Thirty-eight members of a bowling league sign up for a program that claims to improve b scores. The participants bowl a set of five games before the program begins and a set of fi games again at the end to measure their improvement.

The mean number of points improved (over the set of five games) was $\bar { x } = 10$. Assume th standard deviation is $\sigma = 45$ and let $\mu$ be the population mean number of points improved for the set of five games. To determine whether the program is effective, a test is made of the hypotheses $H _ { 0 } : \mu = 0$ versus $H _ { 1 } : \mu 0$.
Compute the value of the test statistic. wling
A) 1.37
B) 0.22
C) 0.0853
D) 9.19

Question

Thirty-eight members of a bowling league sign up for a program that claims to improve b scores. The participants bowl a set of five games before the program begins and a set of fi games again at the end to measure their improvement.

The mean number of points improved (over the set of five games) was $\bar { x } = 10$. Assume th standard deviation is $\sigma = 45$ and let $\mu$ be the population mean number of points improved for the set of five games. To determine whether the program is effective, a test is made of the hypotheses $H _ { 0 } : \mu = 0$ versus $H _ { 1 } : \mu > 0$.
Compute the value of the test statistic. wling
A) 1.37
B) 0.22
C) 0.0853
D) 9.19

Quizplus · Accepted Answer

The test statistic is calculated using the formula $z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$, where $\bar{x} = 10$, $\mu = 0$, $\sigma = 45$, and $n = 38$. Plugging these values into the formula gives $z = \frac{10 - 0}{45 / \sqrt{38}} \approx 1.37$.

Thirty-Eight Members of a Bowling League Sign Up for a Program