Section 10.4: Hypothesis Tests for a Population Standard Deviation

  • 10.1 The Language of Hypothesis Testing
  • 10.2 Hypothesis Tests for a Population Proportion
  • 10.3 Hypothesis Tests for a Population Mean
  • 10.4 Hypothesis Tests for a Population Standard Deviation
  • 10.5 Putting It Together: Which Method Do I Use?

By the end of this lesson, you will be able to...

  • test hypotheses about a population standard deviation

For a quick overview of this section, watch this short video summary:

Before we begin this section, we need a quick refresher of the Χ 2 distribution.

The Chi-Square ( Χ 2 ) distribution

Reminder: "chi-square" is pronounced "kai" as in sky, not "chai" like the tea .

If a random sample size n is obtained from a normally distributed population with mean μ and standard deviation σ , then

has a chi-square distribution with n-1 degrees of freedom.

Properties of the Χ 2 distribution

  • It is not symmetric.
  • The shape depends on the degrees of freedom.
  • As the number of degrees of freedom increases, the distribution becomes more symmetric.
  • Χ 2 ≥0

Finding Probabilities Using StatCrunch

We again have some conditions that need to be true in order to perform the test 

  • the sample was randomly selected, and
  • the population from which the sample is drawn is normally distributed

Note that in the second requirement, the population must be normally distributed. The steps in performing the hypothesis test should be familiar by now.

Performing a Hypothesis Test Regarding σ

Step 1 : State the null and alternative hypotheses.

Step 2 : Decide on a level of significance, α .

Step 4 : Determine the P -value.

Step 5 : Reject the null hypothesis if the P -value is less than the level of significance, α.

Step 6 : State the conclusion.

In Example 2 , in Section 10.2, we assumed that the standard deviation for the resting heart rates of ECC students was 12 bpm. Later, in Example 2 in Section 10.3, we considered the actual sample data below.

( Click here to view the data in a format more easily copied.)

Based on this sample, is there enough evidence to say that the standard deviation of the resting heart rates for students in this class is different from 12 bpm?

Note: Be sure to check that the conditions for performing the hypothesis test are met.

[ reveal answer ]

From the earlier examples, we know that the resting heart rates could come from a normally distributed population and there are no outliers.

Step 1 : H 0 : σ = 12 H 1 : σ ≠ 12

Step 2 : α = 0.05

Step 4 : P -value = 2P( Χ 2 > 15.89) ≈ 0.2159

Step 5 : Since P -value > α , we do not reject H 0 .

Step 6 : There is not enough evidence at the 5% level of significance to support the claim that the standard deviation of the resting heart rates for students in this class is different from 12 bpm.

Hypothesis Testing Regarding σ Using StatCrunch

Let's look at Example 1 again, and try the hypothesis test with technology.

Using DDXL:

Using StatCrunch:

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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

  • State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
  • Collect data in a way designed to test the hypothesis.
  • Perform an appropriate statistical test .
  • Decide whether to reject or fail to reject your null hypothesis.
  • Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

  • H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

  • an estimate of the difference in average height between the two groups.
  • a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

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hypothesis testing of standard deviation

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

  • Normal distribution
  • Descriptive statistics
  • Measures of central tendency
  • Correlation coefficient

Methodology

  • Cluster sampling
  • Stratified sampling
  • Types of interviews
  • Cohort study
  • Thematic analysis

Research bias

  • Implicit bias
  • Cognitive bias
  • Survivorship bias
  • Availability heuristic
  • Nonresponse bias
  • Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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Hypothesis Testing for Population Standard Deviation

A comprehensive guide, to testing hypotheses for standard deviation.

In the realm of analysis it is vital to comprehend the variability in data. The standard deviation serves as a measure of this variability by quantifying how data points deviate from the mean. However what if you need to determine whether a populations standard deviation aligns with a hypothesis? This is where hypothesis testing for deviation becomes relevant. In this article we will explore the concept of hypothesis testing for deviation, its importance and how you can conduct it using a calculator specifically designed for this purpose. Whether you are a student seeking assistance with statistics or a professional researcher this guide will serve as your resource.

Understanding Hypothesis Testing for Standard Deviation

Hypothesis testing for deviation also referred to as variance hypothesis testing, aids in establishing whether a populations standard deviation aligns with a given hypothesis or if there exists a difference. This technique proves valuable when evaluating data variability, which holds significance in various fields such, as quality control, finance and medical research.

To understand hypothesis testing, for deviation it is important to grasp the following concepts;

Null Hypothesis (H0); This is the initial assumption that there is no significant difference between the standard deviation and the hypothesized value.

Alternative Hypothesis (Ha); In contrast to the hypothesis this suggests that there is a difference between the standard deviation and the hypothesized value.

Test Statistic; This numerical value is calculated from your sample data and used to evaluate whether the null hypothesis should be rejected.

Significance Level (α); This represents the probability of making a Type I error, which occurs when we reject the hypothesis incorrectly. Commonly used values for α are 0.05 and 0.01.

To perform hypothesis testing, for deviation you need to follow these steps;

Step 1; Clearly state your hypothesis (H0) and alternative hypothesis (Ha) based on your research question. For example;

H0; The population standard deviation equals a value (σ = x). Ha; The population standard deviation does not equal the value (σ ≠ x).Step 2; Gather a sample from the group of people you’re interested, in studying.

Step 3; Calculate the test statistic, such as the squared statistic using the data from your sample. This will help you evaluate your hypothesis.

Step 4; Find the value either by referring to a table for squared distribution or by utilizing an online calculator designed specifically for testing hypotheses about standard deviation. You can compare this value with your calculated test statistic.

Step 5; Make a decision by comparing your test statistic with the value. If your test statistic falls within the rejection region (for example it is greater than the value in a two tailed test) then you reject the hypothesis. Otherwise if it does not fall within that range you fail to reject the hypothesis.

Step 6; Based on your decision in Step 5 draw conclusions regarding your research question.

Using a calculator specifically designed for performing hypothesis tests about deviation can simplify and expedite this process. These calculators allow you to input information about your sample data and hypothesis details. As a result they provide you with both the test statistic and critical values, for analysis while saving valuable time.

In summary;

Testing the deviation through hypothesis testing is a statistical method used to evaluate data variability. It enables you to make informed decisions by relying on evidence, from your sample. Whether you’re involved in research, quality control or seeking assistance with statistics having a grasp of this technique can greatly benefit you. Don’t forget to utilize resources like the Hypothesis Test, for Standard Deviation Calculator to streamline your analysis and improve your skills.

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Hypothesis Tests for One or Two Variances or Standard Deviations

Chi-Square-tests and F-tests for variance or standard deviation both require that the original population be normally distributed.

Testing a Claim about a Variance or Standard Deviation

To test a claim about the value of the variance or the standard deviation of a population, then the test statistic will follow a chi-square distribution with $n-1$ dgrees of freedom, and is given by the following formula.

The television habits of 30 children were observed. The sample mean was found to be 48.2 hours per week, with a standard deviation of 12.4 hours per week. Test the claim that the standard deviation was at least 16 hours per week.

