hypothesis population mean example

Module 10: Inference for Means

Hypothesis test for a population mean (5 of 5), learning outcomes.

• Interpret the P-value as a conditional probability.

We finish our discussion of the hypothesis test for a population mean with a review of the meaning of the P-value, along with a review of type I and type II errors.

Review of the Meaning of the P-value

At this point, we assume you know how to use a P-value to make a decision in a hypothesis test. The logic is always the same. If we pick a level of significance (α), then we compare the P-value to α.

• If the P-value ≤ α, reject the null hypothesis. The data supports the alternative hypothesis.
• If the P-value > α, do not reject the null hypothesis. The data is not strong enough to support the alternative hypothesis.

In fact, we find that we treat these as “rules” and apply them without thinking about what the P-value means. So let’s pause here and review the meaning of the P-value, since it is the connection between probability and decision-making in inference.

Birth Weights in a Town

Let’s return to the familiar context of birth weights for babies in a town. Suppose that babies in the town had a mean birth weight of 3,500 grams in 2010. This year, a random sample of 50 babies has a mean weight of about 3,400 grams with a standard deviation of about 500 grams. Here is the distribution of birth weights in the sample.

Obviously, this sample weighs less on average than the population of babies in the town in 2010. A decrease in the town’s mean birth weight could indicate a decline in overall health of the town. But does this sample give strong evidence that the town’s mean birth weight is less than 3,500 grams this year?

We now know how to answer this question with a hypothesis test. Let’s use a significance level of 5%.

Let μ = mean birth weight in the town this year. The null hypothesis says there is “no change from 2010.”

• H 0 : μ = 3,500
• H a : μ < 3,500

Since the sample is large, we can conduct the T-test (without worrying about the shape of the distribution of birth weights for individual babies.)

[latex]T\text{}=\text{}\frac{\mathrm{3,400}-\mathrm{3,500}}{\frac{500}{\sqrt{50}}}\text{}\approx \text{}-1.41[/latex]

Statistical software tells us the P-value is 0.082 = 8.2%. Since the P-value is greater than 0.05, we fail to reject the null hypothesis.

Our conclusion: This sample does not suggest that the mean birth weight this year is less than 3,500 grams ( P -value = 0.082). The sample from this year has a mean of 3,400 grams, which is 100 grams lower than the mean in 2010. But this difference is not statistically significant. It can be explained by the chance fluctuation we expect to see in random sampling.

What Does the P-Value of 0.082 Tell Us?

A simulation can help us understand the P-value. In a simulation, we assume that the population mean is 3,500 grams. This is the null hypothesis. We assume the null hypothesis is true and select 1,000 random samples from a population with a mean of 3,500 grams. The mean of the sampling distribution is at 3,500 (as predicted by the null hypothesis.) We see this in the simulated sampling distribution.

In the simulation, we can see that about 8.6% of the samples have a mean less than 3,400. Since probability is the relative frequency of an event in the long run, we say there is an 8.6% chance that a random sample of 500 babies has a mean less than 3,400 if the population mean is 3,500. We can see that the corresponding area to the left of T = −1.41 in the T-model (with df = 49) also gives us a good estimate of the probability. This area is the P-value, about 8.2%.

If we generalize this statement, we say the P-value is the probability that random samples have results more extreme than the data if the null hypothesis is true. (By more extreme, we mean further from value of the parameter, in the direction of the alternative hypothesis.) We can also describe the P-value in terms of T-scores. The P-value is the probability that the test statistic from a random sample has a value more extreme than that associated with the data if the null hypothesis is true.

What Does a P-Value Mean?

Do women who smoke run the risk of shorter pregnancy and premature birth? The mean pregnancy length is 266 days. We test the following hypotheses.

• H 0 : μ = 266
• H a : μ < 266

Suppose a random sample of 40 women who smoke during their pregnancy have a mean pregnancy length of 260 days with a standard deviation of 21 days. The P-value is 0.04.

What probability does the P-value of 0.04 describe? Label each of the following interpretations as valid or invalid.

Review of Type I and Type II Errors

We know that statistical inference is based on probability, so there is always some chance of making a wrong decision. Recall that there are two types of wrong decisions that can be made in hypothesis testing. When we reject a null hypothesis that is true, we commit a type I error. When we fail to reject a null hypothesis that is false, we commit a type II error.

The following table summarizes the logic behind type I and type II errors.

