample test: Used to compare a sample to a known value or a hypothesized value.
• Two-sample test: Compares two independent samples to assess if there is a significant difference between their means or distributions.
• Paired-sample test: Compares two related samples, such as pre-test and post-test data, to evaluate changes within the same subjects over time or under different conditions.
• Chi-square test: Used to analyze categorical data and determine if there is a significant association between variables.
• ANOVA (Analysis of Variance): Compares means across multiple groups to check if there is a significant difference between them.

3. What are the steps of hypothesis testing?

The steps of hypothesis testing are as follows:

• Formulate the hypotheses: State the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
• Set the significance level: Determine the acceptable level of error (alpha) for making a decision.
• Collect and analyze data: Gather and process the sample data.
• Compute test statistic: Calculate the appropriate statistical test to assess the evidence.
• Make a decision: Compare the test statistic with critical values or p-values and determine whether to reject H0 in favor of Ha or not.
• Draw conclusions: Interpret the results and communicate the findings in the context of the research question.

4. What are the 2 types of hypothesis testing?

• One-tailed (or one-sided) test: Tests for the significance of an effect in only one direction, either positive or negative.
• Two-tailed (or two-sided) test: Tests for the significance of an effect in both directions, allowing for the possibility of a positive or negative effect.

The choice between one-tailed and two-tailed tests depends on the specific research question and the directionality of the expected effect.

5. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

• Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
• Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
• Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

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Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

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Stats Hypothesis testing

Hypothesis testing

Hypothesis testing is one of the most widely used approaches of statistical inference.

The idea of hypothesis testing (more formally: null hypothesis significance testing - NHST) is the following: if we have some data observed, and we have a statistical model, we can use this statistical model to specify a fixed hypothesis about how the data did arise. For the example with the plants and music, this hypothesis could be: music has no influence on plants, all differences we see are due to random variation between individuals.

The null hypothesis H0 and the alternative hypothesis H1

Such a scenario is called the null hypothesis H0. Although it is very typical to use the assumption of no effect as null-hypothesis, note that it is really your choice, and you could use anything as null hypothesis, also the assumption: “classical music doubles the growth of plants”. The fact that it’s the analyst’s choice what to fix as null hypothesis is part of the reason why there are are a large number of tests available. We will see a few of them in the following chapter about important hypothesis tests.

The hypothesis that H0 is wrong, or !H0, is usually called the alternative hypothesis, H1

Given a statistical model, a “normal” or “simple” null hypothesis specifies a single value for the parameter of interest as the “base expectation”. A composite null hypothesis specifies a range of values for the parameter.

If we have a null hypothesis, we calculate the probability that we would see the observed data or data more extreme under this scenario. This is called a hypothesis tests, and we call the probability the p-value. If the p-value falls under a certain level (the significance level $\alpha$) we say the null hypothesis was rejected, and there is significant support for the alternative hypothesis. The level of $\alpha$ is a convention, in ecology we chose typically 0.05, so if a p-value falls below 0.05, we can reject the null hypothesis.

Test Statistic

Type I and II error

Significance level, Power

Misinterpretations

A problem with hypothesis tests and p-values is that their results are notoriously misinterpreted. The p-value is NOT the probability that the null hypothesis is true, or the probability that the alternative hypothesis is false, although many authors have made the mistake of interpreting it like that \citep[][]{Cohen-earthisround-1994}. Rather, the idea of p-values is to control the rate of false positives (Type I error). When doing hypothesis tests on random data, with an $\alpha$ level of 0.05, one will get exactly 5\% false positives. Not more and not less.

Further readings

• The Essential Statistics lecture notes
• http://www.stats.gla.ac.uk/steps/glossary/hypothesis_testing.html

Examples in R

Recall statistical tests, or more formally, null-hypothesis significance testing (NHST) is one of several ways in which you can approach data. The idea is that you define a null-hypothesis, and then you look a the probability that the data would occur under the assumption that the null hypothesis is true.

Now, there can be many null hypothesis, so you need many tests. The most widely used tests are given here.