  • The hypotheses are: $H_0: \sigma = 16$ $H_a: \sigma < 16$
  • We shall choose   $\alpha = 0.05$.
  • The test statistic is   $\chi^2 = \dfrac{(n-1)s^2}{\sigma_0^2} = \dfrac{(30-1)12.4^2}{16^2} = 17.418$.
  • The p-value is   $p = \chi^2\text{cdf}(0,17.418,29) = 0.0447$.
  • The variation in television watching was less than 16 hours per week.

Testing a the Difference of Two Variances or Two Standard Deviations

Two equal variances would satisfy the equation   $\sigma_1^2 = \sigma_2^2$,   which is equivalent to   $\dfrac{ \sigma_1^2}{\sigma_2^2} = 1$.   Since sample variances are related to chi-square distributions, and the ratio of chi-square distributions is an F-distribution, we can use the F-distribution to test against a null hypothesis of equal variances. Note that this approach does not allow us to test for a particular magnitude of difference between variances or standard deviations.

Given sample sizes of $n_1$ and $n_2$, the test statistic will have   $n_1-1$   and   $n_2-1$   degrees of freedom, and is given by the following formula.

If the larger variance (or standard deviation) is present in the first sample, then the test is right-tailed. Otherwise, the test is left-tailed. Most tables of the F-distribution assume right-tailed tests, but that requirement may not be necessary when using technology.

Samples from two makers of ball bearings are collected, and their diameters (in inches) are measured, with the following results:

  • Acme: $n_1 = 80$, $s_1 = 0.0395$
  • Bigelow: $n_2 = 120$, $s_2 = 0.0428$
  • The hypotheses are: $H_0: \sigma_1 = \sigma_2$ $H_a: \sigma_1 \neq \sigma_2$
  • The test statistic is   $F = \dfrac{s_1^2}{s_2^2} = \dfrac{0.0395^2}{0.0428^2} = 0.8517$.
  • Since the first sample had the smaller standard deviation, this is a left-tailed test. The p-value is   $p = \operatorname{Fcdf}(0,0.8517,79,119) = 0.2232$.
  • There is insufficient evidence to conclude that the diameters of the ball bearings in the two companies have different standard deviations.

If the two samples had been reversed in our computations, we would have obtained the test statistic   $F = 1.1741$,   and performing a right-tailed test, found the p-value   $p = \operatorname{Fcdf}(1.1741,\infty,119,79) = 0.2232$.   Of course, the answer is the same.

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8.7 Hypothesis Tests for a Population Mean with Unknown Population Standard Deviation

Learning objectives.

  • Conduct and interpret hypothesis tests for a population mean with unknown population standard deviation.

Some notes about conducting a hypothesis test:

  • The null hypothesis [latex]H_0[/latex] is always an “equal to.”  The null hypothesis is the original claim about the population parameter.
  • The alternative hypothesis [latex]H_a[/latex] is a “less than,” “greater than,” or “not equal to.”  The form of the alternative hypothesis depends on the context of the question.
  • If the alternative hypothesis is a “less than”,  then the test is left-tail.  The p -value is the area in the left-tail of the distribution.
  • If the alternative hypothesis is a “greater than”, then the test is right-tail.  The p -value is the area in the right-tail of the distribution.
  • If the alternative hypothesis is a “not equal to”, then the test is two-tail.  The p -value is the sum of the area in the two-tails of the distribution.  Each tail represents exactly half of the p -value.
  • Think about the meaning of the p -value.  A data analyst (and anyone else) should have more confidence that they made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using a significance level of  0.05.  Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (a significance level of 0.05 is less than either number), a data analyst should have more confidence that they made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.
  • The significance level must be identified before collecting the sample data and conducting the test.  Generally, the significance level will be included in the question.  If no significance level is given, a common standard is to use a significance level of 5%.
  • An alternative approach for hypothesis testing is to use what is called the critical value approach .  In this book, we will only use the p -value approach.  Some of the videos below may mention the critical value approach, but this approach will not be used in this book.

Steps to Conduct a Hypothesis Test for a Population Mean with Unknown Population Standard Deviation

  • Write down the null and alternative hypotheses in terms of the population mean [latex]\mu[/latex].  Include appropriate units with the values of the mean.
  • Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
  • Collect the sample information for the test and identify the significance level [latex]\alpha[/latex].

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ df & = & n-1 \\ \\ \end{eqnarray*}[/latex]

  • The results of the sample data are significant. There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
  • The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
  • Write down a concluding sentence specific to the context of the question.

USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON A POPULATION MEAN WITH UNKNOWN POPULATION STANDARD DEVIATION

The p -value for a hypothesis test on a population mean is the area in the tail(s) of the distribution of the sample mean.  When the population standard deviation is unknown, use the [latex]t[/latex]-distribution to find the p -value.

If the p -value is the area in the left-tail:

  • For t-score , enter the value of [latex]t[/latex] calculated from [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex].
  • For degrees of freedom , enter the degrees of freedom for the [latex]t[/latex]-distribution [latex]n-1[/latex].
  • For the logic operator , enter true .  Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.
  • The output from the t.dist function is the area under the [latex]t[/latex]-distribution to the left of the entered [latex]t[/latex]-score.
  • Visit the Microsoft page for more information about the t.dist function.

If the p -value is the area in the right-tail:

  • The output from the t.dist.rt function is the area under the [latex]t[/latex]-distribution to the right of the entered [latex]t[/latex]-score.
  • Visit the Microsoft page for more information about the t.dist.rt function.

If the p -value is the sum of area in the tails:

  • For t-score , enter the absolute value of [latex]t[/latex] calculated from [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex].  Note:  In the t.dist.2t function, the value of the [latex]t[/latex]-score must be a positive number.  If the [latex]t[/latex]-score is negative, enter the absolute value of the [latex]t[/latex]-score into the t.dist.2t function.
  • The output from the t.dist.2t function is the sum of areas in the tails under the [latex]t[/latex]-distribution.
  • Visit the Microsoft page for more information about the t.dist.2t function.

Statistics students believe that the mean score on the first statistics test is 65.  A statistics instructor thinks the mean score is higher than 65.  He samples ten statistics students and obtains the following scores:

The instructor performs a hypothesis test using a 1% level of significance. The test scores are assumed to be from a normal distribution.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu=65  \\ H_a: & & \mu \gt 65  \end{eqnarray*}[/latex]

From the question, we have [latex]n=10[/latex], [latex]\overline{x}=67[/latex], [latex]s=3.1972...[/latex] and [latex]\alpha=0.01[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=3.1972...[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the area in the right-tail of the distribution.

This is a t-distribution curve. The peak of the curve is at 0 on the horizontal axis. The point t is also labeled. A vertical line extends from point t to the curve with the area to the right of this vertical line shaded. The p-value equals the area of this shaded region.

To use the t.dist.rt function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{67-65}{\frac{3.1972...}{\sqrt{10}}} \\ & = & 1.9781... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=10-1=9[/latex].

So the p -value[latex]=0.0396[/latex].

