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)	77	88	76	88	84	80	89	81
Score (new design)	71	92	73	87	91	73	83	78

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 one decimal place.

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

(1) The Claim is that using the neq club reduces the scores. So we use a left tailed test.

The Hypothesis:

H0: = 0

Ha: < 0

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(2) Standard deviation = 5.0

To Calculate the standard deviation

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1,

where SS = SUM(X - Mean)².

#	Difference	Mean	(X-Mean)2
1	-6	-1.88	16.9744
2	4	-1.88	34.5744
3	-3	-1.88	1.2544
4	-1	-1.88	0.7744
5	7	-1.88	78.8544
6	-7	-1.88	26.2144
7	-6	-1.88	16.9744
8	-3	-1.88	1.2544
Total	-15		176.8752

n	8
Sum	-15
Mean	-1.88
SS	176.8752
Variance	25.2679
Std Dev	5.0267

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(3) The Test Statistic: Since sample size is small, and population std. deviation is unknown, we use the students t test.

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(4) The p value (Left tailed) at t = -1.058), df = n - 1 = 7 is 0.1626

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(5) The rejection rule is that if p value is > , then reject H0.

Here since p value (0.1626) is > 0.05, we fail to reject H0.

The Conclusion: There isn't sufficient evidence at the 5% level of significance to support the manufacturers claim that the newly designed golf clubs helps to reduce the scores of golfers.

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