SECTION 8.3 Testing the Difference Between Means (Dependent Samples) 439 Before After d d2 96 89 7 49 95 90 5 25 95 95 0 0 99 94 5 25 87 91 -4 16 104 100 4 16 105 103 2 4 94 100 -6 36 Σ = 13 Σ = 171 The t-Test for the Difference Between Means A golf instruction and club fitting company claims that people who play golf can improve (decrease) their average golf scores by taking lessons from one of their coaches. The average golf scores of eight randomly selected people who play golf are determined before and after taking lessons from one of the company’s coaches. The results are shown in the table. At a = 0.10, is there enough evidence to support the company’s claim? Assume the average golf scores are normally distributed. Person 1 2 3 4 5 6 7 8 Average golf score (before taking lessons) 96 95 95 99 87 104 105 94 Average golf score (after taking lessons) 89 90 95 94 91 100 103 100 SOLUTION Because the samples are random and dependent, and the populations are normally distributed, you can use the t@test. The claim is that “people who play golf can improve (decrease) their average golf scores.” In other words, the company claims that a person’s average golf score before taking the lessons will be greater than the person’s average golf score after taking the lessons. Each difference is given by d = 1score before lessons2 - 1score after lessons2. The null and alternative hypotheses are H0: md … 0 and Ha: md 7 0 (claim). Because the test is a right-tailed test, a = 0.10, and d.f. = 8 - 1 = 7, the critical value is t0 = 1.415. The rejection region is t 7 1.415. Using the table at the left, you can calculate d and sd as shown below. Notice that the alternative formula is used to calculate the standard deviation. d = Σd n = 13 8 = 1.625 sd = HΣd2 - c 1Σd22 n d n - 1 = H171 - 11322 8 8 - 1 ≈ 4.6272 The standardized test statistic is t = d - md sd 1n ≈ 1.625 - 0 4.6272 18 ≈ 0.9933. The figure shows the location of the rejection region and the standardized test statistic t. Because t is not in the rejection region, you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 10% level of significance to support the golf instruction and club fitting company’s claim that people who play golf can improve (decrease) their average golf scores by taking lessons from one of their coaches. = 0.10 1 − = 0.90 −3 −2 −1 0 1 2 3 t α α t0 = 1.415 t ≈ 0.9933 Study Tip To simplify the calculation of t, you can round the values of d and sd to four decimal places, as shown in Examples 1 and 2. See Minitab steps on page 464. EXAMPLE 1 Study Tip You can also use a P@value to perform a hypothesis test for the difference between means. For instance, in Example 1, you can enter the data in Minitab (as shown on page 464) and find P = 0.177. Because P 7 a, you fail to reject the null hypothesis.
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