Conclusion:

Because p -value[latex]=0.0396 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that mean score on the test is greater than 65.

  • The null hypothesis [latex]\mu=65[/latex] is the claim that the mean test score is 65.
  • The alternative hypothesis [latex]\mu \gt 65[/latex] is the claim that the mean test score is greater than 65.
  • Keep all of the decimals throughout the calculation (i.e. in the sample standard deviation, the [latex]t[/latex]-score, etc.) to avoid any round-off error in the calculation of the p -value.  This ensures that we get the most accurate value for the p -value.
  • The p -value is the area in the right-tail of the [latex]t[/latex]-distribution, to the right of [latex]t=1.9781...[/latex].
  • The p -value of 0.0396 tells us that under the assumption that the mean test score is 65 (the null hypothesis), there is a 3.96% chance that the mean test score is 65 or more.  Compared to the 1% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.

A company claims that the average change in the value of their stock is $3.50 per week.  An investor believes this average is too high. The investor records the changes in the company’s stock price over 30 weeks and finds the average change in the stock price is $2.60 with a standard deviation of $1.80.  At the 5% significance level, is the average change in the company’s stock price lower than the company claims?

[latex]\begin{eqnarray*} H_0: & & \mu=$3.50  \\ H_a: & & \mu \lt $3.50  \end{eqnarray*}[/latex]

From the question, we have [latex]n=30[/latex], [latex]\overline{x}=2.6[/latex], [latex]s=1.8[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=1.8.[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the area in the left-tail of the distribution.

his is a t-distribution curve. The peak of the curve is at 0 on the horizontal axis. The point t is also labeled. A vertical line extends from point t to the curve with the area to the left of this vertical line shaded. The p-value equals the area of this shaded region.

To use the t.dist function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{2.6-3.5}{\frac{1.8}{\sqrt{30}}} \\ & = & -1.5699... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=30-1=29[/latex].

So the p -value[latex]=0.0636[/latex].

Because p -value[latex]=0.0636 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that average change in the stock price is lower than $3.50.

  • The null hypothesis [latex]\mu=$3.50[/latex] is the claim that the average change in the company’s stock is $3.50 per week.
  • The alternative hypothesis [latex]\mu \lt $3.50[/latex] is the claim that the average change in the company’s stock is less than $3.50 per week.
  • The p -value is the area in the left-tail of the [latex]t[/latex]-distribution, to the left of [latex]t=-1.5699...[/latex].
  • The p -value of 0.0636 tells us that under the assumption that the average change in the stock is $3.50 (the null hypothesis), there is a 6.36% chance that the average change is $3.50 or less.  Compared to the 5% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the company’s claim that the average change in their stock price is $3.50 per week is most likely correct.

A paint manufacturer has their production line set-up so that the average volume of paint in a can is 3.78 liters.  The quality control manager at the plant believes that something has happened with the production and the average volume of paint in the cans has changed.  The quality control department takes a sample of 100 cans and finds the average volume is 3.62 liters with a standard deviation of 0.7 liters.  At the 5% significance level, has the volume of paint in a can changed?

[latex]\begin{eqnarray*} H_0: & & \mu=3.78 \mbox{ liters}  \\ H_a: & & \mu \neq 3.78 \mbox{ liters}  \end{eqnarray*}[/latex]

From the question, we have [latex]n=100[/latex], [latex]\overline{x}=3.62[/latex], [latex]s=0.7[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=0.7[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p -value is the sum of area in the tails of the distribution.

This is a t distribution curve. The peak of the curve is at 0 on the horizontal axis. The point -t and t are also labeled. A vertical line extends from point t to the curve with the area to the right of this vertical line shaded with the shaded area labeled half of the p-value. A vertical line extends from -t to the curve with the area to the left of this vertical line shaded with the shaded area labeled half of the p-value. The p-value equals the area of these two shaded regions.

To use the t.dist.2t function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{3.62-3.78}{\frac{0.07}{\sqrt{100}}} \\ & = & -2.2857... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=100-1=99[/latex].

So the p -value[latex]=0.0244[/latex].

Because p -value[latex]=0.0244 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that average volume of paint in the cans has changed.

  • The null hypothesis [latex]\mu=3.78[/latex] is the claim that the average volume of paint in the cans is 3.78.
  • The alternative hypothesis [latex]\mu \neq 3.78[/latex] is the claim that the average volume of paint in the cans is not 3.78.
  • Keep all of the decimals throughout the calculation (i.e. in the [latex]t[/latex]-score) to avoid any round-off error in the calculation of the p -value.  This ensures that we get the most accurate value for the p -value.
  • The p -value is the sum of the area in the two tails.  The output from the t.dist.2t function is exactly the sum of the area in the two tails, and so is the p -value required for the test.  No additional calculations are required.
  • The t.dist.2t function requires that the value entered for the [latex]t[/latex]-score is positive .  A negative [latex]t[/latex]-score entered into the t.dist.2t function generates an error in Excel.  In this case, the value of the [latex]t[/latex]-score is negative, so we must enter the absolute value of this [latex]t[/latex]-score into field 1.
  • The p -value of 0.0244 is a small probability compared to the significance level, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.  In other words, the average volume of paint in the cans has most likely changed from 3.78 liters.

Watch this video: Hypothesis Testing: t -test, right tail by ExcelIsFun [11:02]

Watch this video: Hypothesis Testing: t -test, left tail by ExcelIsFun [7:48]

Watch this video: Hypothesis Testing: t -test, two tail by ExcelIsFun [8:54]

Concept Review

The hypothesis test for a population mean is a well established process:

  • Collect the sample information for the test and identify the significance level.
  • When the population standard deviation is unknown, find the p -value (the area in the corresponding tail) for the test using the [latex]t[/latex]-distribution with [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex] and [latex]df=n-1[/latex].
  • Compare the p -value to the significance level and state the outcome of the test.

Attribution

“ 9.6   Hypothesis Testing of a Single Mean and Single Proportion “ in Introductory Statistics by OpenStax  is licensed under a  Creative Commons Attribution 4.0 International License.

Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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8.6: Hypothesis Test of a Single Population Mean with Examples

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Steps for performing Hypothesis Test of a Single Population Mean

Step 1: State your hypotheses about the population mean. Step 2: Summarize the data. State a significance level. State and check conditions required for the procedure

  • Find or identify the sample size, n, the sample mean, \(\bar{x}\) and the sample standard deviation, s .

The sampling distribution for the one-mean test statistic is, approximately, T- distribution if the following conditions are met

  • Sample is random with independent observations .
  • Sample is large. The population must be Normal or the sample size must be at least 30.

Step 3: Perform the procedure based on the assumption that \(H_{0}\) is true

  • Find the Estimated Standard Error: \(SE=\frac{s}{\sqrt{n}}\).
  • Compute the observed value of the test statistic: \(T_{obs}=\frac{\bar{x}-\mu_{0}}{SE}\).
  • Check the type of the test (right-, left-, or two-tailed)
  • Find the p-value in order to measure your level of surprise.

Step 4: Make a decision about \(H_{0}\) and \(H_{a}\)

  • Do you reject or not reject your null hypothesis?

