are M1−α/2and Mα/2. Use Table VI and the relation M1−A=n1(n1+n2+1)−MAto find the critical values.
From Table 10.11, we see that n1=10 and n2=10.
The critical values for a two-tailed test at the 5% signifi- cance level areM1−0.05/2andM0.05/2, that is,M1−0.025 and M0.025. First we use Table VI to find M0.025. We go down the leftmost column, labeledn2, to “10.” Then, going across the row forαlabeled 0.025 to the column labeled “10,” we reach 131; thusM0.025=131. Now we apply the aforementioned relation and the result just ob- tained to getM1−0.025:
M1−0.025 = 10(10+10+1)−M0.025
= 210−131=79. See Fig. 10.14A.
FIGURE 10.14A
Do not reject H0 Reject
H0
Reject H0
M 0.025
79 131
0.025
Step 5 If the value of the test statistic falls in the rejection region, reject H0; otherwise, do not rejectH0.
The value of the test statistic is M =83, as found in Step 3, which does not fall in the rejection region shown in Fig. 10.14A. Thus we do not reject H0. The test re- sults are not statistically significant at the 5% level.
Step 4 Obtain the P-value by using technology.
Using technology, we find that the P-value for the hypothesis test is P=0.104, as shown in Fig. 10.14B.
FIGURE 10.14B
M = 83 P = 0.104
M
Step 5 If P≤α, rejectH0; otherwise, do not rejectH0.
From Step 4,P =0.104. Because the P-value exceeds the specified significance level of 0.05, we do not re- ject H0. The test results are not statistically significant at the 5% level and (see Table 9.8 on page 378) provide at most weak evidence against the null hypothesis.
Step 6 Interpret the results of the hypothesis test.
Interpretation At the 5% significance level, the data do not provide sufficient evidence to conclude that the mean tear strengths differ for the two brands of vinyl floor covering.
Comparing the Mann–Whitney Test and the Pooled t-Test
In Section 10.2, you learned how to perform a pooledt-test to compare two population means when the variable under consideration is normally distributed on each of the two populations and the population standard deviations are equal. Because two normal distributions with equal standard deviations have the same shape, you can also use the Mann–Whitney test to perform such a hypothesis test.
472 CHAPTER 10 Inferences for Two Population Means
Under conditions of normality, the pooledt-test is more powerful than the Mann–
Whitney test but, surprisingly, not much more powerful. However, if the two distribu- tions of the variable under consideration have the same shape but are not normal, the Mann–Whitney test is usually more powerful than the pooledt-test, often consider- ably so.
KEY FACT 10.5 The Mann–Whitney Test Versus the Pooledt-Test
Suppose that the distributions of a variable of two populations have the same shape and that you want to compare, using independent simple random sam- ples, the two population means. When deciding between the pooledt-test and the Mann–Whitney test, follow these guidelines: If you are reasonably sure that the two distributions are normal, use the pooledt-test; otherwise, use the Mann–Whitney test.
Comparing Two Population Medians with the Mann–Whitney Procedure
The Mann–Whitney test can be used to compare two population medians as well as two population means. To use Procedure 10.5 to compare two population medians, simply replaceμ1withη1andμ2withη2.
In some of the exercises at the end of this section, you will be asked to use the Mann–Whitney test to perform hypothesis tests for comparing two population medians.
THE TECHNOLOGY CENTER
Some statistical technologies have programs that automatically perform a Mann–
Whitney test. In this subsection, we present output and step-by-step instructions for such programs. (Note to TI-83/84 Plus users:At the time of this writing, the TI-83/84 Plus does not have a built-in program for conducting a Mann–Whitney test. However, a TI program, MANNWHIT, to help with the calculations is located in the TI Programs folder on the WeissStats CD. See theTI-83/84 Plus Manualfor details.)