It is possible to have some influence over the likelihoods of committing these errors. But decreasing the chance of a type I error increases the chance of a type II error. We have to decide which error is more serious for a given situation. Sometimes a type I error is more serious. Other times a type II error is more serious. Sometimes neither is serious.

Recall that if the null hypothesis is true, the probability of committing a type I error is α. Why is this? Well, when we choose a level of significance (α), we are choosing a benchmark for rejecting the null hypothesis. If the null hypothesis is true, then the probability that we will reject a true null hypothesis is α. So the smaller α is, the smaller the probability of a type I error.

It is more complicated to calculate the probability of a type II error. The best way to reduce the probability of a type II error is to increase the sample size. But once the sample size is set, larger values of α will decrease the probability of a type II error (while increasing the probability of a type I error).

General Guidelines for Choosing a Level of Significance

• If the consequences of a type I error are more serious, choose a small level of significance (α).
• If the consequences of a type II error are more serious, choose a larger level of significance (α). But remember that the level of significance is the probability of committing a type I error.
• In general, we pick the largest level of significance that we can tolerate as the chance of a type I error.

Let’s return to the investigation of the impact of smoking on pregnancy length.

Recap of the hypothesis test: The mean human pregnancy length is 266 days. We test the following hypotheses.

Let’s Summarize

In this “Hypothesis Test for a Population Mean,” we looked at the four steps of a hypothesis test as they relate to a claim about a population mean.

Step 1: Determine the hypotheses.

• The hypotheses are claims about the population mean, µ.
• The null hypothesis is a hypothesis that the mean equals a specific value, µ 0 .
• When [latex]{H}_{a}[/latex] is [latex]μ[/latex] < [latex]{μ}_{0}[/latex] or [latex]μ[/latex] > [latex]{μ}_{0}[/latex] , the test is a one-tailed test.
• When [latex]{H}_{a}[/latex] is [latex]μ[/latex] ≠ [latex]{μ}_{0}[/latex] , the test is a two-tailed test.

Step 2: Collect the data.

Since the hypothesis test is based on probability, random selection or assignment is essential in data production. Additionally, we need to check whether the t-model is a good fit for the sampling distribution of sample means. To use the t-model, the variable must be normally distributed in the population or the sample size must be more than 30. In practice, it is often impossible to verify that the variable is normally distributed in the population. If this is the case and the sample size is not more than 30, researchers often use the t-model if the sample is not strongly skewed and does not have outliers.

Step 3: Assess the evidence.

• If a t-model is appropriate, determine the t-test statistic for the data’s sample mean.

[latex]\frac{\mathrm{sample}\text{}\mathrm{mean}-\mathrm{population}\text{}\mathrm{mean}}{\mathrm{estimated}\text{}\mathrm{standard}\text{}\mathrm{error}}\text{}=\text{}\frac{\stackrel{¯}{x}-μ}{s/\sqrt{n}}[/latex]

• Use the test statistic, together with the alternative hypothesis, to determine the P-value.
• The P-value is the probability of finding a random sample with a mean at least as extreme as our sample mean, assuming that the null hypothesis is true.
• As in all hypothesis tests, if the alternative hypothesis is greater than, the P-value is the area to the right of the test statistic. If the alternative hypothesis is less than, the P-value is the area to the left of the test statistic. If the alternative hypothesis is not equal to, the P-value is equal to double the tail area beyond the test statistic.

Step 4: Give the conclusion.

The logic of the hypothesis test is always the same. To state a conclusion about H 0 , we compare the P-value to the significance level, α.

• If P ≤ α, we reject H 0 . We conclude there is significant evidence in favor of H a .
• If P > α, we fail to reject H 0 . We conclude the sample does not provide significant evidence in favor of H a .
• We write the conclusion in the context of the research question. Our conclusion is usually a statement about the alternative hypothesis (we accept H a or fail to acceptH a ) and should include the P-value.

Other Hypothesis Testing Notes

• Remember that the P-value is the probability of seeing a sample mean at least as extreme as the one from the data if the null hypothesis is true. The probability is about the random sample; it is not a “chance” statement about the null or alternative hypothesis.
• If our test results in rejecting a null hypothesis that is actually true, then it is called a type I error.
• If our test results in failing to reject a null hypothesis that is actually false, then it is called a type II error.
• If rejecting a null hypothesis would be very expensive, controversial, or dangerous, then we really want to avoid a type I error. In this case, we would set a strict significance level (a small value of α, such as 0.01).
• Finally, remember the phrase “garbage in, garbage out.” If the data collection methods are poor, then the results of a hypothesis test are meaningless.

