Continuing problem 1, test, at the 10% level of significance, the hypothesis tha

Question

Continuing problem 1, test, at the 10% level of significance, the hypothesis that it takes mice less time to run the maze on the second trial, on average using the rejection region method.

d-bar = 25.2

sd=35.6609

point estimate (mu)d=25.2

E = 34.00122

a. Identify the claim you wish to test and set up the null and alternative hypotheses.

b. Identify the distribution of the test statistic and compute its value. c. Construct the rejection region

c. Construct the rejection region.

d. State your decision.

e. Write your conclusion.

Explanation / Answer

The basic concepts of hypothesis testing were explained in Chapter 8. With the z, t, and x2 tests, a sample mean, variance, or proportion can be compared to a specific population mean, variance, or proportion to determine whether the null hypothesis should be rejected. There are, however, many instances when researchers wish to compare two sample means, using experimental and control groups. For example, the average lifetimes of two different brands of bus tires might be compared to see whether there is any difference in 432 Chapter 9 Testing the Difference Between Two Means, Two Variances, and Two Proportions 9–2 9–1 Introduction Statistics Today Statistics Today blu49076_ch09.qxd 5/1/2003 8:19 AM Page 432 Section 9–2 Testing the Difference Between Two Means: Large Samples 433 9–3 9–2 Testing the Difference Between Two Means: Large Samples Objective 1. Test the difference between two large sample means, using the z test. tread wear. Two different brands of fertilizer might be tested to see whether one is better than the other for growing plants. Or two brands of cough syrup might be tested to see whether one brand is more effective than the other. In the comparison of two means, the same basic steps for hypothesis testing shown in Chapter 8 are used, and the z and t tests are also used. When comparing two means by using the t test, the researcher must decide if the two samples are independent or dependent. The concepts of independent and dependent samples will be explained in Sections 9–4 and 9–5. Furthermore, when the samples are independent, there are two different formulas that can be used depending on whether or not the variances are equal. To determine if the variances are equal, use the F test shown in Section 9–3. Finally, the z test can be used to compare two proportions, as shown in Section 9–6. Suppose a researcher wishes to determine whether there is a difference in the average age of nursing students who enroll in a nursing program at a community college and those who enroll in a nursing program at a university. In this case, the researcher is not interested in the average age of all beginning nursing students; instead, he is interested in comparing the means of the two groups. His research question is: Does the mean age of nursing students who enroll at a community college differ from the mean age of nursing students who enroll at a university? Here, the hypotheses are H0: m1 m2 H1: m1 m2 where m1 mean age of all beginning nursing students at the community college m2 mean age of all beginning nursing students at the university Another way of stating the hypotheses for this situation is H0: m1