The t -test can be used to test whether one sample is different from a reference value (e.g. 0: one-sample t -test), whether two samples are different (two-sample t -test) or whether two paired samples are different (paired t -test).

The t -test assumes that the data are normally distributed. It can handle samples with same or different variances, but needs to be “told” so.

t-test for 1 sample (PARAMETRIC TEST)

The one-sample t-test compares the MEAN score of a sample to a known value, usually the population MEAN (the average for the outcome of some population of interest).

Our null hypothesis is that the mean of the sample is not less than 2.5 (real example: weight data of 200 lizards collected for a research, we want to compare it with the known average weights available in the scientific literature)

t-test for 1 sample (NON-PARAMETRIC TEST)

One-sample Wilcoxon signed rank test is a non-parametric alternative method of one-sample t-test, which is used to test whether the location (MEDIAN) of the measurement is equal to a specified value

Create fake data log-normally distributed and verify data distribution

Our null hypothesis is that the median of x is not different from 1

Two Independent Samples T-test (PARAMETRIC TEST)

Parametric method for examining the difference in MEANS between two independent populations. The t -test should be preceeded by a graphical depiction of the data in order to check for normality within groups and for evidence of heteroscedasticity (= differences in variance), like so:

Reshape the data:

Now plot them as points (not box-n-whiskers):

The points to the right scatter similar to those on the left, although a bit more asymmetrically. Although we know that they are from a log-normal distribution (right), they don’t look problematic.

If data are not normally distributed, we sometimes succeed making data normal by using transformations, such as square-root, log, or alike (see section on transformations).

While t -tests on transformed data now actually test for differences between these transformed data, that is typically fine. Think of the pH-value, which is only a log-transform of the proton concentration. Do we care whether two treatments are different in pH or in proton concentrations? If so, then we need to choose the right data set. Most likely, we don’t and only choose the log-transform because the data are actually lognormally distributed, not normally.

A non-parametric alternative is the Mann-Whitney-U-test, or, the ANOVA-equivalent, the Kruskal-Wallis test. Both are available in R and explained later, but instead we recommend the following:

Use rank-transformations, which replaces the values by their rank (i.e. the lowest value receives a 1, the second lowest a 2 and so forth). A t -test of rank-transformed data is not the same as the Mann-Whitney-U-test, but it is more sensitive and hence preferable (Ruxton 2006) or at least equivalent (Zimmerman 2012).

To use the rank, we need to employ the “formula”-invokation of t.test! In this case, results are the same, indicating that our hunch about acceptable skew and scatter was correct.

(Note that the original t -test is a test for differences between means, while the rank- t -test becomes a test for general differences in values between the two groups, not specifically of the mean.)

Cars example:

Test the difference in car consumption depending on the transmission type. Check wherever the 2 ‘independent populations’ are normally distributed

Graphic representation

We have two ~normally distributed populations. In order to test for differences in means, we applied a t-test for independent samples.

Any time we work with the t-test, we have to verify whether the variance is equal betwenn the 2 populations or not, then we fit the t-test accordingly. Our Ho or null hypothesis is that the consumption is the same irrespective to transmission. We assume non-equal variances

From the output: please note that CIs are the confidence intervales for differences in means

Same results if you run the following (meaning that the other commands were all by default)

The alternative could be one-sided (greater, lesser) as we discussed earlier for one-sample t-tests

If we assume equal variance, we run the following

Ways to check for equal / not equal variance

1) To examine the boxplot visually

2) To compute the actual variance

There is 2/3 times difference in variance.

3) Levene’s test

Mann-Whitney U test/Wilcoxon rank-sum test for two independent samples (NON-PARAMETRIC TEST)

We change the response variable to hp (Gross horsepower)

The ‘population’ of cars with manual transmission has a hp not normally distributed, so we have to use a test for independent samples - non-parametric

We want to test a difference in hp depending on the transmission Using a non-parametric test, we test for differences in MEDIANS between 2 independent populations

Our null hypothesis will be that the medians are equal (two-sided)

Wilcoxon signed rank test for two dependend samples (NON PARAMETRIC)

This is a non-parametric method appropriate for examining the median difference in 2 populations observations that are paired or dependent one of the other.

