hypothesis testing when population mean is unknown

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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.

Hypothesis Testing Calculator

Related: confidence interval calculator, type ii error.

The first step in hypothesis testing is to calculate the test statistic. The formula for the test statistic depends on whether the population standard deviation (σ) is known or unknown. If σ is known, our hypothesis test is known as a z test and we use the z distribution. If σ is unknown, our hypothesis test is known as a t test and we use the t distribution. Use of the t distribution relies on the degrees of freedom, which is equal to the sample size minus one. Furthermore, if the population standard deviation σ is unknown, the sample standard deviation s is used instead. To switch from σ known to σ unknown, click on $\boxed{\sigma}$ and select $\boxed{s}$ in the Hypothesis Testing Calculator.

Next, the test statistic is used to conduct the test using either the p-value approach or critical value approach. The particular steps taken in each approach largely depend on the form of the hypothesis test: lower tail, upper tail or two-tailed. The form can easily be identified by looking at the alternative hypothesis (H a ). If there is a less than sign in the alternative hypothesis then it is a lower tail test, greater than sign is an upper tail test and inequality is a two-tailed test. To switch from a lower tail test to an upper tail or two-tailed test, click on $\boxed{\geq}$ and select $\boxed{\leq}$ or $\boxed{=}$, respectively.

In the p-value approach, the test statistic is used to calculate a p-value. If the test is a lower tail test, the p-value is the probability of getting a value for the test statistic at least as small as the value from the sample. If the test is an upper tail test, the p-value is the probability of getting a value for the test statistic at least as large as the value from the sample. In a two-tailed test, the p-value is the probability of getting a value for the test statistic at least as unlikely as the value from the sample.

To test the hypothesis in the p-value approach, compare the p-value to the level of significance. If the p-value is less than or equal to the level of signifance, reject the null hypothesis. If the p-value is greater than the level of significance, do not reject the null hypothesis. This method remains unchanged regardless of whether it's a lower tail, upper tail or two-tailed test. To change the level of significance, click on $\boxed{.05}$. Note that if the test statistic is given, you can calculate the p-value from the test statistic by clicking on the switch symbol twice.

In the critical value approach, the level of significance ($\alpha$) is used to calculate the critical value. In a lower tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the lower tail of the sampling distribution of the test statistic. In an upper tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the upper tail of the sampling distribution of the test statistic. In a two-tailed test, the critical values are the values of the test statistic providing areas of $\alpha / 2$ in the lower and upper tail of the sampling distribution of the test statistic.

To test the hypothesis in the critical value approach, compare the critical value to the test statistic. Unlike the p-value approach, the method we use to decide whether to reject the null hypothesis depends on the form of the hypothesis test. In a lower tail test, if the test statistic is less than or equal to the critical value, reject the null hypothesis. In an upper tail test, if the test statistic is greater than or equal to the critical value, reject the null hypothesis. In a two-tailed test, if the test statistic is less than or equal the lower critical value or greater than or equal to the upper critical value, reject the null hypothesis.

When conducting a hypothesis test, there is always a chance that you come to the wrong conclusion. There are two types of errors you can make: Type I Error and Type II Error. A Type I Error is committed if you reject the null hypothesis when the null hypothesis is true. Ideally, we'd like to accept the null hypothesis when the null hypothesis is true. A Type II Error is committed if you accept the null hypothesis when the alternative hypothesis is true. Ideally, we'd like to reject the null hypothesis when the alternative hypothesis is true.

Hypothesis testing is closely related to the statistical area of confidence intervals. If the hypothesized value of the population mean is outside of the confidence interval, we can reject the null hypothesis. Confidence intervals can be found using the Confidence Interval Calculator . The calculator on this page does hypothesis tests for one population mean. Sometimes we're interest in hypothesis tests about two population means. These can be solved using the Two Population Calculator . The probability of a Type II Error can be calculated by clicking on the link at the bottom of the page.

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10.2 - t-test: when population variance is unknown.

Now that, for purely pedagogical reasons, we have the unrealistic situation (of a known population variance) behind us, let's turn our attention to the realistic situation in which both the population mean and population variance are unknown.

Example 10-2 Section  

It is assumed that the mean systolic blood pressure is \(\mu\) = 120 mm Hg. In the Honolulu Heart Study, a sample of \(n=100\) people had an average systolic blood pressure of 130.1 mm Hg with a standard deviation of 21.21 mm Hg. Is the group significantly different (with respect to systolic blood pressure!) from the regular population?

The null hypothesis is \(H_0:\mu=120\), and because there is no specific direction implied, the alternative hypothesis is \(H_A:\mu\ne 120\). In general, we know that if the data are normally distributed, then:

\(T=\dfrac{\bar{X}-\mu}{S/\sqrt{n}}\)

follows a \(t\)-distribution with \(n-1\) degrees of freedom. Therefore, it seems reasonable to use the test statistic:

\(T=\dfrac{\bar{X}-\mu_0}{S/\sqrt{n}}\)

for testing the null hypothesis \(H_0:\mu=\mu_0\) against any of the possible alternative hypotheses \(H_A:\mu \neq \mu_0\), \(H_A:\mu<\mu_0\), and \(H_A:\mu>\mu_0\). For the example in hand, the value of the test statistic is:

\(t=\dfrac{130.1-120}{21.21/\sqrt{100}}=4.762\)

The critical region approach tells us to reject the null hypothesis at the \(\alpha=0.05\) level if \(t\ge t_{0.025, 99}=1.9842\) or if \(t\le t_{0.025, 99}=-1.9842\). Therefore, we reject the null hypothesis because \(t=4.762>1.9842\), and therefore falls in the rejection region:

Again, as always, we draw the same conclusion by using the \(p\)-value approach. The \(p\)-value approach tells us to reject the null hypothesis at the \(\alpha=0.05\) level if the \(p\)-value \(\le \alpha=0.05\). In this case, the \(p\)-value is \(2 \times P(T_{99}>4.762)<2\times P(T_{99}>1.9842)=2(0.025)=0.05\):

As expected, we reject the null hypothesis because \(p\)-value \(\le 0.01<\alpha=0.05\).

