Q:

Suppose a bowler claims that her bowling score is not equal to 150 points, on average. Several of her teammates do not believe her, so the bowler decides to do a hypothesis test, at a 5% significance level, to persuade them. She bowls 22 games. The mean score of the sample games is 157 points. The bowler knows from experience that the standard deviation for her bowling score is 18 points.

Accepted Solution

A:
Answer:The test statistic (z-score) of this one-mean hypothesis test is z=1.823.Step-by-step explanation:The null and alternative hypothesis are:[tex]H_0: \mu=150\\\\H_a: \mu\neq 150[/tex]The significance level is 0.05.The sample size is n=22.The sample mean is M=157.The standard deviation is 18.The z-score can be calculated as:[tex]z=\frac{M-\mu}{\sigma/\sqrt{n}}=\frac{157-150}{18/\sqrt{22}} =\frac{7}{3.84} =1.823[/tex]The test statistic (z-score) of this one-mean hypothesis test is z=1.823.