Question

Question 9 - of 24 Step 5 of 5 02:07:57 A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d = (golf score after using the newly designed golf clubs)-(golf score before using the newly designed golf clubs). Use a significance level of a = 0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. Golfer 1 2 3 4 5 6 7 8 Score (old design) 94 73 90 82 85 76 79 Score (new design) 88 80 75 86 72 80 75 00 88 y Copy Data Step 5 of 5: Draw a conclusion for the hypothesis test. Answer 2 Points I Keypad

Answer 1

Solution

The following table is obtained: Sample 1 Sample 2 Difference = Sample 1 - Sample 2 88 94 -6 80 73 7 88 90 -2 75 82 -7 86 85 1 72 76 -4 80 81 -1 75 79 -4 Average 80.5 82.5 -2 St. Dev. 6.279 6.99 4.472 n 8 8 8 From the sample data, it is found that the corresponding sample means are: X1 = 80.5 X = 82.5

Also, the provided sample standard deviations are: 81 = 6.279 82 = 6.99 and the sample size is n = 8. For the score differences we have D = -2 8D = 4.472 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: up = 0 Ha: up < 0 This corresponds to a left-tailed test, for which a t-test for two paired samples be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 7. Hence, it is found that the critical value for this left-tailed test is te = -1.895, for a = 0.05 and df = 7. The rejection region for this left-tailed test is R= :{t:t< -1.895)

(3) Test Statistics The t-statistic is computed as shown in the following formula: t D 8D/vn -2 4.472/18 = -1.265 (4) Decision about the null hypothesis Since it is observed that t = -1.265 > t = -1.895, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.1232, and since p= 0.1232 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the golfers can lower their scores by using the manufacturer's newly designed golf clubs​, at the 0.05 significance level