Again, we'll learn how to ask Minitab to conduct the t -test for a mean \(\mu\) in a bit, but this is what the Minitab output for this example looks like:

By the way, the decision to reject the null hypothesis is consistent with the one you would make using a 95% confidence interval. Using the data, a 95% confidence interval for the mean \(\mu\) is:

\(\bar{x}\pm t_{0.025,99}\left(\dfrac{s}{\sqrt{n}}\right)=130.1 \pm 1.9842\left(\dfrac{21.21}{\sqrt{100}}\right)\)

which simplifies to \(130.1\pm 4.21\). That is, we can be 95% confident that the mean systolic blood pressure of the Honolulu population is between 125.89 and 134.31 mm Hg. How can a population living in a climate with consistently sunny 80 degree days have elevated blood pressure?!

Anyway, the critical region approach for the \(\alpha=0.05\) hypothesis test tells us to reject the null hypothesis that \(\mu=120\):

if \(t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}\geq 1.9842\) or if \(t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}\leq -1.9842\)

which is equivalent to rejecting:

if \(\bar{x}-\mu_0 \geq 1.9842\left(\dfrac{s}{\sqrt{n}}\right)\) or if \(\bar{x}-\mu_0 \leq -1.9842\left(\dfrac{s}{\sqrt{n}}\right)\)

if \(\mu_0 \leq \bar{x}-1.9842\left(\dfrac{s}{\sqrt{n}}\right)\) or if \(\mu_0 \geq \bar{x}+1.9842\left(\dfrac{s}{\sqrt{n}}\right)\)

which, upon inserting the data for this particular example, is equivalent to rejecting:

if \(\mu_0 \leq 125.89\) or if \(\mu_0 \geq 134.31\)

which just happen to be (!) the endpoints of the 95% confidence interval for the mean. Indeed, the results are consistent!

Hypothesis test for Population Mean, based on the sample mean, σ unknown

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Hypothesis tests about a population mean.

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8.2 A Single Population Mean (Unknown σ)

To find the standard error, we need the population standard deviation, [latex]\sigma[/latex]. Unfortunately, this value isn’t generally unknown. In this situation, the next best thing we can do is to use the sample standard deviation, [latex]s[/latex], as a substitute for [latex]\sigma[/latex]. This substitution works well for large samples but this estimate is off for small sample sizes. In the case of small sample sizes taken from an underlying normal distribution, a different kind of distribution, called a t -distribution gives better results.

Choosing appropriate distribution: What do I use: z or t distribution? Do we know population standard deviation, σ?

• If YES, use normal distribution ( z -distribution)
• If NO, then use t -distribution. Note : You may use normal distribution if sample size is at least 30 ( n ≥ 30) even if σ is unknown. For n ≥ 30, you can use sample standard deviation ( s ) in place of population standard deviation ([latex]\sigma[/latex]). Note that since the t and z -distributions look similar ( see Desmos demo ) for larger sample sizes, probability calculations from these distributions will lead to similar results. For this reason, when dealing with large sample sizes from populations with unknown standard deviations (unknown [latex]\sigma[/latex]), many simply prefer to use t -distribution instead of z .

Finding Critical Values

Use Desmos or StatKey. (Note: If you need to use more than 3 decimal places for the critical value, you may want to skip StatKey)

Critical t -value using Desmos |  Critical t -value using StatKey (no audio)

Please complete the following practice exercise: Finding the critical value t for a desired confidence level

EXAMPLE: MARGIN OF ERROR

A survey of 46 people showed that the respondents spent an average of $31 on their child’s last birthday gift with a standard deviation of $9. Find the critical value for a 95% confidence level, the standard error, and the margin of error. Assume that the population is normally distributed.

The mean and the standard deviations given here are about  a sample , as it says in the question —  a sample  of size 46 with a mean of $31 and a standard deviation of $9 .

Given facts are:

[latex]n=46[/latex]

[latex]\bar x = $31[/latex]

[latex]s = $9[/latex].   This is not σ  (The notation σ represents the population standard deviation. What does s represent?)

Since the population standard deviation is unknown, we use t -distribution. If the sample size is at least 30, the result from using normal distribution is approximately equal to the one from that of using a t -distribution. Therefore, using normal distribution ( z -distribution) wouldn’t be too far off, but t -distribution is there, so why not use it.

Critical value: Use a t -distribution to find the critical value. What would be the degrees of freedom for the t -distribution here?

Standard Error: Standard Error is given by [latex]\sigma_{\bar x} = \dfrac{s}{\sqrt n}[/latex]

Now that we have the pieces sorted out, let’s use the EBM (or Margin or Error, ME) formula to find the margin of error.

Margin of Error = (Critical Value) • (Standard Error)

EXAMPLE: MEAN WITH STATISTICS

The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 40.9 for a sample of size 20 and standard deviation 11.7.

Estimate how much the drug will lower a typical patient’s systolic blood pressure (using a 98% confidence level).

Assume the data is from a normally distributed population .  Round your answers to 3 decimals .

Confidence level, c = 0.98  (for a 98% confidence interval)

Sample info (These include sample statistics):

Sample Mean, [latex]\bar x = 40.9[/latex] Sample Standard Deviation, [latex]s = 11.7[/latex] Sample Size, [latex]n = 20[/latex]

Population standard deviation is not known. Sample size is fewer than 30 and the population is normally distributed. Therefore, use a t-distribution. Note that confidence interval is:

Point Estimate ± Margin of Error and Margin of Error = Critical Value  • Standard Error , where t c is the critical value and the standard error of the mean = [latex]\dfrac{s}{\sqrt n}[/latex].