m2 0 H1: m1 m2 0 If there is no difference in population means, subtracting them will give a difference of zero. If they are different, subtracting will give a number other than zero. Both methods of stating hypotheses are correct; however, the first method will be used in this book. Assumptions for the Test to Determine the Difference Between Two Means 1. The samples must be independent of each other. That is, there can be no relationship between the subjects in each sample. 2. The populations from which the samples were obtained must be normally distributed, and the standard deviations of the variable must be known, or the sample sizes must be greater than or equal to 30. The theory behind testing the difference between two means is based on selecting pairs of samples and comparing the means of the pairs. The population means need not be known. All possible pairs of samples are taken from populations. The means for each pair of samples are computed and then subtracted, and the differences are plotted. If both blu49076_ch09.qxd 5/1/2003 8:19 AM Page 433 434 Chapter 9 Testing the Difference Between Two Means, Two Variances, and Two Proportions 9–4 Figure 9–1 Differences of Means of Pairs of Samples Unusual Stats Unusual Stats Adult children who live with their parents spend more than 2 hours a day doing household chores. According to a study, daughters contribute about 17 hours a week and sons about 14.4 hours. Source: Reprinted with permission from Psychology Today magazine. Copyright © 1995 (Sussex Publishers, Inc.). populations have the same mean, then most of the differences will be zero or close to zero. Occasionally, there will be a few large differences due to chance alone, some positive and others negative. If the differences are plotted, the curve will be shaped like the normal distribution and have a mean of zero, as shown in Figure 9–1. The variance of the difference is equal to the sum of the individual variances of and . That is, where So the standard deviation of is Formula for the z Test for Comparing Two Means from Independent Populations This formula is based on the general format of where is the observed difference, and the expected difference m1 m2 is zero when the null hypothesis is m1 m2, since that is equivalent to m1 m2 0. Finally, the standard error of the difference is In the comparison of two sample means, the difference may be due to chance, in which case the null hypothesis will not be rejected, and the researcher can assume that the means of the populations are basically the same. The difference in this case is not significant. See Figure 9–2(a). On the other hand, if the difference is significant, the null hypothesis is rejected and the researcher can conclude that the population means are different. See Figure 9–2(b). A s2 1 n1 s2 2 n2 X1 X2 Test value observed value expected value standard error z X1 X2 m1 m2 A s2 1 n1 s2 2 n2 A s2 1 n1 s2 2 n2 X1 X2 s2 X1 s2 1 n1 and s2 X2 s2 2 n2 s2 X1 s2 X2 2 X1 X2 s X1 X2 X1 X2 0 Distribution ofX1 –X2 blu49076_ch09.qxd 5/1/2003 8:19 AM Page 434 These tests can also be one-tailed, using the following hypotheses: Right-tailed Left-tailed H0: m1 m2 H0: m1 m2 0 H0: m1 m2 H0: m1 m2 0 H1: m1 m2 or H1: m1 m2 0 H1: m1 m2 or H1: m1 m2 0 The same critical values used in Section 8–3 are used here. They can be obtained from Table E in Appendix C. If and are not known, the researcher can use the variances obtained from each sample and , but both sample sizes must be 30 or more. The formula then is provided that n1 30 and n2 30. When one or both sample sizes are less than 30 and s1 and s2 are unknown, the t test must be used, as shown in Section 9–4. The basic format for hypothesis testing using the traditional method is reviewed here. STEP 1 State the hypotheses and identify the claim. STEP 2 Find the critical value(s). STEP 3 Compute the test value. STEP 4 Make the decision. STEP 5 Summarize the results. A survey found that the average hotel room rate in New Orleans is $88.42 and the average room rate in Phoenix is $80.61. Assume that the data were obtained from two samples of 50 hotels each and that the standard deviations were $5.62 and $4.83, respectively. At a 0.05, can it be concluded that there is a significant difference in the rates? Source: USA TODAY. z X1 X2 m1 m2 A s2 1 n1 s2 2 n2 s2 2 s2 1 s2 s 2 2 1 Section 9–2 Testing the Difference Between Two Means: Large Samples 435 9–5 Sample 1 (a) Difference is not significant (b) Difference is significant X1 Population µ1 = µ2 Sample 2 X2 Sample 2 X2 Sample 1 X1 (b) Reject H0: 1 = 2 sinceX1 –X2 (a) Do not reject H is significant. 