8.2.3  Hypothesis Testing
8.2.3  Hypothesis TestingIn this section we will be comparing one sample mean to one known or hypothesized population value. In Lesson 5 you learned how to conduct randomization tests. Here, you will learn how to conduct a one sample mean \(t\) test and a one sample mean \(z\) test. The \(t\) distribution is used to estimate the sampling distribution when the sample size is large (at least 30) or when the population is known to be normally distributed (but \(\sigma\) is unknown). The \(z\) distribution is used on rare occasions when the population is normal and the population standard deviation is known. Note that for this course the one sample mean \(z\) test is optional; it used only in specific cases where the population is known to be normally distributed and when the population standard deviation (\(\sigma\)) is known. The most commonly used one sample mean test is the "one sample mean \(t\) test" which is also known as a "single sample mean \(t\) test."
8.2.3.1  One Sample Mean t Test, Formulas
8.2.3.1  One Sample Mean t Test, Formulas
Five Step Hypothesis Testing Procedure
Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large (\(\ge\ 30\)) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are:
Research Question  Is the mean different from \( \mu_{0} \)?  Is the mean greater than \(\mu_{0}\)?  Is the mean less than \(\mu_{0}\)? 

Null Hypothesis, \(H_{0}\)  \(\mu=\mu_{0} \)  \(\mu=\mu_{0} \)  \(\mu=\mu_{0} \) 
Alternative Hypothesis, \(H_{a}\)  \(\mu\neq \mu_{0} \)  \(\mu> \mu_{0} \)  \(\mu<\mu_{0} \) 
Type of Hypothesis Test  Twotailed, nondirectional  Righttailed, directional  Lefttailed, directional 
where \( \mu_{0} \) is the hypothesized population mean.
For the test of one group mean we will be using a \(t\) test statistic:
 Test Statistic: One Group Mean

\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}\)
\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size
Note that structure of this formula is similar to the general formula for a test statistic: \(\frac{sample\;statisticnull\;value}{standard\;error}\)
When testing hypotheses about a mean or mean difference, a \(t\) distribution is used to find the \(p\)value. These \(t\) distributions are indexed by a quantity called degrees of freedom, calculated as \(df = n – 1\) for the situation involving a test of one mean or test of mean difference. The \(p\)value can be found using Minitab Express.
If \(p \leq \alpha\) reject the null hypothesis.
If \(p>\alpha\) fail to reject the null hypothesis.
Based on your decision in Step 4, write a conclusion in terms of the original research question.
The new few pages will walk you through examples before giving you the opportunity to do two on your own.
8.2.3.1.1  Video Example: Book Costs
8.2.3.1.1  Video Example: Book CostsResearch question: Does the average Penn State student spend more than \$300 each semester on textbooks?
In a sample of 226 Penn State students, the mean cost of a student’s textbooks was \$344 with a standard deviation of \$106.
8.2.3.1.2 : Example: Pulse Rate
8.2.3.1.2 : Example: Pulse RateA research study measured the pulse rates of 57 college men and found a mean pulse rate of 70.4211 beats per minute with a standard deviation of 9.9480 beats per minute. Researchers want to know if the mean pulse rate for all college men is different from the current standard of 72 beats per minute.
Pulse rates are quantitative. The sampling distribution will be approximately normally distributed because \(n \ge 30\).
This is a twotailed test because we want to know if the mean pulse rate is different from 72.
\(H_{0}:\mu=72 \)
\(H_{a}: \mu\neq 72 \)
 Test Statistic: One Group Mean