Margin of Error = [latex]t_c \cdot \dfrac{s}{\sqrt n}[/latex]

So, confidence interval is:

Point Estimate ± Critical Value • Standard Error Point Estimate [latex]\pm \:{\color{#e03e2d} {t_c}} \cdot \dfrac{s}{\sqrt n}[/latex]. First, recognize that the sample mean is our point estimate. x̅ = 40.9 . And our sample standard deviation is given as well: s = 11.7 , while the sample size n = 20 .

Let’s update our CI:

Point Estimate ± Margin of Error  = Point Estimate [latex]\pm \:{\color{#e03e2d}{ t_c}} \cdot \dfrac{s}{\sqrt {n}}[/latex] = [latex]40.9 \pm \:{\color{#e03e2d} {t_c}} \cdot \dfrac{11.7}{\sqrt {20}}[/latex]

Now, we just need the critical value. With [latex]n = 20[/latex], degrees of freedom, [latex]d.f. = n - 1 = .........[/latex]

Use Desmos or StatKey or TiCalculator to find the critical value, [latex]t_c[/latex], for a 98% confidence level. After you have computed the value, click on Show More below to show critical value and more:

[latex]t_c= 2.53948319062[/latex]

Let’s plug this [latex]t_c[/latex] into the formula above to find the confidence interval: [latex]40.9 \pm 2.53948319062[/latex] • [latex]\dfrac{11.7}{\sqrt {20}}[/latex] = [latex]40.9 \pm 6.64379473907[/latex]

This is our confidence interval in ± notation.

Three Ways to Write Confidence Intervals

1) So, the confidence interval in interval notation: (34.25620530117019, 47.54379469882981)   →  Round to 3 decimals:  (34.256, 47.544)

2) Confidence interval in tri-inequality notation: 34.25620530117019 < [latex]\mu[/latex] < 47.54379469882981 34.256 < [latex]\mu[/latex] < 47.544

3) Confidence interval in plus-minus notation: Margin of Error, ME or EBM =  47.54379469882981 − 40.9 = 6.64379469883 ≈ 6.644 Confidence interval: 40.9 ± 6.644

The SUBEDI calculator gives answers in ± notation, whereas the LibreText calculator ‘s results are in interval notation. Be sure to convert CI in one notation to another. (See Three Ways to Write a Confidence Interval for additional details on notations).

ONLINE CALCULATOR Approach

Go to  Confidence Interval for a Mean calculator  @ rsubedi.com

Confidence Level (in decimal),  [latex]c: \fbox{$\mathstrut \;0.98\;$}[/latex]

Number of Samples

Distribution Type to Use?

Sample Size, [latex]n: \fbox{$\mathstrut \;20\;$}[/latex]

Sample Mean, [latex]\bar x: \fbox{$\mathstrut \;40.9\;$}[/latex]

Sample Standard Deviation, [latex]n: \fbox{$\mathstrut \;11.7\;$}[/latex]

CALCULATE Results show in a panel to the right. CI is displayed in [latex]\pm[/latex] notation.

Go to:  Confidence Interval for a Mean With Statistics  from the list of  online calculators.

Enter the following values and press  Calculate .

Results displayed are:

EXAMPLE: MEAN WITH DATA

You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 11 randomly selected physical therapy patients.

Confidence level, c = 0.90  (for a 90% confidence interval)

Population standard deviation is not known. Sample size is fewer than 30 and the population is normally distributed. Therefore, use a t -distribution.

Confidence Level (in decimal), [latex]c: \fbox{$\mathstrut \;0.90\;$}[/latex]

Enter your data in the spreadsheet column shown for data entry.

Select the above data to copy. Once copied, on SUBEDI Calc click on the first cell of the spreadsheet and paste the data (Control+V or Command + V).

Go to:  Confidence Interval Calculator with Data  from the list of  online calculators Enter the following values and press  Calculate .

Data:    (Separate each value with a comma)

[latex]\fbox{$\mathstrut \quad14,    12,    6,    27,    12,    13,     21,    20,    20,     13,    19\quad$}[/latex]

[latex]\text{CL}: \fbox{$\mathstrut \;0.90\;$}[/latex]

CALCULATE Results displayed are:

Statistics Study Guide Copyright © by Ram Subedi is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.

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Module 10: Inference for Means

Hypothesis test for a population mean (1 of 5), learning outcomes.

• Recognize when to use a hypothesis test or a confidence interval to draw a conclusion about a population mean.
• Under appropriate conditions, conduct a hypothesis test about a population mean. State a conclusion in context.

Introduction

In Inference for Means , our focus is on inference when the variable is quantitative, so the parameters and statistics are means. In “Estimating a Population Mean,” we learned how to use a sample mean to calculate a confidence interval. The confidence interval estimates a population mean. In “Hypothesis Test for a Population Mean,” we learn to use a sample mean to test a hypothesis about a population mean.

We did hypothesis tests in earlier modules. In Inference for One Proportion , each claim involved a single population proportion. In Inference for Two Proportions , the claim was a statement about a treatment effect or a difference in population proportions. In “Hypothesis Test for a Population Mean,” the claims are statements about a population mean. But we will see that the steps and the logic of the hypothesis test are the same. Before we get into the details, let’s practice identifying research questions and studies that involve a population mean.

Cell Phone Data

Cell phones and cell phone plans can be very expensive, so consumers must think carefully when choosing a cell phone and service. This decision is as much about choosing the right cellular company as it is about choosing the right phone. Many people use the data/Internet capabilities of a phone as much as, if not more than, they use voice capability. The data service of a cell company is therefore an important factor in this decision. In the following example, a student named Melanie from Los Angeles applies what she learned in her statistics class to help her make a decision about buying a data plan for her smartphone.

