Question

Jolyn, a golfer, claims that her drive distance is not equal to 222 meters, on average. Several of her friends do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She hits 11 drives. The mean distance of the sample drives is 218 meters. Jolyn knows from experience that the standard deviation for her drive distance is 14 meters. H0: μ=222; Ha: μ≠222 α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? Provide your answer below:

Answer 1

-0.95

Explanation:

The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=218. The sample the golfer uses is 11 drives, so n=11. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 222 meters. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic:

z0=x¯−μ0σn√=218−2221411√≈−44.221≈−0.95

So, the test statistic for this hypothesis test is z0=−0.95.

Answer 2

0.29

Explanation:

z-score is given by the formula:

z=(μ-M)/σ where

• μ is the hypothesis drive distance, which is 222
• M is the mean sample drive distance
• σ is the standard deviation of the sample drive distance

Therefore

z=(222-218)/14≈0.29