Question

One of the top golf camps in the country advertises that a week with its coaches will lower your average golf score by more than two points. One disgruntled customer claims that the camp does not live up to its advertised claim. To test the customer's claim, eight randomly selected men attending the camp agree to participate in a study, and their pre-camp and post-camp average scores are listed in the following table. (These are their average scores on a par-72 golf course.) Test the customer's claim at a 0.10 level of significance.

Before Camp	75	74	76	75	76	76	75	78
After Camp	73	72	73	74	74	72	74	75

1. State the hypotheses for this test.

1. Compute the value of the test statistic.

1. Determine the p-value for this test.

Answer 1

Before Camp	After Camp	Difference
75	73	2
74	72	2
76	73	3
75	74	1
76	74	2
76	72	4
75	74	1
78	75	3

∑d = 18

∑d² = 48

n = 8

Mean , x̅d = Ʃd/n = 18/8 = 2.25

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(48-(18)²/8)/(8-1)] = 1.0351

Null and Alternative hypothesis:

Ho : µd ≤ 2 ; H1 : µd > 2

Test statistic:

t = (x̅d- µd)/(sd/√n) = (2.25-2)/(1.0351/√8) = 0.6831

df = n-1 = 7

p-value = T.DIST.RT(0.6831, 7) = 0.2582

Decision:

p-value > α, Do not reject the null hypothesis

Conclusion:

There is not enough evidence to conclude that a week with its coaches will lower your average golf score by more than two points.