This is a dataset about some water measurements taken at different levels of a river: ‘up’ and ‘down’ are water quality measurements of the same river taken before and after a water treatment filter, respectively

The line you see in the plot corresponds to x=y, that is, the same water measuremets before and after the water treatment (it seems to be true in 2 rivers only, 5 and 15)

Our null hypothesis is that the median before and after the treatment are not different

the assumption of normality is certainly not met for the measurements after the treatment

Paired T-test for two dependend samples test. (PARAMETRIC)

This parametric method examinates the difference in means for two populations that are paired or dependent one of the other

This is a dataset about the density of a fish prey species (fish/km2) in 121 lakes before and after removing a non-native predator

changing the order of variables, we have a change in the sign of the t-test estimated mean of differences

low p ->reject Ho, means are equal

Testing for normality

The normal distribution is the most important and most widely used distribution in statistics. We can say that a distribution is normally distributed when: 1) is symmetric around their mean. 2) the mean, median, and mode of a normal distribution are equal. 3) the area under the normal curve is equal to 1.0. 4) distributions are denser in the center and less dense in the tails. 5) distributions are defined by two parameters, the mean and the standard deviation (sd). 6) 68% of the area of a normal distribution is within one standard deviation of the mean. 7) Approximately 95% of the area of a normal distribution is within two standard deviations of the mean.

Normal distribution

Load example data

Visualize example data

Visual Check for Normality: quantile-quantile plot

This one plots the ranked samples from our distribution against a similar number of ranked quantiles taken from a normal distribution. If our sample is normally distributed then the line will be straight. Exceptions from normality show up different sorts of non-linearity (e.g. S-shapes or banana shapes).

Normality test: the shapiro.test

As an example we will create a fake data log-normally distributed and verify the assumption of normality

Correlations tests

Correlation tests measure the relationship between variables. This relationship can goes from +1 to -1, where 0 means no relation. Some of the tests that we can use to estimate this relationship are the following:

-Pearson’s correlation is a parametric measure of the linear association between 2 numeric variables (PARAMETRIC TEST)

-Spearman’s rank correlation is a non-parametric measure of the monotonic association between 2 numeric variables (NON-PARAMETRIC TEST)

-Kendall’s rank correlation is another non-parametric measure of the associtaion, based on concordance or discordance of x-y pairs (NON-PARAMETRIC TEST)

Compute the three correlation coefficients

Test the null hypothesis, that means that the correlation is 0 (there is no correlation)

When we have non-parametric data and we do not know which correlation method to choose, as a rule of thumb, if the correlation looks non-linear, Kendall tau should be better than Spearman Rho.

Further handy functions for correlations

Plot all possible combinations with “pairs”

To make it simpler we select what we are interested

Building a correlation matrix

• Ruxton, G. D. (2006). The unequal variance t-test is an underused alternative to Student??????s t-test and the Mann-Whitney U test. Behavioral Ecology, 17, 688-690.
• Zimmerman, D. W. (2012). A note on consistency of non-parametric rank tests and related rank transformations. British Journal of Mathematical and Statistical Psychology, 65, 122-44.
• http://www.uni-kiel.de/psychologie/rexrepos/rerDescriptive.html

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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, \(H_{0}\), and the alternative hypothesis. \(H_{a}\), in terms of the appropriate parameter \((\mu \text{or} p)\).

• The mean number of years Americans work before retiring is 34.
• At most 60% of Americans vote in presidential elections.
• The mean starting salary for San Jose State University graduates is at least $100,000 per year.
• Twenty-nine percent of high school seniors get drunk each month.
• Fewer than 5% of adults ride the bus to work in Los Angeles.
• The mean number of cars a person owns in her lifetime is not more than ten.
• About half of Americans prefer to live away from cities, given the choice.
• Europeans have a mean paid vacation each year of six weeks.
• The chance of developing breast cancer is under 11% for women.
• Private universities' mean tuition cost is more than $20,000 per year.
• \(H_{0}: \mu = 34; H_{a}: \mu \neq 34\)
• \(H_{0}: p \leq 0.60; H_{a}: p > 0.60\)
• \(H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000\)
• \(H_{0}: p = 0.29; H_{a}: p \neq 0.29\)
• \(H_{0}: p = 0.05; H_{a}: p < 0.05\)
• \(H_{0}: \mu \leq 10; H_{a}: \mu > 10\)
• \(H_{0}: p = 0.50; H_{a}: p \neq 0.50\)
• \(H_{0}: \mu = 6; H_{a}: \mu \neq 6\)
• \(H_{0}: p ≥ 0.11; H_{a}: p < 0.11\)
• \(H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000\)

