Question

A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously tepid offense historically managed. After a few months, the team owner looks at the data to test the coach’s claim. He takes a sample of 36 of the team’s games under the new coach and finds that they scored an average of 101 points with a standard deviation of 6 points. Over the past 10 years, the team had averaged 99 points. What is the value of the appropriate test statistic to test the new coach’s claim at the 1% significance level? a. z = 2.00 b. t35 = 2.00 c. t35 = 0.33 d. z = 0.33

Answer 1

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

The statistic value under the null hypothesis is:

Z= X[bar] - μ  = 101 - 99 = 2

σ/√n 6/√36

I don't have σ, but since this is an approximation I can use the value of S instead.

I hope it helps!