This lesson introduces the rejection region approach to hypothesis testing and compares it to the pvalue approach. One sample ttest for population mean is introduced. The lesson is concluded by a discussion of computation of power and sample size for one sample ttest.
Lesson 9 Objectives 
Upon successful completion of this lesson, you will be able to:

Unit Summary 

Reading Assignment
An Introduction to Statistical Methods and Data Analysis, chapters 10.2, 5.6 and 5.7. Chapters 5.6 and 5.7 are mainly for Lesson 9.2 and 9.3, though.
Let's start out here by having Dr. Wiesner walk through a comparison of the pvalue approach with the rejection region approach to hypothesis testing.
One can perform hypothesis testing using the pvalue approach, or one can perform hypothesis testing using a rejection region approach. The conclusions from the two approaches are exactly the same.
There are six parts of a test when using the rejection region approach:
Test statistic: The sample statistic one uses to either reject H_{o} (and conclude H_{a}) or not to reject H_{o}.
Critical values: The values of the test statistic that separate the rejection and nonrejection regions.
Rejection region: the set of values for the test statistic that leads to rejection of H_{o}.
Nonrejection region: the set of values not in the rejection region that leads to nonrejection of H_{o}.
As mentioned in lesson 8, the logic of hypothesis testing is to reject the null hypothesis if the sample data are not consistent with the null hypothesis. Thus, one rejects the null hypothesis if the observed test statistic is more extreme in the direction of the alternative hypothesis than one can tolerate. The critical values are the boundary values obtained corresponding to the preset α level.
Oneproportion Ztest for π
Step 0. Check the conditions for the oneproportion ztest to be valid:
 nπ_{o} ≥ 5
 n(1  π_{o}) ≥ 5
Step 1. Set up the hypotheses as one of:
Twotailed Righttailed Lefttailed H_{o }: π = π_{o} ORH_{o }: π = π_{o} ORH_{o} : π = π_{o} H_{a} : π ≠ π_{o} H_{a }: π > π_{o} H_{a} : π < π_{o}
Step 2. Decide on the significance level, α .
Step 3. Compute the value of the test statistic:
\[z=\frac{\hat{\pi}\pi_0}{\sqrt{\frac{\pi_0(1\pi_0)}{n}}}\]
Step 4. Find the appropriate critical values for the tests using the ztable. Write down clearly the rejection region for the problem.
Step 5. Check to see if the value of the test statistic falls in the rejection region. If it does, then reject H_{o} (and conclude H_{a}). If it does not fall in the rejection region, do not reject H_{o}.
Step 6. State the conclusion in words.
Some expert claims that the probability of each person being lefthanded is 0.25. It is observed that out of 30 randomly sampled people, 10 are lefthanded. Using α = 0.05, is there sufficient evidence to conclude that the population proportion is different from 0.25?
a. Use the rejection region approach to perform the testing.
Step 0. Can we use the oneproportion ztest?
The answer is yes since the hypothesized value π_{o} is 0.25 and we can check that:
nπ_{o} = 30 · 0.25 = 7.5 ≥ 5,
n(1  π_{o}) = 30 · (1  0.25) = 22.5 ≥ 5.Step 1. Set up the hypotheses (since the research hypothesis is to check whether the proportion is different from 0.25, we set it up as a twotailed test):
H_{o}: π = 0.25
H_{a}: π ≠ 0.25Step 2. Decide on the significance level, α .
According to the question, α = 0.05.
Step 3. Compute the value of the test statistic:
\[z=\frac{\hat{\pi}\pi_0}{\sqrt{\frac{\pi_0(1\pi_0)}{n}}}=\frac{10/300.25}{\sqrt{\frac{0.25(10.25)}{30}}}=1.054\]
Step 4. Find the appropriate critical values for the test using the ztable. Write down clearly the rejection region for the problem. We can use Table 2 to find the value of Z_{0.025} since the row for df = ∞ (infinite) refers to the zvalue.
From Table 2, Z_{0.025} is found to be 1.96 and thus the critical values are ± 1.96. The rejection region for the twotailed test is given by:
z > 1.96 or z < 1.96
Step 5. Check whether the value of the test statistic falls in the rejection region. If it does, then reject H_{o} (and conclude H_{a}). If it does not fall in the rejection region, do not reject H_{o}.
The observed zvalue is 1.05 and will be denoted as z*. Since z* does not fall within the rejection region, we do not reject H_{o}.
Step 6. State the conclusion in words.
Based on the observed data, there is not enough evidence to conclude that the population proportion of lefthanded people is different from 0.25.
b. Use the pvalue approach to perform the testing.