Step 5: Make a conclusion

  • What does this mean in the context of the data?

The following examples illustrate a left-, right-, and two-tailed test.

Example \(\pageindex{1}\).

\(H_{0}: \mu = 5, H_{a}: \mu < 5\)

Test of a single population mean. \(H_{a}\) tells you the test is left-tailed. The picture of the \(p\)-value is as follows:

Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.

Exercise \(\PageIndex{1}\)

\(H_{0}: \mu = 10, H_{a}: \mu < 10\)

Assume the \(p\)-value is 0.0935. What type of test is this? Draw the picture of the \(p\)-value.

left-tailed test

alt

Example \(\PageIndex{2}\)

\(H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2\)

This is a test of a single population proportion. \(H_{a}\) tells you the test is right-tailed . The picture of the p -value is as follows:

Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.

Exercise \(\PageIndex{2}\)

\(H_{0}: \mu \leq 1, H_{a}: \mu > 1\)

Assume the \(p\)-value is 0.1243. What type of test is this? Draw the picture of the \(p\)-value.

right-tailed test

alt

Example \(\PageIndex{3}\)

\(H_{0}: \mu = 50, H_{a}: \mu \neq 50\)

This is a test of a single population mean. \(H_{a}\) tells you the test is two-tailed . The picture of the \(p\)-value is as follows.

Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.

Exercise \(\PageIndex{3}\)

\(H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5\)

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the \(p\)-value.

two-tailed test

alt

Full Hypothesis Test Examples

Example \(\pageindex{4}\).

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65  H_{a}: \mu > 65\)

Since the instructor thinks the average score is higher, use a "\(>\)". The "\(>\)" means the test is right-tailed.

Determine the distribution needed:

Random variable: \(\bar{X} =\) average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Use \(t_{df}\). Therefore, the distribution for the test is \(t_{9}\) where \(n = 10\) and \(df = 10 - 1 = 9\).

The sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Calculate the \(p\)-value using the Student's \(t\)-distribution:

\[t_{obs} = \dfrac{\bar{x}-\mu_{\bar{x}}}{\left(\dfrac{s}{\sqrt{n}}\right)}=\dfrac{67-65}{\left(\dfrac{3.1972}{\sqrt{10}}\right)}\]

Use the T-table or Excel's t_dist() function to find p-value:

\(p\text{-value} = P(\bar{x} > 67) =P(T >1.9782 )= 1-0.9604=0.0396\)

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.

Compare \(\alpha\) and the \(p-\text{value}\):

Since \(α = 0.05\) and \(p\text{-value} = 0.0396\). \(\alpha > p\text{-value}\).

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\).

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The \(p\text{-value}\) can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq: . Arrow down to \(\mu\): and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p\text{-value}\) (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 65\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(t = 1.9781\) (test statistic) and \(p = 0.0396\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Exercise \(\PageIndex{4}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors.

  • \(H_{0}: \mu = 5\)
  • \(H_{a}: \mu < 5\)
  • \(p = 0.0082\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

  • Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
  • Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{5}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

  • \(H_{0}: \mu \leq 1\)
  • \(H_{a}: \mu > 1\)
  • Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
  • Do the calculations : \(p\text{-value} ( = 0.036)\)

4. State the Conclusions : Since the \(p\text{-value} (= 0.036)\) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

The hypothesis test itself has an established process. This can be summarized as follows:

  • Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
  • Determine the random variable.
  • Determine the distribution for the test.
  • Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A t -score is an example of test statistics.)
  • Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

  • Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
  • Data from Bloomberg Businessweek . Available online at www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
  • Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
  • Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
  • Data from Growing by Degrees by Allen and Seaman.
  • Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
  • Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
  • Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
  • Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
  • Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
  • Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
  • Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
  • Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
  • Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
  • Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
  • “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
  • Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
  • Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).
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Lesson 21: Hypothesis Testing with Known Standard Deviation

Hi everyone! Read through the material below, watch the videos, work on the Excel lecture and follow up with your instructor if you have questions.

Learning Outcomes.

  • Determine if the data supports a hypothesis at a given significance level using known distributions.

Topic . This lesson covers: Hypothesis Testing with Known Standard Deviation

  • 9.1 Null and Alternative Hypotheses
  • 9.2 Outcomes and the Type I and Type II Errors
  • Introductory Statistics by Sheldon Ross, 3rd edition: Section 9.2 – 9.3.1
  • Statistics with Microsoft Excel by Beverly J. Dretzke, 5th ed.,  P. 131 – 153

WeBWorK . Sets 9.1 & 9.2

The Applied View

Watch the video Tests of Significance .

  • In the 1970s, statistician Ron Thisted did a statistical analysis of Shakespeare’s vocabulary. Based on his analysis he created a computer program. What could his program tell you about a Shakespearean poem?
  • In analyzing a poem to see whether or not it was authored by Shakespeare, Thisted set up a null hypothesis and an alternative hypothesis. State those hypotheses in words.
  • What was the approximate distribution of the number of unique words per poem in Shakespeare’s poems?
  • Thisted observed 10 unique words in the newly discovered poem. Was that sufficient evidence to conclude that Shakespeare did not write the poem?
  • Which is better evidence against the null hypothesis, a large p-value or a small p-value?

Exit Ticket

The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether the meat inspector’s claim is true. When will a type I or type II error occur? Which is more serious?

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Module 11: The Chi Square Distribution

Hypothesis test for variance, learning outcomes.

  • Conduct a hypothesis test on one variance and interpret the conclusion in context

Recall: STANDARD DEVIATION AND VARIANCE

The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.

To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations [latex](x- \overline{x})[/latex] values for a sample, or the [latex]x – μ[/latex] values for a population). The symbol [latex]\sigma ^2[/latex] represents the population variance; the population standard deviation [latex]σ[/latex] is the square root of the population variance. The symbol [latex]s^2[/latex] represents the sample variance; the sample standard deviation [latex]s[/latex] is the square root of the sample variance.

The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.

A test of a single variance assumes that the underlying distribution is normal . The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is:

[latex]\displaystyle\dfrac{\left(n-1\right)s^2}{\sigma^2}[/latex]

  • [latex]n[/latex] = the total number of data
  • [latex]s^2[/latex] = sample variance
  • [latex]\sigma^2[/latex] = population variance

You may think of [latex]s[/latex] as the random variable in this test. The number of degrees of freedom is [latex]df=n-1[/latex]. A test of a single variance may be right-tailed, left-tailed, or two-tailed.  The example below will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average.

Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?

Even though we are given the population standard deviation, we can set up the test using the population variance as follows.

H 0 : σ 2 = 5 2 H a : σ 2 > 5 2

A scuba instructor wants to record the collective depths of each of his students’ dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be?

Recall: ORDER OF OPERATIONS

To calculate the test statistic follow the following steps:.

1st find the numerator:

Step 1: Calculate [latex](n-1)[/latex] by reading the problem or counting the total number of data points and then subtract [latex]1[/latex].

Step 2: Calculate [latex]s^2[/latex], and find the variance from the sample. This can be given to you in the problem or can be calculated with the following formula described in Module 2.