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

• \(p < 0.30\)
• \(p \leq 0.30\)
• \(p \geq 0.30\)
• \(p > 0.30\)

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

• \(p = 0.20\)
• \(p > 0.20\)
• \(p < 0.20\)
• \(p \leq 0.20\)

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

• \(H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5\)
• \(H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5\)
• \(H_{0}: \mu = 4.75, H_{a}: \mu > 4.75\)
• \(H_{0}: \mu = 4.5, H_{a}: \mu > 4.5\)

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

• The mean number of cars a person owns in his or her lifetime is not more than ten.
• Private universities mean tuition cost is more than $20,000 per year.
• Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
• Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
• Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
• Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
• Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
• Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
• Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
• Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
• Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
• Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

• State a consequence of committing a Type I error.
• State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

• To conclude the drug is safe when in, fact, it is unsafe.
• Not to conclude the drug is safe when, in fact, it is safe.
• To conclude the drug is safe when, in fact, it is safe.
• Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

• at least 20%, when in fact, it is less than 20%.
• 20%, when in fact, it is 20%.
• less than 20%, when in fact, it is at least 20%.
• less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

• is more than seven hours.
• is at most seven hours.
• is at least seven hours.
• is less than seven hours.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:

• to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
• to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
• to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
• to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is \(\bar{X} \sim\) ________________

• \(N\left(7.24, \frac{1.93}{\sqrt{22}}\right)\)
• \(N\left(7.24, 1.93\right)\)

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

• Is this a test of one mean or proportion?
• State the null and alternative hypotheses. \(H_{0}\) : ____________________ \(H_{a}\) : ____________________
• Is this a right-tailed, left-tailed, or two-tailed test?
• What symbol represents the random variable for this test?
• In words, define the random variable for this test.
• \(x =\) ________________
• \(n =\) ________________
• \(p′ =\) _____________
• Calculate \(\sigma_{x} =\) __________. Show the formula set-up.
• State the distribution to use for the hypothesis test.
• Find the \(p\text{-value}\).
• Reason for the decision:
• Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's \(t\) - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using \(\alpha = 0.05\), is the data highly inconsistent with the claim?

• \(H_{0}: \mu \geq 50,000\)
• \(H_{a}: \mu < 50,000\)
• Let \(\bar{X} =\) the average lifespan of a brand of tires.
• normal distribution
• \(z = -2.315\)
• \(p\text{-value} = 0.0103\)
• Check student’s solution.
• alpha: 0.05
• Decision: Reject the null hypothesis.
• Reason for decision: The \(p\text{-value}\) is less than 0.05.
• Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
• \((43,537, 49,463)\)

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

• \(H_{0}: \mu = $1.00\)
• \(H_{a}: \mu \neq $1.00\)
• Let \(\bar{X} =\) the average cost of a daily newspaper.
• \(z = –0.866\)
• \(p\text{-value} = 0.3865\)
• \(\alpha: 0.01\)
• Decision: Do not reject the null hypothesis.
• Reason for decision: The \(p\text{-value}\) is greater than 0.01.
• Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
• \(($0.84, $1.06)\)

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let \(x =\) the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

• \(H_{0}: \mu = 10\)
• \(H_{a}: \mu \neq 10\)
• Let \(\bar{X}\) the mean number of sick days an employee takes per year.
• Student’s t -distribution
• \(t = –1.12\)
• \(p\text{-value} = 0.300\)
• \(\alpha: 0.05\)
• Reason for decision: The \(p\text{-value}\) is greater than 0.05.
• Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
• \((4.9443, 11.806)\)