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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.

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.

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.

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.

Module 10: Inference for Means

Hypothesis Test for a Population Mean (5 of 5)

Learning outcomes.

• Interpret the P-value as a conditional probability.

We finish our discussion of the hypothesis test for a population mean with a review of the meaning of the P-value, along with a review of type I and type II errors.

Review of the Meaning of the P-value

At this point, we assume you know how to use a P-value to make a decision in a hypothesis test. The logic is always the same. If we pick a level of significance (α), then we compare the P-value to α.

• If the P-value ≤ α, reject the null hypothesis. The data supports the alternative hypothesis.
• If the P-value > α, do not reject the null hypothesis. The data is not strong enough to support the alternative hypothesis.

In fact, we find that we treat these as “rules” and apply them without thinking about what the P-value means. So let’s pause here and review the meaning of the P-value, since it is the connection between probability and decision-making in inference.

Birth Weights in a Town

Let’s return to the familiar context of birth weights for babies in a town. Suppose that babies in the town had a mean birth weight of 3,500 grams in 2010. This year, a random sample of 50 babies has a mean weight of about 3,400 grams with a standard deviation of about 500 grams. Here is the distribution of birth weights in the sample.

Obviously, this sample weighs less on average than the population of babies in the town in 2010. A decrease in the town’s mean birth weight could indicate a decline in overall health of the town. But does this sample give strong evidence that the town’s mean birth weight is less than 3,500 grams this year?

We now know how to answer this question with a hypothesis test. Let’s use a significance level of 5%.

Let μ = mean birth weight in the town this year. The null hypothesis says there is “no change from 2010.”

• H 0 : μ < 3,500
• H a : μ = 3,500

Since the sample is large, we can conduct the T-test (without worrying about the shape of the distribution of birth weights for individual babies.)

T = [latex]\frac{3,400-3,500}{\frac{500}{\sqrt{50}}}[/latex]  ≈  – 1.41

Statistical software tells us the P-value is 0.082 = 8.2%. Since the P-value is greater than 0.05, we fail to reject the null hypothesis.

Our conclusion: This sample does not suggest that the mean birth weight this year is less than 3,500 grams ( P -value = 0.082). The sample from this year has a mean of 3,400 grams, which is 100 grams lower than the mean in 2010. But this difference is not statistically significant. It can be explained by the chance fluctuation we expect to see in random sampling.

What Does the P-Value of 0.082 Tell Us?

A simulation can help us understand the P-value. In a simulation, we assume that the population mean is 3,500 grams. This is the null hypothesis. We assume the null hypothesis is true and select 1,000 random samples from a population with a mean of 3,500 grams. The mean of the sampling distribution is at 3,500 (as predicted by the null hypothesis.) We see this in the simulated sampling distribution.

In the simulation, we can see that about 8.6% of the samples have a mean less than 3,400. Since probability is the relative frequency of an event in the long run, we say there is an 8.6% chance that a random sample of 500 babies has a mean less than 3,400 if the population mean is 3,500. We can see that the corresponding area to the left of T = −1.41 in the T-model (with df = 49) also gives us a good estimate of the probability. This area is the P-value, about 8.2%.

If we generalize this statement, we say the P-value is the probability that random samples have results more extreme than the data if the null hypothesis is true. (By more extreme, we mean further from value of the parameter, in the direction of the alternative hypothesis.) We can also describe the P-value in terms of T-scores. The P-value is the probability that the test statistic from a random sample has a value more extreme than that associated with the data if the null hypothesis is true.

What Does a P-Value Mean?

Do women who smoke run the risk of shorter pregnancy and premature birth? The mean pregnancy length is 266 days. We test the following hypotheses.

• H 0 : μ = 266
• H a : μ < 266

Suppose a random sample of 40 women who smoke during their pregnancy have a mean pregnancy length of 260 days with a standard deviation of 21 days. The P-value is 0.04.

What probability does the P-value of 0.04 describe? Label each of the following interpretations as valid or invalid.

Review of Type I and Type II Errors

We know that statistical inference is based on probability, so there is always some chance of making a wrong decision. Recall that there are two types of wrong decisions that can be made in hypothesis testing. When we reject a null hypothesis that is true, we commit a type I error. When we fail to reject a null hypothesis that is false, we commit a type II error.

