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)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05α=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)	93	85	78	75	84	75	81	77
Score (new design)	90	87	75	71	91	71	79	75

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

Step 2: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

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

Step 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5: Make the decision for the hypothesis test.

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.895
since our test is left-tailed
reject Ho, if to < -1.895
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 1.125
We have d = 1.125
pooled variance = calculate value of Sd= √S^2 = sqrt [ 111-(9^2/8 ] / 7 = 3.796
to = d/ (S/√n) = 0.838
critical Value
the value of |t α| with n-1 = 7 d.f is 1.895
we got |t o| = 0.838 & |t α| =1.895
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < 0.8382 ) = 0.7852
hence value of p0.05 < 0.7852,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: 0.838
critical value: reject Ho, if to < -1.895
decision: Do not Reject Ho
p-value: 0.7852
we do not have enough evidence to support the claim that golfers can lower their scores by using the manufacturer's newly designed golf clubs.