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

• \(H_{0}: p \geq 0.6\)
• \(H_{a}: p < 0.6\)
• Let \(P′ =\) the proportion of students who feel more enriched as a result of taking Elementary Statistics.
• normal for a single proportion
• \(p\text{-value} = 0.1308\)
• Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is \((0.411, 0.648)\)

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

• \(H_{0}: \mu = 4\)
• \(H_{a}: \mu \neq 4\)
• Let \(\bar{X}\) the average I.Q. of a set of brown trout.
• two-tailed Student's t-test
• \(t = 1.95\)
• \(p\text{-value} = 0.076\)
• Reason for decision: The \(p\text{-value}\) is greater than 0.05
• Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
• \((3.8865,5.9468)\)

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

• \(H_{a}: p < 0.13\)
• Let \(P′ =\) the proportion of Americans who have seen or sensed angels
• –2.688
• \(p\text{-value} = 0.0036\)
• Reason for decision: The \(p\text{-value}\)e is less than 0.05.
• Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

• \(H_{0}: \mu \geq 129\)
• \(H_{a}: \mu < 129\)
• Let \(\bar{X} =\) the average time in seconds that Terri finishes Lap 4.
• Student's t -distribution
• \(t = 1.209\)
• Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
• \((128.63, 130.37)\)

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

• \(H_{0}: p = 0.60\)
• \(H_{a}: p < 0.60\)
• Let \(P′ =\) the proportion of family members who shed tears at a reunion.
• –1.71
• Reason for decision: \(p\text{-value} < \alpha\)
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the \(p\text{-value}\) and alpha are quite close, so other tests should be done.
• We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. \((0.3829, 0.6171)\). The“plus-4s” confidence interval (see chapter 8) is \((0.3861, 0.6139)\)

Note that here the “large-sample” \(1 - \text{PropZTest}\) provides the approximate \(p\text{-value}\) of 0.0438. Whenever a \(p\text{-value}\) based on a normal approximation is close to the level of significance, the exact \(p\text{-value}\) based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

• \(H_{0}: \mu \geq 22\)
• \(H_{a}: \mu < 22\)
• Let \(\bar{X} =\) the mean number of bubbles per blow.
• –2.667
• \(p\text{-value} = 0.00486\)
• Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
• \((18.501, 21.499)\)

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

• 5 stores @ $2.02
• 15 stores @ $0.25
• 3 stores @ $1.29
• 6 stores @ $0.35
• 4 stores @ $2.27
• 7 stores @ $1.50
• 5 stores @ $1.89
• 8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Let \(\bar{X} =\) the mean cost in dollars of macaroni and cheese in a certain town.
• Student's \(t\)-distribution
• \(t = 0.340\)
• \(p\text{-value} = 0.36756\)
• Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
• \((0.8291, 1.241)\)

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

• HAMLET, Prince of Denmark and student of Statistics
• POLONIUS, Hamlet’s tutor
• HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

• \(H_{0}: p = 0.01\)
• \(H_{a}: p > 0.01\)
• Let \(P′ =\) the proportion of errors generated
• Normal for a single proportion
• Decision: Reject the null hypothesis
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is \((0.004, 0.144)\).

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

• \(H_{0}: p = 0.50\)
• \(H_{a}: p < 0.50\)
• Let \(P′ =\) the proportion of friends that has a pierced ear.
• –1.70
• \(p\text{-value} = 0.0448\)
• Reason for decision: The \(p\text{-value}\) is less than 0.05. (However, they are very close.)
• Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
• Confidence Interval: \((0.245, 0.515)\): The “plus-4s” confidence interval is \((0.259, 0.519)\).

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

• \(H_{0}: p = 0.40\)
• \(H_{a}: p < 0.40\)
• Let \(P′ =\) the proportion of schoolmates who fear public speaking.
• –1.01
• \(p\text{-value} = 0.1563\)
• Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
• Confidence Interval: \((0.3241, 0.4240)\): The “plus-4s” confidence interval is \((0.3257, 0.4250)\).