The following table summarizes the logic behind type I and type II errors.

It is possible to have some influence over the likelihoods of committing these errors. But decreasing the chance of a type I error increases the chance of a type II error. We have to decide which error is more serious for a given situation. Sometimes a type I error is more serious. Other times a type II error is more serious. Sometimes neither is serious.

Recall that if the null hypothesis is true, the probability of committing a type I error is α. Why is this? Well, when we choose a level of significance (α), we are choosing a benchmark for rejecting the null hypothesis. If the null hypothesis is true, then the probability that we will reject a true null hypothesis is α. So the smaller α is, the smaller the probability of a type I error.

It is more complicated to calculate the probability of a type II error. The best way to reduce the probability of a type II error is to increase the sample size. But once the sample size is set, larger values of α will decrease the probability of a type II error (while increasing the probability of a type I error).

General Guidelines for Choosing a Level of Significance

• If the consequences of a type I error are more serious, choose a small level of significance (α).
• If the consequences of a type II error are more serious, choose a larger level of significance (α). But remember that the level of significance is the probability of committing a type I error.
• In general, we pick the largest level of significance that we can tolerate as the chance of a type I error.

Let’s return to the investigation of the impact of smoking on pregnancy length.

Recap of the hypothesis test: The mean human pregnancy length is 266 days. We test the following hypotheses.

Let’s Summarize

In this “Hypothesis Test for a Population Mean,” we looked at the four steps of a hypothesis test as they relate to a claim about a population mean.

Step 1: Determine the hypotheses.

• The hypotheses are claims about the population mean, µ.
• The null hypothesis is a hypothesis that the mean equals a specific value, µ 0 .
• When H a is μ < μ 0 or μ > μ 0 , the test is a one-tailed test.
• When H a is μ ≠ μ 0 , the test is a two-tailed test.

Step 2: Collect the data.

Since the hypothesis test is based on probability, random selection or assignment is essential in data production. Additionally, we need to check whether the t-model is a good fit for the sampling distribution of sample means. To use the t-model, the variable must be normally distributed in the population or the sample size must be more than 30. In practice, it is often impossible to verify that the variable is normally distributed in the population. If this is the case and the sample size is not more than 30, researchers often use the t-model if the sample is not strongly skewed and does not have outliers.

Step 3: Assess the evidence.

• If a t-model is appropriate, determine the t-test statistic for the data’s sample mean.

[latex]\frac{sample mean-populationmean}{estimated standarderror}=\frac{\bar{x}-\mu}{s/\sqrt{n}}[/latex]

• Use the test statistic, together with the alternative hypothesis, to determine the P-value.
• The P-value is the probability of finding a random sample with a mean at least as extreme as our sample mean, assuming that the null hypothesis is true.
• As in all hypothesis tests, if the alternative hypothesis is greater than, the P-value is the area to the right of the test statistic. If the alternative hypothesis is less than, the P-value is the area to the left of the test statistic. If the alternative hypothesis is not equal to, the P-value is equal to double the tail area beyond the test statistic.

Step 4: Give the conclusion.

The logic of the hypothesis test is always the same. To state a conclusion about H 0 , we compare the P-value to the significance level, α.

• If P ≤ α, we reject H 0 . We conclude there is significant evidence in favor of H a .
• If P > α, we fail to reject H 0 . We conclude the sample does not provide significant evidence in favor of H a .
• We write the conclusion in the context of the research question. Our conclusion is usually a statement about the alternative hypothesis (we accept H a or fail to accept H a ) and should include the P-value.

Other Hypothesis Testing Notes

• Remember that the P-value is the probability of seeing a sample mean at least as extreme as the one from the data if the null hypothesis is true. The probability is about the random sample; it is not a “chance” statement about the null or alternative hypothesis.
• If our test results in rejecting a null hypothesis that is actually true, then it is called a type I error.
• If our test results in failing to reject a null hypothesis that is actually false, then it is called a type II error.
• If rejecting a null hypothesis would be very expensive, controversial, or dangerous, then we really want to avoid a type I error. In this case, we would set a strict significance level (a small value of α, such as 0.01).
• Finally, remember the phrase “garbage in, garbage out.” If the data collection methods are poor, then the results of a hypothesis test are meaningless.
• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

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