[latex]s^2= \frac{\sum (x- \overline{x})^2}{n-1}[/latex]. Note if you are performing a test of a single standard deviation,

Step 3: Multiply the values you got in Step 1 and Step 2.

Note: if you are performing a test of a single standard deviation, in step 2, calculate the standard deviation, [latex]s[/latex], by taking the square root of the variance.

2nd find the denominator: If you are performing a test of a single variance, read the problem or calculate the population variance with the data. If you are performing a test of a single standard deviation, read the problem or calculate the population standard deviation with the data.

Formula for the Population Variance: [latex]\sigma ^2 = \frac{\sum (x- \mu)^2}{N}[/latex]

Formula for the Population Standard Deviation: [latex]\sigma = \sqrt{\frac{\sum (x- \mu)^2}{N}}[/latex]

3rd take the numerator and divide by the denominator.

With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.

With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers .

Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ 2 , or the population standard deviation, σ .

Random Variable: The sample standard deviation, s , is the random variable. Let s = standard deviation for the waiting times.

H 0 : σ 2 = 7.22 H a : σ 2 < 7.22

The word “less” tells you this is a left-tailed test.

Distribution for the test:  [latex]X^2_{24}[/latex], where:

  • n = the number of customers sampled
  • df = n – 1 = 25 – 1 = 24

Calculate the test statistic:

X 2 = [latex]\dfrac{\left ( n-1 \right )s^2}{\sigma^2}[/latex] = [latex]\dfrac{\left ( 25-1 \right )\left ( 3.5 \right )^2}{7.2^2}[/latex] = 5.67

where n = 25, s = 3.5, and σ = 7.2.

This is a nonsymmetrical chi-square curve with values of 0 and 5.67 labeled on the horizontal axis. The point 5.67 lies to the left of the peak of the curve. A vertical upward line extends from 5.67 to the curve and the region to the left of this line is shaded. The shaded area is equal to the p-value.

Probability statement: p -value = P (χ 2 < 5.67) = 0.000042

Compare α and the p -value:

α = 0.05; p -value = 0.000042; α > p -value

Make a decision: Since α > p -value, reject H 0 . This means that you reject σ 2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes.

Using a calculator:

In 2nd DISTR , use 7:χ2cdf . The syntax is (lower, upper, df) for the parameter list. For this example, χ 2 cdf(-1E99,5.67,24) . The p -value = 0.000042.

The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August 2012, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom and the test statistic, sketch the graph of the p -value, and draw a conclusion. Test at the 1% significance level.

  • Introductory Statistics. Authored by : Barbara Illowsky, Susan Dean. Provided by : OpenStax. Located at : https://openstax.org/books/introductory-statistics/pages/1-introduction . License : CC BY: Attribution . License Terms : Access for free at https://openstax.org/books/introductory-statistics/pages/1-introduction

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  • 10.2 Two Population Means with Known Standard Deviations
  • Introduction
  • 1.1 Definitions of Statistics, Probability, and Key Terms
  • 1.2 Data, Sampling, and Variation in Data and Sampling
  • 1.3 Frequency, Frequency Tables, and Levels of Measurement
  • 1.4 Experimental Design and Ethics
  • 1.5 Data Collection Experiment
  • 1.6 Sampling Experiment
  • Chapter Review
  • Bringing It Together: Homework
  • 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
  • 2.2 Histograms, Frequency Polygons, and Time Series Graphs
  • 2.3 Measures of the Location of the Data
  • 2.4 Box Plots
  • 2.5 Measures of the Center of the Data
  • 2.6 Skewness and the Mean, Median, and Mode
  • 2.7 Measures of the Spread of the Data
  • 2.8 Descriptive Statistics
  • Formula Review
  • 3.1 Terminology
  • 3.2 Independent and Mutually Exclusive Events
  • 3.3 Two Basic Rules of Probability
  • 3.4 Contingency Tables
  • 3.5 Tree and Venn Diagrams
  • 3.6 Probability Topics
  • Bringing It Together: Practice
  • 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
  • 4.2 Mean or Expected Value and Standard Deviation
  • 4.3 Binomial Distribution (Optional)
  • 4.4 Geometric Distribution (Optional)
  • 4.5 Hypergeometric Distribution (Optional)
  • 4.6 Poisson Distribution (Optional)
  • 4.7 Discrete Distribution (Playing Card Experiment)
  • 4.8 Discrete Distribution (Lucky Dice Experiment)
  • 5.1 Continuous Probability Functions
  • 5.2 The Uniform Distribution
  • 5.3 The Exponential Distribution (Optional)
  • 5.4 Continuous Distribution
  • 6.1 The Standard Normal Distribution
  • 6.2 Using the Normal Distribution
  • 6.3 Normal Distribution—Lap Times
  • 6.4 Normal Distribution—Pinkie Length
  • 7.1 The Central Limit Theorem for Sample Means (Averages)
  • 7.2 The Central Limit Theorem for Sums (Optional)
  • 7.3 Using the Central Limit Theorem
  • 7.4 Central Limit Theorem (Pocket Change)
  • 7.5 Central Limit Theorem (Cookie Recipes)
  • 8.1 A Single Population Mean Using the Normal Distribution
  • 8.2 A Single Population Mean Using the Student's t-Distribution
  • 8.3 A Population Proportion
  • 8.4 Confidence Interval (Home Costs)
  • 8.5 Confidence Interval (Place of Birth)
  • 8.6 Confidence Interval (Women's Heights)
  • 9.1 Null and Alternative Hypotheses
  • 9.2 Outcomes and the Type I and Type II Errors
  • 9.3 Distribution Needed for Hypothesis Testing
  • 9.4 Rare Events, the Sample, and the Decision and Conclusion
  • 9.5 Additional Information and Full Hypothesis Test Examples
  • 9.6 Hypothesis Testing of a Single Mean and Single Proportion
  • 10.1 Two Population Means with Unknown Standard Deviations
  • 10.3 Comparing Two Independent Population Proportions
  • 10.4 Matched or Paired Samples (Optional)
  • 10.5 Hypothesis Testing for Two Means and Two Proportions
  • 11.1 Facts About the Chi-Square Distribution
  • 11.2 Goodness-of-Fit Test
  • 11.3 Test of Independence
  • 11.4 Test for Homogeneity
  • 11.5 Comparison of the Chi-Square Tests
  • 11.6 Test of a Single Variance
  • 11.7 Lab 1: Chi-Square Goodness-of-Fit
  • 11.8 Lab 2: Chi-Square Test of Independence
  • 12.1 Linear Equations
  • 12.2 The Regression Equation
  • 12.3 Testing the Significance of the Correlation Coefficient (Optional)
  • 12.4 Prediction (Optional)
  • 12.5 Outliers
  • 12.6 Regression (Distance from School) (Optional)
  • 12.7 Regression (Textbook Cost) (Optional)
  • 12.8 Regression (Fuel Efficiency) (Optional)
  • 13.1 One-Way ANOVA
  • 13.2 The F Distribution and the F Ratio
  • 13.3 Facts About the F Distribution
  • 13.4 Test of Two Variances
  • 13.5 Lab: One-Way ANOVA
  • A | Appendix A Review Exercises (Ch 3–13)
  • B | Appendix B Practice Tests (1–4) and Final Exams
  • C | Data Sets
  • D | Group and Partner Projects
  • E | Solution Sheets
  • F | Mathematical Phrases, Symbols, and Formulas
  • G | Notes for the TI-83, 83+, 84, 84+ Calculators

Even though this situation is not likely (knowing the population standard deviations), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal, and both populations must be normal. The random variable is X ¯ 1 – X ¯ 2 X ¯ 1 – X ¯ 2 . The normal distribution has the following format: Normal distribution is

Example 10.6

Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distribution. The data are recorded in Table 10.8 .