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

• \(H_{0}: p = 0.14\)
• \(H_{a}: p < 0.14\)
• Let \(P′ =\) the proportion of NYC residents that smoke.
• –0.2756
• \(p\text{-value} = 0.3914\)
• At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
• Confidence Interval: \((0.0502, 0.2070)\): The “plus-4s” confidence interval (see chapter 8) is \((0.0676, 0.2297)\).

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

• \(H_{0}: \mu = 69,110\)
• \(H_{0}: \mu > 69,110\)
• Let \(\bar{X} =\) the mean salary in dollars for California registered nurses.
• \(t = 1.719\)
• \(p\text{-value}: 0.0466\)
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
• \(($68,757, $73,485)\)

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?

After conducting the test, your decision and conclusion are

• Reject \(H_{0}\): There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
• Do not reject \(H_{0}\): There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
• Do not reject \(H_{0}\): There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
• Reject \(H_{0}\): There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.

At a 1% level of significance, an appropriate conclusion is:

• There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
• There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
• There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
• There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

At a significance level of \(a = 0.05\), what is the correct conclusion?

• There is enough evidence to conclude that the mean number of hours is more than 4.75
• There is enough evidence to conclude that the mean number of hours is more than 4.5
• There is not enough evidence to conclude that the mean number of hours is more than 4.5
• There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the \(p\text{-value}\).

State \(\alpha\).

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

• \(H_{0}: p = 0.488\) \(H_{a}: p \neq 0.488\)
• \(p\text{-value} = 0.0114\)
• \(\alpha = 0.05\)
• Reject the null hypothesis.
• At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
• The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \(\alpha = 0.05\), is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \(\alpha = 0.05\) level in Kentucky? Are the results applicable across the country? Why?

• \(H_{0}: p = 0.517\) \(H_{0}: p \neq 0.517\)
• \(p\text{-value} = 0.9203\).
• \(\alpha = 0.05\).
• Do not reject the null hypothesis.
• At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
• However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use \(\alpha = 0.01\) level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \(\alpha = 0.05 level\), can it be concluded that the mean rainfall was below the reported average? What if \(\alpha = 0.01\)? Assume the amount of summer rainfall follows a normal distribution.

• \(H_{0}: \mu \geq 11.52\) \(H_{a}: \mu < 11.52\)
• \(p\text{-value} = 0.000002\) which is almost 0.
• At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
• We would make the same conclusion if alpha was 1% because the \(p\text{-value}\) is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \(\alpha = 0.10\) level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the \(\alpha = 0.05\) level can it be concluded that the sample mean is higher than 5.8 visits per year?

• \(H_{0}: \mu \leq 5.8\) \(H_{a}: \mu > 5.8\)
• \(p\text{-value} = 0.9987\)
• At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At \(\alpha = 0.05\) level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

• \(H_{0}: \mu \geq 150\) \(H_{0}: \mu < 150\)
• \(p\text{-value} = 0.0622\)
• \(\alpha = 0.01\)
• At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
• The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion

Hypothesis testing

• PMID: 8900794
• DOI: 10.1097/00002800-199607000-00009

Hypothesis testing is the process of making a choice between two conflicting hypotheses. The null hypothesis, H0, is a statistical proposition stating that there is no significant difference between a hypothesized value of a population parameter and its value estimated from a sample drawn from that population. The alternative hypothesis, H1 or Ha, is a statistical proposition stating that there is a significant difference between a hypothesized value of a population parameter and its estimated value. When the null hypothesis is tested, a decision is either correct or incorrect. An incorrect decision can be made in two ways: We can reject the null hypothesis when it is true (Type I error) or we can fail to reject the null hypothesis when it is false (Type II error). The probability of making Type I and Type II errors is designated by alpha and beta, respectively. The smallest observed significance level for which the null hypothesis would be rejected is referred to as the p-value. The p-value only has meaning as a measure of confidence when the decision is to reject the null hypothesis. It has no meaning when the decision is that the null hypothesis is true.

• Data Interpretation, Statistical*
• Nursing Research
• Reproducibility of Results

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