\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}\)
\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size
\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}=\frac{70.421172}{\frac{9.9480}{\sqrt{57}}}=1.198\)
Our \(t\) test statistic is 1.198
\(df=n1=571=56\)
\(p=0.117981+0.117981=0.235962\)
Given that the null hypothesis is true and \(\mu=72\), the probability of taking a random sample of \(n=57\) and finding a sample mean this or more extremely different is 0.235962. This is our pvalue.
\(p>.05\), therefore we fail to reject the null hypothesis.
There is not sufficient evidence to state that the mean pulse of college men is different from 72.
8.2.3.1.3  Example: Coffee
8.2.3.1.3  Example: CoffeeIn the population of Americans who drink coffee, the average daily consumption is 3 cups per day. A university wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 50 students how many cups of coffee they drink each day and found \(\overline{x}=3.8\) and \(s=1.5\). Do they have evidence that their students drink more than the national average?
Amount of coffee consumed is a quantitative variable. We are given that random sampling methods were employed. Because \(n \ge 30\), we can approximate the sampling distribution using a t distribution.
This is a righttailed test because we want to know if the mean in the sample is greater than the national average.
\(H_{0}:\mu= 3\)
\(H_{a}:\mu>3\)
 Test Statistic: One Group Mean
\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}\)
\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size
\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}=\frac{3.83}{\frac{1.5}{\sqrt{50}}}=3.771\)
Our \(t\) test statistic is 3.771
\(df=n1=501=49\)
Using Minitab Express, we can find that \(P(t > 3.771) =0.0002191\)
pvalue = 0.0002191
If \(\mu=3\), then the probability of taking a random sample of \(n=50\) and finding \(\overline{x} \geq 3\) is 0.0002191
\(p\leq.05\), therefore we reject the null hypothesis.
There is evidence to state the mean number of coffees consumed in the population of all students at this university is greater than 3.
8.2.3.1.4  Example: Transportation Costs
8.2.3.1.4  Example: Transportation CostsAccording to CNN, in 2011, the average American spent \$16,803 on housing. A suburban community wants to know if their residents spent less than this national average. In a survey of 30 randomly selected residents, they found that they spent an annual average of \$15,800 with a standard deviation of \$2,600.
Housing costs are quantitative. Because \(n \ge 30\), the sampling distribution can be approximated using the \(t\) distribution.
This is a lefttailed test because we want to know if residents of this community spent less than the national average.
\(H_{0}:\mu=16803\)
\(H_{a}:\mu<16803\)
 Test Statistic: One Group Mean

\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}\)
\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size
\(t=\frac{\overline{x}\mu_0}{\frac{s}{\sqrt{n}}}=\frac{1580016803}{\frac{2600}{\sqrt{30}}}=2.113\)
Our t test statistics is 2.113
\(df=n1=301=29\)
This is a lefttailed test so we want to know the probability of \(t < 2.113\)
Using Minitab Express we can find that \(p=0.0216634\)
\(p\leq .05\), therefore we reject the null hypothesis.
There is evidence to state that on average residents of this community spent less than the national average on housing in 2011.
8.2.3.2  Minitab Express: One Sample Mean t Tests
8.2.3.2  Minitab Express: One Sample Mean t TestsA hypothesis test for one group mean can be conducted in Minitab Express using raw data or summarized data.
 If you have a data file with every individual's observation, then you have raw data.
 If you do not have each individual's observation, but rather have the sample mean, sample standard deviation, and sample size, then you have summarized data.
The next two pages will show you how to use Minitab Express to conduct a one sample mean t test using either raw data or summarized data. There is also one example of using Minitab Express to conduct a one sample mean z test which is only performed if the population is known to be normally distributed and the population standard deviation (\(\sigma\)) is available.
8.2.3.2.1  Minitab Express: 1 Sample Mean t Test, Raw Data
8.2.3.2.1  Minitab Express: 1 Sample Mean t Test, Raw DataMinitabExpress – One Sample Mean t Test Using Raw Data
Research question: Is the mean GPA in the population different from 3.0?
Null hypothesis: \(\mu\) = 3.0
Alternative hypothesis: \(\mu\) ≠ 3.0
The GPAs of \(n) = 226 students are available.
A one sample mean \(t\) test should be performed because the shape of the population is unknown, however the sample size is large (\(n\) ≥ 30).
To perform a one sample mean \(t\) test in Minitab Express using raw data:
 Open Minitab data set:
 On a PC: Select STATISTICS > One Sample > t
On a Mac: Select Statistics > 1Sample Inference > t  Doubleclick on the variable GPA to insert it into the Sample box
 Check the box Perform a hypothesis test
 For the Hypothesized mean enter 3
 Click the Options tab
 Use the default Alternative hypothesis of Mean ≠ hypothesized value
 Use the default Confidence level of 95
 Click OK
This should result in the following output:
N  Mean  StDev  SE Mean  95% CI for \(\mu\) 

226  3.23106  0.51040  0.03395  (3.16416, 3.29796) 
\(\mu\): mean of GPA
Null hypothesis  H_{0}: \(\mu\) = 3 