Melanie read an advertisement from the Cell Phone Giants (CPG, for short, and yes, we’re using a fictitious company name) that she thinks is too good to be true. The CPG ad states that customers in Los Angeles get average data download speeds of 4 Mbps. With this speed, the ad claims, it takes, on average, only 12 seconds to download a typical 3-minute song from iTunes.

Only 12 seconds on average to download a 3-minute song from iTunes! Melanie has her doubts about this claim, so she gathers data to test it. She asks a friend who uses the CPG plan to download a song, and it takes 13 seconds to download a 3-minute song using the CPG network. Melanie decides to gather more evidence. She uses her friend’s phone and times the download of the same 3-minute song from various locations in Los Angeles. She gets a mean download time of 13.5 seconds for her sample of downloads.

What can Melanie conclude? Her sample has a mean download time that is greater than 12 seconds. Isn’t this evidence that the CPG claim is wrong? Why is a hypothesis test necessary? Isn’t the conclusion clear?

Let’s review the reason Melanie needs to do a hypothesis test before she can reach a conclusion.

Why should Melanie do a hypothesis test?

Melanie’s data (with a mean of 13.5 seconds) suggest that the average download time overall is greater than the 12 seconds claimed by the manufacturer. But wait. We know that samples will vary. If the CPG claim is correct, we don’t expect all samples to have a mean download time exactly equal to 12 seconds. There will be variability in the sample means. But if the overall average download time is 12 seconds, how much variability in sample means do we expect to see? We need to determine if the difference Melanie observed can be explained by chance.

We have to judge Melanie’s data against random samples that come from a population with a mean of 12. For this reason, we must do a simulation or use a mathematical model to examine the sampling distribution of sample means. Based on the sampling distribution, we ask, Is it likely that the samples will have mean download times that are greater than 13.5 seconds if the overall mean is 12 seconds? This probability (the P-value) determines whether Melanie’s data provides convincing evidence against the CPG claim.

Now let’s do the hypothesis test.

Step 1: Determine the hypotheses.

As always, hypotheses come from the research question. The null hypothesis is a hypothesis that the population mean equals a specific value. The alternative hypothesis reflects our claim. The alternative hypothesis says the population mean is “greater than” or “less than” or “not equal to” the value we assume is true in the null hypothesis.

Melanie’s hypotheses:

• H 0 : It takes 12 seconds on average to download Melanie’s song from iTunes with the CPG network in Los Angeles.
• H a : It takes more than 12 seconds on average to download Melanie’s song from iTunes using the CPG network in Los Angeles.

We can write the hypotheses in terms of µ. When we do so, we should always define µ. Here μ = the average number of seconds it takes to download Melanie’s song on the CPG network in Los Angeles.

• H 0 : μ = 12
• H a : μ > 12

Step 2: Collect the data.

To conduct a hypothesis test, Melanie knows she has to use a t-model of the sampling distribution. She thinks ahead to the conditions required, which helps her collect a useful sample.

Recall the conditions for use of a t-model.

• There is no reason to think the download times are normally distributed (they might be, but this isn’t something Melanie could know for sure). So the sample has to be large (more than 30).
• The sample has to be random. Melanie decides to use one phone but randomly selects days, times, and locations in Los Angeles.

Melanie collects a random sample of 45 downloads by using her friend’s phone to download her song from iTunes according to the randomly selected days, times, and locations.

Melanie’s sample of size 45 downloads has an average download time of 13.5 seconds. The standard deviation for the sample is 3.2 seconds. Now Melanie needs to determine how unlikely this data is if CPG’s claim is actually true.

Step 3: Assess the evidence.

Assuming the average download time for Melanie’s song is really 12 seconds, what is the probability that 45 random downloads of this song will have a mean of 13.5 seconds or more?

This is a question about sampling variability. Melanie must determine the standard error. She knows the standard error of random sample means is [latex]\sigma \text{}/\sqrt{n}[/latex]. Since she has no way of knowing the population standard deviation, σ, Melanie uses the sample standard deviation, s = 3.2, as an approximation. Therefore, Melanie approximates the standard error of all sample means ( n = 45) to be

[latex]s\text{}/\sqrt{n}\text{}=\text{}3.2\text{}/\sqrt{45}\text{}=\text{}0.48[/latex]

Now she can assess how far away her sample is from the claimed mean in terms of standard errors. That is, she can compute the t-score of her sample mean.

[latex]T\text{}=\text{}\frac{\mathrm{statistic}-\mathrm{parameter}}{\mathrm{standard}\text{}\mathrm{error}}\text{}=\text{}\frac{\stackrel{¯}{x}-μ}{s\text{}/\sqrt{n}}\text{}=\text{}\frac{13.5-12}{0.48}\text{}=\text{}3.14[/latex]

The sample mean for Melanie’s random sample is approximately 3.14 standard errors above the overall mean of 12. We know from previous experience that a sample mean this far above µ is very unlikely. With a t-score this large, the P-value is very small. We use a simulation of the t-model for 44 degrees of freedom to verify this.

We want the probability that the sample mean is greater than 13.5. This corresponds to the probability that T is greater than 3.14. The P-value is 0.0015.

Step 4: State a conclusion.

Here the logic is the same as for other hypothesis tests. We use the P-value to make a decision. The P-value helps us determine if the difference we see between the data and the hypothesized value of µ is statistically significant or due to chance. One of two outcomes can occur:

• One possibility is that results similar to the actual sample are extremely unlikely. This means the data does not fit with results from random samples selected from the population described by the null hypothesis. In this case, it is unlikely that the data came from this population. The probability as measured by the P-value is small, so we view this as strong evidence against the null hypothesis. We reject the null hypothesis in favor of the alternative hypothesis.
• The other possibility is that results similar to the actual sample are fairly likely (not unusual). This means the data fits with typical results from random samples selected from the population described by the null hypothesis. The probability as measured by the P-value is large. In this case, we do not have evidence against the null hypothesis, so we cannot reject it in favor of the alternative hypothesis.

