Question

02:11:54 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 on 7 8 73 6 76 9082 Score (old design) 94 Score (new design) 88 85 8 79 y 80 88 75 86 72 80 75 Copy Data Step 1 of 5: State the null and alternative hypotheses for the test. Answer 2 Points Keypad Hop 0

Answer 1

Solution: The data provided in the question statement can be tabulated as below. Calculating a few additional columns: Golfer Score (Old) X1 Score (New) X2 Difference d = X2-X1 (d - d)2 1 94 88 -6 16 2 73 80 7 81 3 90 88 -2 0 4 82 75 -7 25 5 85 86 1 9 6 76 72 -4 4 7 81 80 -1 8 79 75 -4 4 Id=-16 £ (d – d) 2 = 140 d = 2d /n=-16/8 = -2.00 sa = £ (d – d)/(n − 1) = /140/(8 – 1) = 4.472 Level of significance = a = 0.05 Null Hypothesis: Ho: Hd = 0 Alternate Hypothesis: Hi: ud < 0 t = Test Statistic: (ū – Ma) / (sa/ vn) = t = (-2.00 - 0)/(4.472/18) t = -1.265

Degrees of freedom = df = n - 1 = 8 -1 = 7 From t-distribution table; the area in the left tail corresponding to (t = -1.265) and for (df = 7), is given by = 0.1232 Hence, p-value = 0.1232 The p-value is greater than a, hence, we fail to reject the null hypothesis. There is not sufficient evidence to support the manufacturer's claim that golfers can lower their scores by using the newly designed golf clubs.