Does the data indicate that Wax 1 is more effective than Wax 2? Test at a 5 percent level of significance.

This is a test of two independent groups, two population means, population standard deviations known.

Random Variable : X ¯ 1 –  X ¯ 2 X ¯ 1 –  X ¯ 2 = difference in the mean number of months the competing floor waxes last.

H 0 : μ 1 ≤ μ 2

H a : μ 1 > μ 2

The words is more effective says that Wax 1 lasts longer than Wax 2 , on average. Longer is a > symbol and goes into H a . Therefore, this is a right-tailed test.

Distribution for the test: The population standard deviations are known, so the distribution is normal. Using the formula, the distribution is

X ¯ 1 – X ¯ 2 ~ N ( 0 , 0.33 2 20 + 0.36 2 20 ) . X ¯ 1 – X ¯ 2 ~ N ( 0 , 0.33 2 20 + 0.36 2 20 ) .

Since μ 1 ≤ μ 2 , then μ 1 – μ 2 ≤ 0 and the mean for the normal distribution is zero.

Calculate the p value using the normal distribution: p value = 0.1799

X ¯ 1 X ¯ 1 – X ¯ 2 X ¯ 2 = 3 – 2.9 = 0.1

Compare α and the p value: α = 0.05 and p value = 0.1799. Therefore, α < p value.

Make a decision: Since α < p value, do not reject H 0 .

Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time Wax 1 lasts is longer (Wax 1 is more effective) than the mean time Wax 2 lasts.

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT . Arrow over to TESTS and press 3:2-SampZTest . Arrow over to Stats and press ENTER . Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ 1: and arrow to > μ 2 . Press ENTER . Arrow down to Calculate and press ENTER . The p value is p = 0.1799, and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw .

Try It 10.6

The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.9 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5 percent level of significance.

Example 10.7

An interested citizen wanted to know if Democratic U.S. senators are older than Republican U.S. senators, on average. On May 26, 2013, the mean age of 30 randomly selected Republican senators was 61 years 247 days (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days (61.704 years) with a standard deviation of 9.55 years.

Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5 percent level of significance.

This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s- t distribution. Subscripts: 1: Democratic senators; 2: Republican senators

Random variable: X ¯ 1  –  X ¯ 2 X ¯ 1  –  X ¯ 2 = difference in the mean age of Democratic and Republican U.S. senators.

H 0 : µ 1 ≤ µ 2   H 0 : µ 1 – µ 2 ≤ 0

H a : µ 1 > µ 2   H a : µ 1 – µ 2 > 0

The words older than translates as a > symbol and goes into H a . Therefore, this is a right-tailed test.

Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is

Since µ 1 ≤ µ 2 , µ 1 – µ 2 ≤ 0 and the mean for the normal distribution is zero.

Calculating the p value using the normal distribution gives p value = 0.4040.

Compare α and the p value: α = 0.05 and p value = 0.4040. Therefore, α < p value.

Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators.

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Hypothesis Test for Population Standard Deviation for normal population

hypothesis testing of standard deviation

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8.3: Hypothesis Testing of Single Mean

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Learning Objectives

  • To learn how to apply the five-step test procedure for test of hypotheses concerning a population mean when the sample size is small.

In the previous section hypotheses testing for population means was described in the case of large samples. The statistical validity of the tests was insured by the Central Limit Theorem, with essentially no assumptions on the distribution of the population. When sample sizes are small, as is often the case in practice, the Central Limit Theorem does not apply. One must then impose stricter assumptions on the population to give statistical validity to the test procedure. One common assumption is that the population from which the sample is taken has a normal probability distribution to begin with. Under such circumstances, if the population standard deviation is known, then the test statistic

\[\frac{(\bar{x}-\mu _0)}{\sigma /\sqrt{n}} \nonumber \]

still has the standard normal distribution, as in the previous two sections. If \(\sigma\) is unknown and is approximated by the sample standard deviation \(s\), then the resulting test statistic

\[\dfrac{(\bar{x}-\mu _0)}{s/\sqrt{n}} \nonumber \]

follows Student’s \(t\)-distribution with \(n-1\) degrees of freedom.

Standardized Test Statistics for Small Sample Hypothesis Tests Concerning a Single Population Mean

 If \(\sigma\) is known: \[Z=\frac{\bar{x}-\mu _0}{\sigma /\sqrt{n}} \nonumber \]

If \(\sigma\) is unknown: \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}} \nonumber \]

  • The first test statistic (\(\sigma\) known) has the standard normal distribution.
  • The second test statistic (\(\sigma\) unknown) has Student’s \(t\)-distribution with \(n-1\) degrees of freedom.
  • The population must be normally distributed.

The distribution of the second standardized test statistic (the one containing \(s\)) and the corresponding rejection region for each form of the alternative hypothesis (left-tailed, right-tailed, or two-tailed), is shown in Figure \(\PageIndex{1}\). This is just like Figure 8.2.1 except that now the critical values are from the \(t\)-distribution. Figure 8.2.1 still applies to the first standardized test statistic (the one containing (\(\sigma\)) since it follows the standard normal distribution.

ecf5f771ca148089665859c88d8679df.jpg

The \(p\)-value of a test of hypotheses for which the test statistic has Student’s \(t\)-distribution can be computed using statistical software, but it is impractical to do so using tables, since that would require \(30\) tables analogous to Figure 7.1.5, one for each degree of freedom from \(1\) to \(30\). Figure 7.1.6 can be used to approximate the \(p\)-value of such a test, and this is typically adequate for making a decision using the \(p\)-value approach to hypothesis testing, although not always. For this reason the tests in the two examples in this section will be made following the critical value approach to hypothesis testing summarized at the end of Section 8.1, but after each one we will show how the \(p\)-value approach could have been used.

Example \(\PageIndex{1}\)

The price of a popular tennis racket at a national chain store is \(\$179\). Portia bought five of the same racket at an online auction site for the following prices:

\[155\; 179\; 175\; 175\; 161 \nonumber \]

Assuming that the auction prices of rackets are normally distributed, determine whether there is sufficient evidence in the sample, at the \(5\%\) level of significance, to conclude that the average price of the racket is less than \(\$179\) if purchased at an online auction.

