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=0.05 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)	94	73	90	82	85	76	81	79
Score (new design)	88	80	88	75	86	72	80	75

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

Claim : golfers can lower their scores by using the manufacturer's newly designed golf clubs

Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)

Step 1 ) H₀: µ_d = 0    vs H_a: µ_d < 0

We can use TI-84 calculator to perform paired t test.

First find the differences d = New design - old design

First enter the differences d in the column L1.

Then press STAT ---> scroll to TESTS ---> Select T test and hit enter.

Select Data and hit enter , plug the values accordingly.

For List : L1 ( Press 2ND key then 1 )

Select sign under H_a

Scroll to calculate and hit enter.

2) standard deviation of the paired differences( Sx ) = 4.5

3) test statistic = -1.265

4) p - value = 0.1232

5) As p value is greater than  α=0.05 , We do not reject H_0.

So we can conclude that there is no significant evidence that golfers can lower their scores by using the manufacturer's newly designed golf clubs.