0: 1 = 2 sinceX1 –X2 is not significant. Population 2 µ2 Population 1 µ1 Figure 9–2 Hypothesis-Testing Situations in the Comparison of Means Example 9–1 blu49076_ch09.qxd 5/1/2003 8:19 AM Page 435 Solution STEP 1 State the hypotheses and identify the claim. H0: m1 m2 and H1: m1 m2 (claim) STEP 2 Find the critical values. Since a 0.05, the critical values are 1.96 and 1.96. STEP 3 Compute the test value. STEP 4 Make the decision. Reject the null hypothesis at a 0.05, since 7.45 1.96. See Figure 9–3. STEP 5 Summarize the results. There is enough evidence to support the claim that the means are not equal. Hence, there is a significant difference in the rates. The P-values for this test can be determined by using the same procedure shown in Section 8–3. For example, if the test value for a two-tailed test is 1.40, then the P-value obtained from Table E is 0.1616. This value is obtained by looking up the area for z 1.40, which is 0.4192. Then 0.4192 is subtracted from 0.5000 to get 0.0808. Finally, this value is doubled to get 0.1616 since the test is two-tailed. If a 0.05, the decision would be to not reject the null hypothesis, since P-value a. The P-value method for hypothesis testing for this chapter also follows the same format as stated in Chapter 8. The steps are reviewed here. STEP 1 State the hypotheses and identify the claim. STEP 2 Compute the test value. STEP 3 Find the P-value. STEP 4 Make the decision. STEP 5 Summarize the results. Example 9–2 illustrates these steps. A researcher hypothesizes that the average number of sports that colleges offer for males is greater than the average number of sports that colleges offer for females. A sample of the number of sports offered by colleges is shown. At a 0.10, is there enough evidence to support the claim? –1.96 0 +7.45 +1.96 z X1 X2 m1 m2 A s2 1 n1 s2 2 n2 88.42 80.61 0 A 5.622 50 4.833 50 7.45 436 Chapter 9 Testing the Difference Between Two Means, Two Variances, and Two Proportions 9–6 Figure 9–3 Critical and Test Values for Example 9–1 Example 9–2 blu49076_ch09.qxd 5/1/2003 8:19 AM Page 436 Males Females 6 11 11 8 15 6 8 11 13 8 6 14 8 12 18 7 5 13 14 6 6956 9 6 557 6 6 9 18 7 6 10 7 6 5 5 15 6 11 5 5 16 10 7 8 5 9955 8 7 556 5 8 9 6 11 6 9 18 13 7 10 9 5 11 5 8 7 8 5 7 6 7 7 5 10 7 11 4 6 8 7 10 7 10 8 11 14 12 5 8 5 Source: USA TODAY. Solution STEP 1 State the hypotheses and identify the claim: H0: m1 m2 H1: m1 m2 (claim) STEP 2 Compute the test value. Using a calculator or the formulas in Chapter 3, find the mean and standard deviation for each data set. For the males 8.6 and s1 3.3 For the females 7.9 and s2 3.3 Substitute in the formula STEP 3 Find the P-value. For z 1.06, the area is 0.3554, and 0.5000 0.3554 0.1446 or a P-value of 0.1446. STEP 4 Make the decision. Since the P-value is larger than a (that is, 0.1446 0.10), the decision is to not reject the null hypothesis. See Figure 9–4. STEP 5 Summarize the results. There is not enough evidence to support the claim that colleges offer more sports for males than they do for females. *Note: Calculator results may differ due to rounding. Sometimes, the researcher is interested in testing a specific difference in means other than zero. For example, he or she might hypothesize that the nursing students at a 0 0.10 0.1446 z X1 X2 m1 m2 A s2 1 n1 s2 2 n2 8.6 7.9 0 A 3.32 50 3.32 50 1.06* X2 X1 Section 9–2 Testing the Difference Between Two Means: Large Samples 437 9–7 Figure 9–4 P-Value and a Value for Example 9–2 blu49076_ch09.qxd 5/1/2003 8:19 AM Page 437 438 Chapter 9 Testing the Difference Between Two Means, Two Variances, and Two Proportions 9–8 Example 9–3 community college are, on the average, 3.2 years older than those at a university. In this case, the hypotheses are H0: m1 m2 3.2 and H1: m1 m2 3.2 The formula for the z test is still where m1 m2 is the hypothesized difference or expected value. In this case, m1 m2 3.2. Confidence intervals for the difference between two means can also be found. When one is hypothesizing a difference of 0, if the confidence interval contains 0, the null hypothesis is not rejected. If the confidence interval does not contain 0, the null hypothesis is rejected. Confidence intervals for the difference between two means can be found by using this formula: Formula for Confidence Interval for Difference Between Two Means: Large Samples When n1 30 and n2 30, and can be used in place of and .

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