  • Step 1 . The assertion for which evidence must be provided is that the average online price \(\mu\) is less than the average price in retail stores, so the hypothesis test is \[H_0: \mu =179\\ \text{vs}\\ H_a: \mu <179\; @\; \alpha =0.05 \nonumber \]
  • Step 2 . The sample is small and the population standard deviation is unknown. Thus the test statistic is \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}} \nonumber \] and has the Student \(t\)-distribution with \(n-1=5-1=4\) degrees of freedom.
  • Step 3 . From the data we compute \(\bar{x}=169\) and \(s=10.39\). Inserting these values into the formula for the test statistic gives \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}}=\frac{169-179}{10.39/\sqrt{5}}=-2.152 \nonumber \]
  • Step 4 . Since the symbol in \(H_a\) is “\(<\)” this is a left-tailed test, so there is a single critical value, \(-t_\alpha =-t_{0.05}[df=4]\). Reading from the row labeled \(df=4\) in Figure 7.1.6 its value is \(-2.132\). The rejection region is \((-\infty ,-2.132]\).
  • Step 5 . As shown in Figure \(\PageIndex{2}\) the test statistic falls in the rejection region. The decision is to reject \(H_0\). In the context of the problem our conclusion is:

The data provide sufficient evidence, at the \(5\%\) level of significance, to conclude that the average price of such rackets purchased at online auctions is less than \(\$179\).

Rejection Region and Test Statistic

To perform the test in Example \(\PageIndex{1}\) using the \(p\)-value approach, look in the row in Figure 7.1.6 with the heading \(df=4\) and search for the two \(t\)-values that bracket the unsigned value \(2.152\) of the test statistic. They are \(2.132\) and \(2.776\), in the columns with headings \(t_{0.050}\) and \(t_{0.025}\). They cut off right tails of area \(0.050\) and \(0.025\), so because \(2.152\) is between them it must cut off a tail of area between \(0.050\) and \(0.025\). By symmetry \(-2.152\) cuts off a left tail of area between \(0.050\) and \(0.025\), hence the \(p\)-value corresponding to \(t=-2.152\) is between \(0.025\) and \(0.05\). Although its precise value is unknown, it must be less than \(\alpha =0.05\), so the decision is to reject \(H_0\).

Example \(\PageIndex{2}\)

A small component in an electronic device has two small holes where another tiny part is fitted. In the manufacturing process the average distance between the two holes must be tightly controlled at \(0.02\) mm, else many units would be defective and wasted. Many times throughout the day quality control engineers take a small sample of the components from the production line, measure the distance between the two holes, and make adjustments if needed. Suppose at one time four units are taken and the distances are measured as

Determine, at the \(1\%\) level of significance, if there is sufficient evidence in the sample to conclude that an adjustment is needed. Assume the distances of interest are normally distributed.

  • Step 1 . The assumption is that the process is under control unless there is strong evidence to the contrary. Since a deviation of the average distance to either side is undesirable, the relevant test is \[H_0: \mu =0.02\\ \text{vs}\\ H_a: \mu \neq 0.02\; @\; \alpha =0.01 \nonumber \] where \(\mu\) denotes the mean distance between the holes.
  • Step 2 . The sample is small and the population standard deviation is unknown. Thus the test statistic is \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}} \nonumber \] and has the Student \(t\)-distribution with \(n-1=4-1=3\) degrees of freedom.
  • Step 3 . From the data we compute \(\bar{x}=0.02075\) and \(s=0.00171\). Inserting these values into the formula for the test statistic gives \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}}=\frac{0.02075-0.02}{0.00171\sqrt{4}}=0.877 \nonumber \]
  • Step 4 . Since the symbol in \(H_a\) is “\(\neq\)” this is a two-tailed test, so there are two critical values, \(\pm t_{\alpha/2} =-t_{0.005}[df=3]\). Reading from the row in Figure 7.1.6 labeled \(df=3\) their values are \(\pm 5.841\). The rejection region is \((-\infty ,-5.841]\cup [5.841,\infty )\).
  • Step 5 . As shown in Figure \(\PageIndex{3}\) the test statistic does not fall in the rejection region. The decision is not to reject \(H_0\). In the context of the problem our conclusion is:

The data do not provide sufficient evidence, at the \(1\%\) level of significance, to conclude that the mean distance between the holes in the component differs from \(0.02\) mm.

Rejection Region and Test Statistic

To perform the test in "Example \(\PageIndex{2}\)" using the \(p\)-value approach, look in the row in Figure 7.1.6 with the heading \(df=3\) and search for the two \(t\)-values that bracket the value \(0.877\) of the test statistic. Actually \(0.877\) is smaller than the smallest number in the row, which is \(0.978\), in the column with heading \(t_{0.200}\). The value \(0.978\) cuts off a right tail of area \(0.200\), so because \(0.877\) is to its left it must cut off a tail of area greater than \(0.200\). Thus the \(p\)-value, which is the double of the area cut off (since the test is two-tailed), is greater than \(0.400\). Although its precise value is unknown, it must be greater than \(\alpha =0.01\), so the decision is not to reject \(H_0\).

Key Takeaway

  • There are two formulas for the test statistic in testing hypotheses about a population mean with small samples. One test statistic follows the standard normal distribution, the other Student’s \(t\)-distribution.
  • The population standard deviation is used if it is known, otherwise the sample standard deviation is used.
  • Either five-step procedure, critical value or \(p\)-value approach, is used with either test statistic.

Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

  • z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
  • t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
  • \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

  • One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
  • Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

  • One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
  • Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Right Tail Hypothesis Testing

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Left Tail Hypothesis Testing

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Two Tail Hypothesis Testing

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

  • Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
  • Step 2: Set up the alternative hypothesis.
  • Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
  • Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
  • Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

  • Probability and Statistics
  • Data Handling

Important Notes on Hypothesis Testing

  • Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
  • It involves the setting up of a null hypothesis and an alternate hypothesis.
  • There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
  • Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

  • Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
  • Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
  • Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

IMAGES

  1. Hypothesis Test, Two Variances (Standard Deviations)

    hypothesis testing of standard deviation

  2. Chapter 7 Hypothesis Testing with One Sample Larson Farber

    hypothesis testing of standard deviation

  3. Hypothesis Testing for Variance and Standard Deviation using a P-Value

    hypothesis testing of standard deviation

  4. Hypothesis Testing for Variance and Standard Deviation, Normal Distribution

    hypothesis testing of standard deviation

  5. 7.5 Hypothesis Tests for Variance and Standard Deviation

    hypothesis testing of standard deviation

  6. Hypothesis Testing- One-sample Z test with Known Standard Deviation

    hypothesis testing of standard deviation

VIDEO

  1. MATH 1342

  2. Finding Critical Values for Hypothesis Testing

  3. One Sample Hypothesis (Known Standard Deviation)

  4. Hypothesis Testing for Standard Deviation Lecture Part 2

  5. Hypothesis Testing, Difference of Means, Sigma Unknown, Critical Region(Traditional) Method

  6. Session 10: Part 2

COMMENTS

  1. 8.4: Hypothesis Test on a Single Standard Deviation

    A test of a single standard deviation assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population standard deviation (or population variance). The test statistic is: χ2 = (n − 1)s2 σ2 (8.4.1) (8.4.1) χ 2 = ( n − 1) s 2 σ 2. where:

  2. Section 10.4: Hypothesis Tests for a Population Standard Deviation

    test hypotheses about a population standard deviation For a quick overview of this section, watch this short video summary: Hypothesis Testing - One Sample Variance Watch on Before we begin this section, we need a quick refresher of the Χ2 distribution. The Chi-Square ( Χ2) distribution

  3. 8.3: Hypothesis Test Examples for Means with Unknown Standard Deviation

    Exercise 8.3.6 8.3. 6. It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won't grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2.