Step 0  Step 3. The first few steps (Step 0  Step 3) are exactly the same as the rejection region approach.
Step 4. In Step 4, we need to compute the pvalue. Since it is a twotailed test:
\(pvalue=2\times P(z>\frac {\hat{\pi}\pi_0}{\sqrt{\frac{\pi_0 (1\pi_0)}{n}}})\)
\(=2 \times P(z>\frac{10/300.25}{\sqrt{\frac{0.25(10.25)}{30}}})\)
\(=2\times P(z>1.054)=0.2938\)Step 5. Since pvalue = 0.2938 > 0.05 (the α value), we cannot reject the null hypothesis.
Step 6. Conclusion in words:
Based on the observed data, there is insufficient evidence to conclude that the population proportion of lefthanded people is different from 0.25.
Both approaches will ensure the same conclusion and either one will work. However, using the pvalue approach has the following advantages:
Unit Summary 

Reading Assignment
An Introduction to Statistical Methods and Data Analysis, chapters 5.6 and 5.7.
Null and Alternative Hypothesis for Testing a Population Mean
When the parameter that we want to test is the population mean, the test statistic for the population mean when σ is unknown has a tdistribution.
Onesample tTest for the Population Mean μ
Step 0. Conditions for the onesample ttest to be valid for testing one population mean:
Data follows a normal distribution or the sample size is large.
Step 1. Set up the hypotheses as one of:
Twotailed Righttailed Lefttailed H_{o} : μ = μ_{0} ORH_{o} : μ = μ_{0} OR H_{o} : μ = μ_{0}H_{a} : μ ≠ μ_{0} H_{a} : μ > μ_{0} H_{a} : μ < μ_{0}
Step 2. Decide on the significance level, α .
Step 3. Compute the value of the test statistic with the one sample ttest:
\[t=\frac{\bar{x}\mu_0}{s/\sqrt{n}}\]
Step 4. Find the appropriate critical values for the tests using the ttable. Write down clearly the rejection region for the problem. Alternatively, compute the pvalue if you are using the pvalue approach.
Step 5. Check to see if the value of the test statistic falls in the rejection region. If it does, then reject H_{o} (and conclude H_{a}). If it does not fall in the rejection region, do not reject H_{o}. Alternatively, compare the pvalue to α if you are using the pvalue approach. If pvalue ≤ α, reject H_{o} (and conclude H_{a}). If the pvalue > α , do not reject H_{o}.
Step 6. State the conclusion in words.
Performing a tTest by Rejection Region Approach
The mean length of certain construction lumber is supposed to be 8.5 feet. A random sample of 81 pieces of such lumbers gives a sample mean of 8.3 feet and a sample standard deviation of 1.2 feet. A builder claims that the mean of the lumber is different from 8.5 feet. Does the data support the builder's claim at α = 0.05? Use the rejection region approach.
Step 0. Can we use the onesample ttest?
The answer is yes since the sample size is 81. We don't need to check to see if the data follow a normal distribution.
Note: It is also okay to use a onesample ztest for population mean here as suggested by our textbook, since Z is close to t when sample size is large.
Step 1. Set up the hypotheses:
H_{o }: μ = 8.5
H_{a} : μ ≠ 8.5Step 2. Decide on the significance level, α.
α = 0.05
Step 3. Compute the value of the test statistic:
\[t=\frac{\bar{x}\mu_0}{s/\sqrt{n}}=\frac{8.38.5}{1.2/\sqrt{81}}=1.5\]
The observed value of the t statistic is 1.5. We can denote it as t*.
Step 4. Find the appropriate critical values for the tests using the ttable. Write down clearly the rejection region for the problem.
Since n = 81, degrees of freedom = 80, and the critical values are ± t_{0.025}. The value from Minitab is t_{0.025} = 1.99.
The critical values are ±1.99 and the rejection region for the twotailed test is given by:
t > 1.99 or t < 1.99.
Step 5. Check whether the value of the test statistic falls in the rejection region.
Since 1.5 does not fall in the rejection region, we cannot reject H_{o} at α = 0.05.
Step 6. State the conclusion in words:
At α = 0.05, the data does not provide sufficient evidence to conclude that the mean length of the construction lumber is different from 8.5 feet.
Performing a tTest by the PValue Approach
Since the ttable is not as detailed as the ztable, we can only estimate the pvalue of a ttest using the ttable. In order to obtain the exact pvalue of a ttest, one has to use a statistical package such as Minitab.
Minitab commands to obtain P(t > t*):
 Calc > Probability Distributions > tdistribution
 Choose the cumulative distribution to find P(t ≤ t*) and obtain P(t > t*) as 1  P(t ≤ t*).