EXAMPLE 10.13 Using Technology to Conduct a Mann–Whitney Test
Computer-System Training Table 10.9 on page 465 shows the times, in minutes, required to learn how to use a computer system for independent samples of employ- ees without and with computer-system experience. Use Minitab or Excel to decide, at the 5% significance level, whether the data provide sufficient evidence to con- clude that the mean learning time for employees without experience exceeds that for employees with experience.
Solution Letμ1andμ2denote the mean learning times for all employees without and with experience, respectively. We want to perform the hypothesis test
H0:μ1=μ2(mean time for inexperienced employees is not greater) Ha:μ1> μ2(mean time for inexperienced employees is greater) at the 5% significance level. Note that the hypothesis test is right tailed.
We applied the Mann–Whitney test programs to the data, resulting in Out- put 10.3. Steps for generating that output are presented in Instructions 10.3. Note that, like many statistical technologies, both Minitab and Excel give the results of the Mann–Whitney test in terms of medians, but those results can also be interpreted in terms of means.
OUTPUT 10.3 Mann-Whitney test on the learning-time data
MINITAB
EXCEL
As shown in Output 10.3, the P-value for the hypothesis test is 0.02. Because the P-value is less than the specified significance level of 0.05, we reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that the mean learning time for employees without experience exceeds that for employees with experience.
INSTRUCTIONS 10.3
Steps for generating Output 10.3 MINITAB
1 Store the data from Table 10.9 in columns named WITHOUT and WITH
2 ChooseStat➤Nonparametrics➤ Mann-Whitney. . .
3 Specify WITHOUT in theFirst Sampletext box
4 Specify WITH in theSecond Sampletext box
5 Click the arrow button at the right of theAlternativedrop-down list box and selectgreater than
6 ClickOK
EXCEL
1 Store the data from Table 10.9 in ranges named WITHOUT and WITH 2 ChooseDDXL➤Nonparametric
Tests
3 SelectMann Whitney Rank Sumfrom theFunction typedrop-down box 4 Specify WITHOUT in the
1st Quantitative Variabletext box 5 Specify WITH in the
2nd Quantitative Variabletext box 6 ClickOK
7 Click the0.05button 8 Click theRight Tailedbutton 9 Click theComputebutton
474 CHAPTER 10 Inferences for Two Population Means
Exercises 10.4
Understanding the Concepts and Skills
10.94 State the conditions that are required for using the Mann–
Whitney test.
10.95 Suppose that, for two populations, the distributions of the variable under consideration have the same shape. Further sup- pose that you want to perform a hypothesis test based on inde- pendent random samples to compare the two population means.
In each case, decide whether you would use the pooledt-test or the Mann–Whitney test and give a reason for your answer. You know that the distributions of the variable are
a. normal. b. not normal.
10.96 Part of conducting a Mann–Whitney test involves ranking all the data from both samples combined. Explain how to deal with ties.
10.97 Why do two normal distributions that have equal standard deviations have the same shape?
10.98 The Mann–Whitney test can be used to compare two pop- ulation means. That test can also be used to compare two popula-
tion .
Exercises10.99–10.102pertain to critical values for a Mann–
Whitney test. Use Table VI in Appendix A to determine the critical value(s) in each case. For a left-tailed or two-tailed test, you will also need the relation M1−A=n1(n1+n2+1)−MA.
10.99 n1=8,n2=9; Significance level=0.05 a. Right-tailed test b. Left-tailed test c. Two-tailed test
10.100 n1=8,n2=9; Significance level=0.01 a. Right-tailed test b. Left-tailed test c. Two-tailed test
10.101 n1=9,n2=8; Significance level=0.10 a. Right-tailed test b. Left-tailed test c. Two-tailed test
10.102 n1=9,n2=8; Significance level=0.05 a. Right-tailed test b. Left-tailed test c. Two-tailed test
In each of Exercises 10.103–10.108, the null hypothesis is H0:μ1=μ2 and the alternative hypothesis is as specified. We have provided data from independent simple random samples from the two populations under consideration. In each case, use the Mann–Whitney test to perform the required hypothesis test at the 10% significance level.
10.103 Ha:μ1> μ2