Melanie’s data is very unlikely if µ = 12. The probability is essentially zero (P-value = 0.0015). This means we will rarely see sample means greater than 13.5 if µ = 12. So we reject the null and accept the alternative hypothesis. In other words, this sample provides strong evidence that CPG has overstated the speed of its data download capability.

The following activities give you an opportunity to practice parts of the hypothesis testing process for a population mean. Later you will have the opportunity to practice the hypothesis test from start to finish.

For the following scenarios, give the null and alternative hypotheses and state in words what µ represents in your hypotheses. A good definition of µ describes both the variable and the population.

In the previous example, Melanie did not state a significance level for her test. If she had, the logic is the same as we used for hypothesis tests in Modules 8 and 9. To come to 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 .

Use this simulation when needed to answer questions below.

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• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

8.5 Hypothesis Tests for One Population Mean μ

Recall that there are two different procedures used to construct confidence intervals for one population mean [latex]\mu[/latex]: the one-sample Z -interval (used when the population standard deviation [latex]\sigma[/latex] is known) and the one-sample t-interval (used when [latex]\sigma[/latex] is unknown). In a similar vein, there are two different procedures for hypothesis tests for one population mean: the one-sample Z -test is used when [latex]\sigma[/latex] is known and the one-sample t -test is used when [latex]\sigma[/latex] is unknown.

8.5.1 One-Sample Z- Test When σ is Known

Assumptions :

• A simple random sample (SRS)
• Normal population or large sample size ([latex]n \geq 30[/latex])
• The population standard deviation [latex]\sigma[/latex] is known
• State the significance level [latex]\alpha[/latex].
• Compute the value of the test statistic: [latex]z_o = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}[/latex].
• Reject the null [latex]H_0[/latex] if P-value [latex]\leq \alpha[/latex] or [latex]z_o[/latex] falls in the rejection region.
• Conclusion.

Example: One-Sample Z Test

One-Sample Z Test

A machine fills beer into bottles whose volume is supposed to be 341 ml, but the exact amount varies from bottle to bottle. We randomly picked 100 bottles and obtained the sample mean volume of 339 ml. Assume the population standard deviation [latex]\sigma = 5[/latex] ml. Test at the 5% significance level whether the machine is NOT working properly.

Check the assumptions :

• We have a simple random sample (SRS).
• We do not know whether the population is normal or not, but the sample size is large with [latex]n = 100 \geq 30[/latex].
• [latex]\sigma = 5[/latex] ml is known.
• Set up the hypotheses: [latex]H_0: \mu = 341[/latex] ml versus [latex]H_a: \mu \neq 341[/latex] ml. This is a two-tailed test. If the machine works properly, the population mean volume [latex]\mu = 341[/latex] ml.
• The significance level is [latex]\alpha = 0.05[/latex].
• Compute the value of the test statistic:

[latex]z_o = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{339 - 341}{5 / \sqrt{100}} = \frac{-2}{0.5} = -4.[/latex]

• Decision: Since the P-value [latex]\approx 0 \leq 0.05(\alpha)[/latex], reject the null hypothesis [latex]H_0[/latex].
• Conclusion: At the 5% significance level, the data provide sufficient evidence that the machine is NOT working properly.

If using the critical value approach, steps 1-3 are the same, steps 4-6 become:

• Decision: Since the observed value [latex]z_o= -4 <-1.96[/latex] falls in the rejection region, we reject the null hypothesis [latex]H_0[/latex].

P-value approach is preferred for the following reasons:

• It is more professional. P-value is required to be reported for all hypothesis tests in academia.
• The P-value approach provides more information: it not only tells whether we should reject the null or not but also shows how strong the evidence is. However, the critical value approach only tells us whether we should reject the null or not.
• The computer output only provides the P-value; no critical value is provided.

8.5.2 One-Sample t-Test  When σ is Unknown

• The population standard deviation [latex]\sigma[/latex] is unknown
• Compute the value of the test statistic: [latex]t_o = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}[/latex] with a degree of freedom [latex]df = n-1[/latex].
• Reject the null [latex]H_0[/latex] if P-value [latex]\leq \alpha[/latex] or [latex]t_o[/latex] falls in the rejection region.

Example: One-Sample t Test

A computer company claims that the average lifetime of its laptop is about 4 years. A simple random sample of 36 laptops yields an average lifetime of 3.5 years with a sample standard deviation of 4.2 years. Test at the 1% significance level whether the mean lifetime of this brand of laptops is less than 4 years.

• We do not know whether the population is normal or not, but the sample size is large with [latex]n = 36 \geq 30[/latex].
• [latex]\sigma[/latex] is unknown and estimated by [latex]s = 4.2[/latex].
• Set up the hypotheses: [latex]H_0: \mu \geq 4[/latex] years versus [latex]H_a: \mu < 4[/latex] years.
• The significance level is [latex]\alpha = 0.01[/latex].
• Compute the value of the test statistic: [latex]t_o = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{3.5 - 4}{4.2 / \sqrt{36}} = \frac{-0.5}{0.7} = -0.714[/latex] with [latex]df = n -1 = 36 -1 = 35[/latex].

• Decision: Since the P-value [latex]\: \gt \: 0.2 \: \gt \: 0.01(\alpha)[/latex], we can not reject the null [latex]H_0[/latex].
• Conclusion: At the 1% significance level, we do not have sufficient evidence that the mean lifetime of this brand of laptops is less than 4 years.

If we use the critical value approach, steps 1-3 are the same, and steps 4-6 become:

• Decision: Since the observed value [latex]t_o = -0.714 \: \gt \: - 2.438[/latex] falls in the non-rejection region, we can not reject the null hypothesis [latex]H_0[/latex].
• Conclusion: At the 1% significance level, the data do not provide sufficient evidence that the mean lifetime of this brand of laptops is less than 4 years.