Use the pvalue approach to draw a conclusion from the previous example.
Since it is a twotailed test, the pvalue is:
pvalue = 2 · P(t > 1.5) = 2 · P(t > 1.5)
The output of Minitab will give the value:
Cumulative Distribution Function
Student's t distribution with 80 DF
x P ( X ≤ x ) 1.5 0.931225pvalue = 2 · P(t > 1.5) = 2 · (1  P(t 1.5))
= 2 · (1  0.931225) = 0.13755
Since the computed pvalue is larger than 0.05, we cannot reject the null hypothesis at level 0.05.
Click on the 'Minitab Movie' icon to view a display of 'Comparing a tValue to a Critical tValue'.
Using a Confidence Interval to Draw a Conclusion About a Twotailed Test
For the twotailed test:
H_{o }: μ = μ_{0}
H_{a} : μ ≠ μ_{0}
The null hypothesis will be rejected at level α if and only if the value μ_{0} does not fall within the (1  α) confidence interval for μ .
Let's use the previous example to show how to use a confidence interval to draw a conclusion about a twotailed test. A 95% confidence interval for the mean lumber length is:
(8.03, 8.57)
For the twotailed test:
H_{o }: μ = 8.5
H_{a} : μ ≠ 8.5Since 8.5 falls within the 95% confidence interval, we cannot reject the null hypothesis at level 0.05.
It is possible to use a onesided confidence bound to draw a conclusion about a onesided test, but you have to be very careful about obtaining the onesided confidence bound.
Unit Summary 

Reading Assignment
An Introduction to Statistical Methods and Data Analysis, chapter 5.5, and Minitab Help on sample size computation.
When the data indicate that one cannot reject the null hypothesis, does it mean that one can accept the null hypothesis? For example, when the pvalue computed from the data is 0.12, one fails to reject the null hypothesis at α = 0.05. Can we say that the data support the null hypothesis?
Answer: When you perform hypothesis testing, you only set the size of Type I error and guard against it. Thus, we can only present the strength of evidence against the null hypothesis. One can sidestep the concern about Type II error if the conclusion never mentions that the null hypothesis is accepted. When the null hypothesis cannot be rejected, there are two possible cases: 1) one can accept the null hypothesis, 2) the sample size is not large enough to either accept or reject the null hypothesis. To make the distinction, one has to check β. If β at a likely value of the parameter is small, then one accepts the null hypothesis. If the β is large, then one cannot accept the null hypothesis.
The relationship between α and β :
If the sample size is fixed, then decreasing α will increase β . If one wants both to decrease, then one has to increase the sample size.
Power = the probability of correctly rejecting a false null hypothesis = 1  β .
Refer to page 218 (edition 5) or pg 243 (edition 6) of our textbook to see the graphs that show the probability of Type II error.
Usually, acceptable values of power are larger than 0.7. One usually sets the power to be 0.8 or 0.85. Again, the acceptable values of power depend on the problem just as the value of α depends on the problem.
The following are interrelated: Power (which is 1  β), sample size, α , and the distance between the actual mean and the mean specified in the null hypothesis.
To calculate the smallest sample size needed for specified α , β , μ_{a} (μ_{a} is the likely value of μ at which you want to evaluate the power; μ_{a} is chosen subjectively to reflect the likely value of μ from the user's prior knowledge):
OneTailed test:
\[n=\sigma^2 \frac{(t_\alpha + t_\beta)^2}{(\mu_0\mu_a)^2}\]
TwoTailed test:
\[n=\sigma^2 \frac{(t_{\alpha/2 }+ t_\beta)^2}{(\mu_0\mu_a)^2}\]
Note: The above two formulas are included for your reference only. When you need to compute the sample size, you can simply use Minitab. The formula given in our book are the approximation to the above formula, replacing t by z.
Using Minitab to compute the sample size or the power:
Stat > power and sample size > 1sample t
Note: The minimum difference referred to in Minitab is the difference between μ_{0} and μ_{a}.
Note: Onesample ttests are used to perform hypothesis tests of the mean. To calculate power or sample size for these tests, you need to determine the minimum difference (effect) that you consider to be meaningful. Then, you can determine the power or the sample size you need to be able to refhect the null hypothesis when the true value differs from the hypothesized value by this minimum difference.
Weight change in pounds of 14 female subjects after taking an exercise program for six weeks are recorded:
17 7 4 18 2 9 12 9 12 9 18 14 18 20Is there sufficient evidence that the average weight change is different from 0? (set α = 0.05)
a. State the null and alternative hypothesis:
H_{o }: μ = 0
H_{a} : μ ≠ 0b. Use Minitab to check whether the onesample ttest may be used.