  4. Hypothesis Testing

    Step 1: State your null and alternate hypothesis Step 2: Collect data Step 3: Perform a statistical test Step 4: Decide whether to reject or fail to reject your null hypothesis Step 5: Present your findings Other interesting articles Frequently asked questions about hypothesis testing Step 1: State your null and alternate hypothesis

  5. Hypothesis Testing

    Need help with a homework problem? Check out our tutoring page! What is a Hypothesis? Andreas Cellarius hypothesis, showing the planetary motions. A hypothesis is an educated guess about something in the world around you. It should be testable, either by experiment or observation. For example: A new medicine you think might work.

  6. 3.2: Hypothesis Test about the Population Mean when the Population

    Hypothesis Test about the Population Mean (μ) when the Population Standard Deviation (σ) is Known. We are going to examine two equivalent ways to perform a hypothesis test: the classical approach and the p-value approach. The classical approach is based on standard deviations. This method compares the test statistic (Z-score) to a critical ...

  7. 8.6 Hypothesis Tests for a Population Mean with Known Population

    Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed. Collect the sample information for the test and identify the significance level. When the population standard deviation is known, find the p-value (the area in the corresponding tail) for the test using the normal distribution.

  8. Hypothesis Testing for the Standard Deviation

    Why is important to perform a hypothesis test about a standard deviation? What are the two conditions that must be met when performing a hypothesis test about the standard deviation? What test can be used to determine if two samples have similar variances? What causes the null hypothesis to be rejected in an F-test?

  9. Hypothesis testing and p-values (video)

    This is the mean. If I did 1 standard deviation, 2 standard deviations, 3 standard deviations-- that's in the positive direction. Actually let me draw it a little bit different than that. This wasn't a nicely drawn bell curve, but I'll do 1 standard deviation, 2 standard deviation, and then 3 standard deviations in the positive direction.

  10. Statistical hypothesis testing

    A statistical hypothesis test is a method of statistical inference used to decide whether the data at hand sufficiently support a particular hypothesis. More generally, hypothesis testing allows us to make probabilistic statements about population parameters. Thus, it is one way of making decisions under uncertainty.

  11. Hypothesis Testing For Standard Deviation

    Hypothesis testing for deviation also referred to as variance hypothesis testing, aids in establishing whether a populations standard deviation aligns with a given hypothesis or if there exists a difference.

  12. Hypothesis Tests for One or Two Variances or Standard Deviations

    Bigelow: n2 = 120 n 2 = 120, s2 = 0.0428 s 2 = 0.0428. Assuming that the diameters of the bearings from both companies are normally distributed, test the claim that there is no difference in the variation of the diameters between the two companies. The hypotheses are: H0: σ1 = σ2 H 0: σ 1 = σ 2. Ha: σ1 ≠ σ2 H a: σ 1 ≠ σ 2.

  13. 8.7 Hypothesis Tests for a Population Mean with Unknown Population

    Conduct and interpret hypothesis tests for a population mean with unknown population standard deviation. Some notes about conducting a hypothesis test: The null hypothesis H 0 H 0 is always an "equal to." The null hypothesis is the original claim about the population parameter.

  14. 9.4: Two Variance or Standard Deviation F-Test

    Based on a sample of 28 university students, the sample standard deviation 10, and for a sample of 25 community college students, the sample standard deviation 12. Test the claim using the traditional method of hypothesis testing with a level of significance \(\alpha\) = 0.05. Assume that IQ scores are normally distributed.

  15. Z Test: Uses, Formula & Examples

    Population standard deviation is known. As I mention in the Z test vs T test section, use a Z test when you know the population standard deviation. However, when n > 30, the difference between the analyses becomes trivial. Related post: Standard Deviations. Z Test Formula. These Z test formulas allow you to calculate the test statistic.

  16. 11.3: Two Population Means with Known Standard Deviations

    This is a test of two independent groups, two population means, population standard deviations known. Random Variable: X¯1 −X¯2 = X ¯ 1 − X ¯ 2 = difference in the mean number of months the competing floor waxes last. H0: μ1 ≤ μ2 H 0: μ 1 ≤ μ 2. Ha: μ1 > μ2 H a: μ 1 > μ 2. The words "is more effective" says that wax 1 lasts ...

  17. 8.6: Hypothesis Test of a Single Population Mean with Examples

    Full Hypothesis Test Examples. Example 8.6.4 8.6. 4. Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71.

  18. Lesson 21: Hypothesis Testing with Known Standard Deviation

    Determine if the data supports a hypothesis at a given significance level using known distributions. Topic. This lesson covers: Hypothesis Testing with Known Standard Deviation. Openstax Introductory Statistics: 9.1 Null and Alternative Hypotheses. 9.2 Outcomes and the Type I and Type II Errors. Introductory Statistics by Sheldon Ross, 3rd ...

  19. Hypothesis Test for Variance

    Conduct a hypothesis test on one variance and interpret the conclusion in context Recall: STANDARD DEVIATION AND VARIANCE The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.

  20. 10.2 Two Population Means with Known Standard Deviations

    This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student's-t distribution.Subscripts: 1: Democratic senators; 2: Republican senators

  21. Hypothesis Test for Population Standard Deviation for normal population

    The alternative hypothesis is that the true population standard deviation is not equal to 3.25 . We want to test the null hypothesis, H0: σ = 3.25 , against the alternative hypothesis, H1: σ ≠ 3.25 , at the 0.0333 level of significance . Note that this is a two-tailed test .

  22. 8.3: Hypothesis Testing of Single Mean

    Thus the test statistic is. T = x¯ −μ0 s/ n−−√ T = x ¯ − μ 0 s / n. and has the Student t t -distribution with n − 1 = 5 − 1 = 4 n − 1 = 5 − 1 = 4 degrees of freedom. Step 3. From the data we compute x¯ = 169 x ¯ = 169 and s = 10.39 s = 10.39. Inserting these values into the formula for the test statistic gives.

  23. Hypothesis Testing

    Get Started Learn Practice Download Hypothesis Testing Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy.

  24. PDF Math 145

    Z-Test for a Population Mean (˙is known): To test the hypothesis H 0: = 0 based on a SRS of size nfrom a population with unknown mean and known standard deviation ˙, compute the test statistic Z= X 0 ˙= p n ˘N(0;1): (1) Steps to do Hypothesis Testing: 1. Formulate the Null hypothesis (H 0) and Alternative hypothesis (H a). 2.