Exercise: P-value for One sample t-Test

Use the same setting of the previous example (one-sample t-test  with df = 35) to find the P-values of the following hypothesis tests.

• [latex]H_0: \mu = 4 \text{ years versus } H_a: \mu \neq 4 \text{ years}[/latex], with the observed test statistic [latex]t_o = 1.5[/latex].
• [latex]H_0: \mu \geq 4 \text{ years versus } H_a: \mu < 4 \text{ years}[/latex], with the observed test statistic [latex]t_o=-2.5[/latex].
• [latex]H_0: \mu \leq 4 \text{ years versus } H_a: \mu \: \gt \: 4 \text{ years}[/latex], with the observed test statistic [latex]t_o = 3.5[/latex].
• For a two-tailed test, the P-value is twice the area to the right of the absolute value of the observed test statistic [latex]t_o[/latex]. Note that the probability is the area under the density curve of the t-distribution with 35 degrees of freedom.[latex]P[/latex]-value=[latex]2P(t\ge |t_o|)=2P(t\ge 1.5)[/latex]. Since [latex]1.306 (t_{0.1})<1.5<1.690 (t_{0.05})[/latex], we have [latex]0.05 \lt P(t\ge 1.5) \lt 0.1 \Longrightarrow 2\times 0.05 \lt 2 P(t\ge 1.5) \lt 2\times 0.1 \Longrightarrow 0.1 \lt \mbox{P-value} \lt 0.2.[/latex] If use R commander, [latex]2 P(t\ge 1.5)=2\times 0.07129092=0.1425818[/latex].
• For a left-tailed test, the P-value is the area to the left of the observed test statistic [latex]t_o[/latex]. [latex]P[/latex]-value=[latex]P(t \le t_o )=P(t \le -2.5)=P(t \ge 2.5)[/latex]. Since[latex]2.438(t_{0.01})<2.5<2.558(t_{0.0075}) \Longrightarrow 0.0075<\mbox{P-value}< 0.01.[/latex] If use R commander, [latex]P(t \ge 2.5)=0.008627872[/latex].
• For a right-tailed test, the P-value is the area to the right of the observed test statistic [latex]t_o[/latex].[latex]P[/latex]-value=[latex]P(t\ge t_o )=P(t\ge 3.5)[/latex]. Since[latex](t_{0.0025})2.996

Exercise: One-sample t-Test

The number of cell phone users has increased dramatically since 1997. Suppose the mean local monthly bill was $50 for cell phone users in the United States in 2006. A simple random sample of 50 cell phone users was obtained in 2019, and the sample mean local monthly bill was [latex]\bar{x} = 55[/latex] with a sample standard deviation [latex]s = $25[/latex].

• At the 5% significance level, do the data provide sufficient evidence to conclude that the mean local monthly bill for cell phone users in 2019 has changed from the 2006 mean of $50?
• Obtain a 95% confidence interval for the 2019 mean local monthly bill for all cell phone users. Interpret the confidence interval.
• Are the results in parts (a) and (b) consistent with each other? Explain why.
• We do not know whether the population is normal or not since we do not have the data, but the sample size is large with [latex]n = 50 \geq 30[/latex].
• [latex]\sigma[/latex] is unknown and estimated by [latex]s = $25[/latex].
• Set up the hypotheses: [latex]H_0: \mu = 50[/latex] versus [latex]H_a: \mu \neq 50[/latex].
• Compute the value of the test statistic: [latex]t_o = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{55 - 50}{25 / \sqrt{50}} = 1.414[/latex] with [latex]df = n-1 = 50 -1 = 49[/latex].
• Find the P-value. For a two-tailed test, the P-value is twice the area to the right of the observed test statistic [latex]t_o[/latex]. P-value=[latex]2P(t \geq t_o) = 2P(t \geq 1.414)[/latex]. Since [latex]1.299(t_{0.1}) < 1.414 < 1.677(t_{0.05})[/latex], [latex]2\times 0.05 < \text{P-value} <2\times 0.1 \Longrightarrow 0.1<\text{P-value}<0.2.[/latex]
• Decision: Since the P-value [latex]\: \gt \: 0.1>0.05 (\alpha)[/latex], we can not reject the null [latex]H_0[/latex].
• Conclusion: At the 5% significance level, we do not have sufficient evidence that the 2019 mean local monthly bill for cell phone users has changed from the 2006 mean of $50.
• Find [latex]t_{\alpha / 2}[/latex]: [latex]n = 50, df = n-1 = 50-1 =49[/latex]. [latex]1 - \alpha = 0.95 \Longrightarrow \alpha = 0.05 \Longrightarrow \alpha / 2 = 0.025 \Longrightarrow t_{\alpha / 2} = t_{0.025} = 2.010[/latex].
• Interval: [latex]\bar{x} \pm t_{\alpha / 2}\frac{s}{\sqrt{n}} = 55 \pm 2.010 \times \frac{25}{\sqrt{50}} = (47.894, 62.106)[/latex].
• Yes, they are consistent. We cannot reject [latex]H_0: \mu = 50[/latex] and hence can not claim [latex]\mu \neq 50[/latex] in the hypothesis test in part (a). The interval in part (b) contains 50; there is no sufficient evidence that the population mean differs from 50. We cannot reject [latex]H_0: \mu=50[/latex] and claim [latex]\mu \neq  50[/latex]. Therefore, they are consistent.

Introduction to Applied Statistics Copyright © 2024 by Wanhua Su is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Statistics Made Easy

Introduction to Hypothesis Testing

A statistical hypothesis is an assumption about a population parameter .

For example, we may assume that the mean height of a male in the U.S. is 70 inches.

The assumption about the height is the statistical hypothesis and the true mean height of a male in the U.S. is the population parameter .

A hypothesis test is a formal statistical test we use to reject or fail to reject a statistical hypothesis.

