You have 40 golfers who come to you for lessons and you are trying to determine

Question

You have 40 golfers who come to you for lessons and you are trying to determine if standardizing equipment for your students will improve their scores. You had each of the 40 players play 5 rounds of golf with their non-standardized equipment and each player only reported the average of their 5 rounds. The average of all the scores was 79 strokes.
You gave them all the same set of clubs and the same brand golf balls and after some time to adjust to the new clubs, you had the same 40 players play 5 rounds of golf woth their standardized equipment and each player only reported the aberage of their 5 rounds. The average 76.5 and the standard deviation was 2.3 strokes. At the 1% statistical significance level, did standardizing equipment improve the players scores? You have 40 golfers who come to you for lessons and you are trying to determine if standardizing equipment for your students will improve their scores. You had each of the 40 players play 5 rounds of golf with their non-standardized equipment and each player only reported the average of their 5 rounds. The average of all the scores was 79 strokes.
You gave them all the same set of clubs and the same brand golf balls and after some time to adjust to the new clubs, you had the same 40 players play 5 rounds of golf woth their standardized equipment and each player only reported the aberage of their 5 rounds. The average 76.5 and the standard deviation was 2.3 strokes. At the 1% statistical significance level, did standardizing equipment improve the players scores?
You gave them all the same set of clubs and the same brand golf balls and after some time to adjust to the new clubs, you had the same 40 players play 5 rounds of golf woth their standardized equipment and each player only reported the aberage of their 5 rounds. The average 76.5 and the standard deviation was 2.3 strokes. At the 1% statistical significance level, did standardizing equipment improve the players scores?

Explanation / Answer

Solution:

Here, we have to use the one sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

H0: µ = 79 versus Ha: µ > 79

This is a upper tailed test.

We are given,

Level of significance = alpha = 0.01

Sample size = n = 40

Sample mean = Xbar = 76.5

Sample standard deviation = S = 2.3

Degrees of freedom = n - 1 = 40 – 1 = 39

Critical t values = 2.4258

Test statistic = t = (Xbar - µ) / [S/sqrt(n)]

Test statistic = t = (76.5 – 79) / [2.3/sqrt(40)] = -6.8745

P-value = 1.00

Alpha value = 0.01

P-value > Alpha value

So, we do not reject the null hypothesis at the significance level of = 0.01.

At the 1% statistical significance level, standardizing equipment does not improve the players’ scores.

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