Now, the sample size is only 14 and thus we need to use the normal probability plot to check whether the data may come from a normal distribution.
Minitab > Graph > probability plot
We can see that we can use the ttest since the normal probability plot indicates that there is no evidence to suggest that the data do not come from a normal distribution.
c. Use Minitab to perform the test and draw a conclusion using the pvalue.
Stat > Basic Statistics > 1Sample t
Dialog box items:
 Variables: Select the column(s) containing the variable(s) that you want to perform the hypothesis test.
 Test mean: Choose to perform a onesample ttest by checkind the box for Perform hypothesis test; then specify the null hypothesis test value by entering this value into the text box for Hypothesized mean. For this example, enter the value 0
Click on Options... in the Confidence Level text box, type your desired confidence level (for this example, use 95). In the Alternative hypothesis text box, select the desired alternative hypothesis from: mean ≠ hypothesized mean, mean < hypothesized mean, mean > hypothesized mean. For this example, select mean ≠ hypothesized mean.
OneSample T: C1
Test of μ = 0 vs ≠ 0
Variable N Mean StDev SE MeanC1 14 4.07 13.08 3.50Variable 95.0% CI T PC1 ( 11.62, 3.48) 1.16 0.265Using Minitab, we see that the observed tvalue is 1.16 and the pvalue is 0.265 which is greater than α = 0.05. We conclude that we cannot reject the null hypothesis.
There are two possible reasons for the failure of rejection of the null hypothesis:
 the null hypothesis is reasonable, or
 there's an insufficient sample size to achieve a powerful test.
We do not yet know which one is the real reason and thus proceed to compute the power of the test.
d. Use Minitab to compute the power (1  β) of the test at the likely value μ_{a} = 5.0. Based on the computed power, would you accept the null hypothesis?
n = 14, α = 0.05
Difference to detect is = 0  (5) = 5.
Using Minitab > Stat > power and sample size, we find that power = 0.2635. The power is very low and we cannot accept the null hypothesis since the possible Type II error is β = 1  0.2635 = 0.7365. The possible Type II error is too high.
e. Use Minitab to find how large a sample size is needed. Suppose we want α = 0.05, power = 0.8, and the minimum detectable difference = 5?
From the Minitab output of the onesample ttest, we see that the standard deviation is 13.08. We can thus estimate σ by 13.08 for the sample size computation problem:
The answer is given by:
Power and Sample Size
1Sample t Test
Testing mean = null (versus no = null)
Calculating power for mean = null + difference
Alpha = 0.05 Sigma = 13.08
Sample Target Actual Difference Size Power Power 5 56 0.8000 0.8024Thus, we get that 56 samples need to be collected in order to draw meaningful results about the hypothesis testing problem.
Words of Caution: Critics of hypothesistesting procedures have observed that a population mean is rarely exactly equal to the value in the null hypothesis and hence, by obtaining a large enough sample, virtually any null hypothesis can be rejected. Thus, it is important to distinguish between statistical significance and practical significance.
Statistical significance is concerned with whether an observed effect is due to chance and practical significance means that the observed effect is large enough to be useful in the real world.
Last Words from Minitab:
Power & Sample Size Tools
Gathering data is like tasting fine wine—you need the right amount. With wine, too small a sip keeps you from accurately assessing a subtle bouquet, but too large a sip overwhelms the palate.
We can’t tell you how big a sip to take at a winetasting event, but when it comes to collecting data, Minitab Statistical Software’s Power and Sample Size tools can tell you how much data you need to be sure about your results.
1. A dealer in recycled paper places empty trailers at various sites. The trailers are gradually filled by individuals who bring in old newspapers and magazines, and are picked up on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1,600 cubic feet per 2week period. The dealer’s records for eighteen 2week periods show the following volumes (in cubic feet) at a particular site (recycled_paper.txt) where \(\hat{y}\) = 1,718.3 and s = 137.8.
a. Place a 95% confidence interval of \(\mu\).
b. Compute the pvalue for the test statistic. Is there strong evidence that \(\mu\) is greater than 1,600?
2. The undergraduate GPA of 18 students from a large MBA class of 800 students is selected. The data are given as (mba_student_gpa.txt).
Use the data in the file above to test the research hypothesis that the average undergraduate GPA of the MBA class is more than 3.5. Give the level of significance of your test. Use the pvalue approach to perform the test at a default level of significance.
ASK! If you have a question about any part of these practice problems, please post your question to the discussion forum in ANGEL.
Now, find Homework 9 in ANGEL and submit it to the Dropbox by the due date.
If there are data referred to in the homework problems, you will also find these data files in ANGEL.