The Two Types of Statistical Hypotheses

To test whether a statistical hypothesis about a population parameter is true, we obtain a random sample from the population and perform a hypothesis test on the sample data.

There are two types of statistical hypotheses:

The null hypothesis , denoted as H 0 , is the hypothesis that the sample data occurs purely from chance.

The alternative hypothesis , denoted as H 1 or H a , is the hypothesis that the sample data is influenced by some non-random cause.

Hypothesis Tests

A hypothesis test consists of five steps:

1. State the hypotheses. 

State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false.

2. Determine a significance level to use for the hypothesis.

Decide on a significance level. Common choices are .01, .05, and .1. 

3. Find the test statistic.

Find the test statistic and the corresponding p-value. Often we are analyzing a population mean or proportion and the general formula to find the test statistic is: (sample statistic – population parameter) / (standard deviation of statistic)

4. Reject or fail to reject the null hypothesis.

Using the test statistic or the p-value, determine if you can reject or fail to reject the null hypothesis based on the significance level.

The p-value  tells us the strength of evidence in support of a null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis.

5. Interpret the results. 

Interpret the results of the hypothesis test in the context of the question being asked. 

The Two Types of Decision Errors

There are two types of decision errors that one can make when doing a hypothesis test:

Type I error: You reject the null hypothesis when it is actually true. The probability of committing a Type I error is equal to the significance level, often called  alpha , and denoted as α.

Type II error: You fail to reject the null hypothesis when it is actually false. The probability of committing a Type II error is called the Power of the test or  Beta , denoted as β.

One-Tailed and Two-Tailed Tests

A statistical hypothesis can be one-tailed or two-tailed.

A one-tailed hypothesis involves making a “greater than” or “less than ” statement.

For example, suppose we assume the mean height of a male in the U.S. is greater than or equal to 70 inches. The null hypothesis would be H0: µ ≥ 70 inches and the alternative hypothesis would be Ha: µ < 70 inches.

A two-tailed hypothesis involves making an “equal to” or “not equal to” statement.

For example, suppose we assume the mean height of a male in the U.S. is equal to 70 inches. The null hypothesis would be H0: µ = 70 inches and the alternative hypothesis would be Ha: µ ≠ 70 inches.

Note: The “equal” sign is always included in the null hypothesis, whether it is =, ≥, or ≤.

Related:   What is a Directional Hypothesis?

Types of Hypothesis Tests

There are many different types of hypothesis tests you can perform depending on the type of data you’re working with and the goal of your analysis.

The following tutorials provide an explanation of the most common types of hypothesis tests:

Introduction to the One Sample t-test Introduction to the Two Sample t-test Introduction to the Paired Samples t-test Introduction to the One Proportion Z-Test Introduction to the Two Proportion Z-Test

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

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10.2: Two Population Means with Unknown Standard Deviations

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• The two independent samples are simple random samples from two distinct populations.
• if the sample sizes are small, the distributions are important (should be normal)
• if the sample sizes are large, the distributions are not important (need not be normal)

The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch \(t\)-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, \(\bar{X}_{1} - \bar{X}_{2}\), and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error , of the difference in sample means , \(\bar{X}_{1} - \bar{X}_{2}\).

The standard error is:

\[\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}\]

The test statistic ( t -score) is calculated as follows:

\[\dfrac{(\bar{x}-\bar{x}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}\]

• \(s_{1}\) and \(s_{2}\), the sample standard deviations, are estimates of \(\sigma_{1}\) and \(\sigma_{1}\), respectively.
• \(\sigma_{1}\) and \(\sigma_{2}\) are the unknown population standard deviations.
• \(\bar{x}_{1}\) and \(\bar{x}_{2}\) are the sample means. \(\mu_{1}\) and \(\mu_{2}\) are the population means.

The number of degrees of freedom (\(df\)) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The \(df\) are not always a whole number. The test statistic calculated previously is approximated by the Student's t -distribution with \(df\) as follows:

Degrees of freedom

\[df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1}-1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2} + \left(\dfrac{1}{n_{2}-1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}\]

When both sample sizes \(n_{1}\) and \(n_{2}\) are five or larger, the Student's t approximation is very good. Notice that the sample variances \((s_{1})^{2}\) and \((s_{2})^{2}\) are not pooled. (If the question comes up, do not pool the variances.)

It is not necessary to compute the degrees of freedom by hand. A calculator or computer easily computes it.

Example \(\PageIndex{1}\): Independent groups

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table \(\PageIndex{1}\). Each populations has a normal distribution.

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, \(\mu_{g}\) is the population mean for girls and \(\mu_{b}\) is the population mean for boys. This is a test of two independent groups, two population means.

Random variable: \(\bar{X}_{g} - \bar{X}_{b} =\) difference in the sample mean amount of time girls and boys play sports each day.

• \(H_{0}: \mu_{g} = \mu_{b}\)  
• \(H_{0}: \mu_{g} - \mu_{b} = 0\)
• \(H_{a}: \mu_{g} \neq \mu_{b}\)  
• \(H_{a}: \mu_{g} - \mu_{b} \neq 0\)

The words "the same" tell you \(H_{0}\) has an "=". Since there are no other words to indicate \(H_{a}\), assume it says "is different." This is a two-tailed test.

Distribution for the test: Use \(t_{df}\) where \(df\) is calculated using the \(df\) formula for independent groups, two population means. Using a calculator, \(df\) is approximately 18.8462. Do not pool the variances.

Calculate the p -value using a Student's t -distribution: \(p\text{-value} = 0.0054\)

\[s_{g} = 0.866\]

\[s_{b} = 1\]

\[\bar{x}_{g} - \bar{x}_{b} = 2 - 3.2 = -1.2\]

Half the \(p\text{-value}\) is below –1.2 and half is above 1.2.

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\). This means you reject \(\mu_{g} = \mu_{b}\). The means are different.