Alternative hypothesis  H_{1}: \(\mu\) ≠ 3 
TValue  PValue 

6.81  <0.0001 
Select your operating system below to see a stepbystep guide for this example.
We could summarize these results using the five step hypothesis testing procedure:
We do not know if the population is normally distributed, however the sample size is large (\(n \ge 30\)) so we can perform a one sample mean t test.
\(H_0: \mu = 3.0\)
\(H_a: \mu \ne 3.0\)
\(t (225) = 6.81\)
\(p < 0.0001\)
\(p \le \alpha\), reject the null hypothesis
There is evidence that the mean GPA in the population is different from 3.0
8.2.3.2.2  Minitab Express: 1 Sample Mean t Test, Summarized Data
8.2.3.2.2  Minitab Express: 1 Sample Mean t Test, Summarized DataMinitabExpress – One Sample Mean t Test Using Summarized Data
Here we are testing \(H_{a}:\mu\neq72\) and are given \(n=35\), \(\bar{x}=76.8\), and \(s=11.62\).
We do not know the shape of the population, however the sample size is large (\(n \ge 30\)) therefore we can conduct a one sample mean \(t\) test.
 On a PC: Select STATISTICS > One Sample > t
On a Mac: Select Statistics > 1Sample Inference > t  Change Sample data in column to Summarized data
 The Sample size is 35
 The Sample mean is 76.8
 The Sample standard deviation is 11.62
 Check the box Perform a hypothesis test
 For the Hypothesized mean enter 72
 Click the Options tab
 Use the default Alternative hypothesis of Mean ≠ hypothesized value
 Use the default Confidence level of 95
 Click OK
This should result in the following output:
N  Mean  StDev  SE Mean  95% CI for \(\mu\) 

35  76.800  11.620  1.964  (72.808, 80.792) 
\(\mu\) : mean of Sample
Null hypothesis  H_{0}: \(\mu\) = 72 

Alternative hypothesis  H_{1}: \(\mu\) ≠ 72 
TValue  PValue 

2.44  0.0199 
Select your operating system below to see a stepbystep guide for this example.
We could summarize these results using the five step hypothesis testing procedure:
The shape of the population distribution is unknown, however with \(n \ge 30\) we can perform a one sample mean t test.
\(H_0: \mu = 72\)
\(H_a: \mu \ne 72\)
\(t (34) = 2.44\)
\(p = 0.0199\)
\(p \le \alpha\), reject the null hypothesis
There is evidence that the population mean is different from 72.
8.2.3.3  One Sample Mean z Test (Optional)
8.2.3.3  One Sample Mean z Test (Optional)A one sample mean \(z\) test is used when the population is known to be normally distributed and when the population standard deviation (\(\sigma\)) is known. This most frequently occurs in the social sciences when standardized measures are used such as IQ, SAT, ACT, or GRE scores, for which the population parameters are known.
The formula for computing a \(z\) test statistic for one sample mean is identical to that of computing a \(t\) test statistic for one sample mean, except now the population standard deviation is known and can be used in computing the standard error.
 z Test Statistic: One Group Mean
 \(z=\dfrac{\overline{x}\mu_0}{\frac{\sigma}{\sqrt{n}}}\)

\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size
The other primary difference between the one sample mean \(t\) test and the one sample mean \(z\) test is the latter uses the standard normal distribution (i.e., \(z\) distribution) in determining the \(p\)value. Below are the directions for conducting a one sample mean \(z\) test in Minitab Express.
MinitabExpress – Performing a One Sample Mean z Test
Research question: Are the IQ scores of students at one collegeprep school above the national average?
Scores on one American IQ test are normed to have a mean of 100 and standard deviation of 15. In a simple random sample of 25 students at this school the mean was 110.
To perform a one sample mean z test in Minitab Express using summarized data:
 Open Minitab Express without data
 On a PC: Select STATISTICS > One Sample > z
On a Mac: Select Statistics > 1Sample Inference > z  Change Sample data in column to Summarized data
 The Sample size is 25
 The Sample mean is 110
 The Known standard deviation is 15
 Check the box Perform a hypothesis test
 For the Hypothesized mean enter 100
 Click the Options tab
 Change Alternative hypothesis to Mean > hypothesized value
 Use the default Confidence level of 95
 Click OK
This should result in the following output:
N  Mean  SE Mean  95% Lower Bound for \(\mu\) 

25  110.000  3.000  105.065 
\(\mu\) : mean of Sample
Known standard deviation = 15
Null hypothesis  H_{0}: \(\mu_d\) = 100 

Alternative hypothesis  H_{1}: \(\mu_d\) > 100 
ZValue  PValue 

3.33  0.0004 
Select your operating system below to see a stepbystep guide for this example.
We could summarize these results using the five step hypothesis testing procedure:
The population is known to be normally distributed and the population standard deviation is known to be 15. With these two conditions met we can conduct a one sample mean z test
\(H_0: \mu = 100\)
\(H_a: \mu > 100\)
From the Minitab Express output, \(z = 3.33\)
From the Minitab Express output, \(p = 0.0004\)
\(p \le \alpha\), reject the null hypothesis
There is evidence that the mean IQ score of all students at this school is greater than 100.