Question

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) 77 88 76 8884 80 89 81 Score (new design) 71 92 73 87 91 73 83 78 Copy Data

Step 1 of 5:State the null and alternative hypotheses for the test.

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.

Step 3 of 5:Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Find the p-value for the hypothesis test. Round your answer to four decimal places.

Step 5 of 5: Draw a conclusion for the hypothesis test.

Answer 1

step 1:

null Hypothesis:μd		=	0
alternate Hypothesis: μd		<	0

sep 2:

S. No	New design	Old design	diff:(d)=x1-x2	d2
1	71	77	-6	36.00
2	92	88	4	16.00
3	73	76	-3	9.00
4	87	88	-1	1.00
5	91	84	7	49.00
6	73	80	-7	49.00
7	83	89	-6	36.00
8	78	81	-3	9.00
	total	=	Σd=-15	Σd2=205
	mean dbar=	d̅     =	-1.8750
	degree of freedom =n-1                            =		7
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		5.03

Step 3 of 5:

std error=Se=SD/√n=		1.7784
test statistic            =	(d̅-μd)/Se         =	-1.054

Step 4 of 5:

p value

=

0.1634

from excel: tdist(1.054,7,1)

step 5:

fail to reject the null hypothesis