Press STAT . Arrow over to TESTS and press 4:2-SampTTest . Arrow over to Stats and press ENTER . Arrow down and enter 2 for the first sample mean, \(\sqrt{0.866}\) for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER . Arrow down to Pooled: and No . Press ENTER . Arrow down to Calculate and press ENTER . The \(p\text{-value}\) is \(p = 0.0054\), the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).

Exercise \(\PageIndex{1}\)

Two samples are shown in Table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

The \(p\text{-value}\) is \(0.4125\), which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.

When the sum of the sample sizes is larger than \(30 (n_{1} + n_{2} > 30)\) you can use the normal distribution to approximate the Student's \(t\).

Example \(\PageIndex{2}\)

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.

• Is this a test of two means or two proportions?
• Are the populations standard deviations known or unknown?
• Which distribution do you use to perform the test?
• What is the random variable?
• What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
• Is this test right-, left-, or two-tailed?
• What is the \(p\text{-value}\)?
• Do you reject or not reject the null hypothesis?
• Student's t
• \(\bar{X}_{A} - \bar{X}_{B}\)
• \(H_{0}: \mu_{A} \leq \mu_{B}\) and \(H_{a}: \mu_{A} > \mu_{B}\)

• h. Do not reject.
• i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

Exercise \(\PageIndex{2}\)

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

• Are the population standard deviations known?
• Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?
• They are unknown.
• The \(p\text{-value} = 0.0878\). At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.

Example \(\PageIndex{3}\)

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table \(\PageIndex{3}\) and Table \(\PageIndex{4}\).

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

• Are the population standard deviations known or unknown?
• What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
• Is this test right, left, or two tailed?
• At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(See the conclusion in Example, and write yours in a similar fashion)

Be careful not to mix up the information for Group 1 and Group 2!

• Student's \(t\)
• \(\bar{X}_{1} - \bar{X}_{2}\)
• \(H_{0}: \mu_{1} = \mu_{2}\) Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
• \(H_{a}: \mu_{1} < \mu_{2}\) Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
• left-tailed

Figure \(\PageIndex{3}\).

• Reject the null hypothesis

At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.

First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to \(\mu_{1}\): and arrow to \(\neq \mu_{1}\) (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.

Cohen's Standards for Small, Medium, and Large Effect Sizes

Cohen's \(d\) is a measure of effect size based on the differences between two means. Cohen’s \(d\), named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.

Cohen's \(d\) is the measure of the difference between two means divided by the pooled standard deviation: \(d = \dfrac{\bar{x}_{2}-\bar{x}_{2}}{s_{\text{pooled}}}\) where \(s_{pooled} = \sqrt{\dfrac{(n_{1}-1)s^{2}_{1} + (n_{2}-1)s^{2}_{2}}{n_{1}+n_{2}-2}}\)

Example \(\PageIndex{4}\)

Calculate Cohen’s d for Example. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

\(\mu_{1} = 4 s_{1} = 1.5 n_{1} = 11\)

\(\mu_{2} = 3.5 s_{2} = 1 n_{2} = 9\)

\(d = 0.384\)

The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.

Example \(\PageIndex{5}\)

Calculate Cohen’s \(d\) for Example. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.

\(d = 0.834\); Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.

Example 10.2.6

Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in Table and Table, respectively.

Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:

• Calculate Cohen’s d and interpret it.
• Student’s-t
• \(H_{0}: \mu_{1} = \mu_{2}\) Null hypothesis: the means of the weighted alphas are equal.
• \(H_{a}: \mu_{1} \neq \mu_{2}\) Alternative hypothesis : the means of the weighted alphas are not equal.
• \(p\text{-value} = 0.8787\)
• Do not reject the null hypothesis

Figure \(\PageIndex{4}\).

• \(d = 0.040\), Very small, because 0.040 is less than Cohen’s value of 0.2 for small effect size. The size of the difference of the means of the weighted alphas for the two regions of banks is small indicating that there is not a significant difference between their trends in stocks.
• Data from Graduating Engineer + Computer Careers. Available online at www.graduatingengineer.com
• Data from Microsoft Bookshelf .
• Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013).
• “List of current United States Senators by Age.” Wikipedia. Available online at en.Wikipedia.org/wiki/List_of...enators_by_age (accessed June 17, 2013).
• “Sectoring by Industry Groups.” Nasdaq. Available online at www.nasdaq.com/markets/barcha...&base=industry (accessed June 17, 2013).
• “Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013).
• “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013).

Two population means from independent samples where the population standard deviations are not known

• Random Variable: \(\bar{X}_{1} - \bar{X}_{2} =\) the difference of the sampling means
• Distribution: Student's t -distribution with degrees of freedom (variances not pooled)

Formula Review

Standard error: \[SE = \sqrt{\dfrac{(s_{1}^{2})}{n_{1}} + \dfrac{(s_{2}^{2})}{n_{2}}}\]

Test statistic ( t -score): \[t = \dfrac{(\bar{x}_{1}-\bar{x}_{2}) - (\mu_{1}-\mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}\]

Degrees of freedom:

\[df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1} - 1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2}} + \left(\dfrac{1}{n_{2} - 1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}\]

• \(s_{1}\) and \(s_{2}\) are the sample standard deviations, and n 1 and n 2 are the sample sizes.
• \(x_{1}\) and \(x_{2}\) are the sample means.

Cohen’s \(d\) is the measure of effect size:

\[d = \dfrac{\bar{x}_{1} - \bar{x}_{2}}{s_{\text{pooled}}}\]

\[s_{\text{pooled}} = \sqrt{\dfrac{(n_{1} - 1)s^{2}_{1} + (n_{2} - 1)s^{2}_{2}}{n_{1} + n_{2} - 2}}\]

• The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if \(X =\) hair color, then the domain is {black, blond, gray, green, orange}.
• We can tell what specific value x of the random variable \(X\) takes only after performing the experiment.