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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 49 of the team’s games under the new coach and finds that they scored an average of 95 points with a standard deviation of 9 points. Over the past 10 years, the team had averaged 93 points. What is the value of the appropriate test statistic to test the new coach’s claim at the 1% significance level?Multiple Choicez = 0.33z = 1.56t48 = 0.33t48 = 1.56

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 49 of the team’s games under the new coach and finds that they scored an average of 95 points with a standard deviation of 9 points. Over the past 10 years, the team had averaged 93 points. What is the value of the appropriate test statistic to test the new coach’s claim at the 1% significance level?Multiple Choicez = 0.33z = 1.56t48 = 0.33t48 = 1.56

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Solution

To answer this question, we need to use the formula for the t-test statistic:

t = (X̄ - μ) / (s/√n)

where: X̄ = sample mean = 95 μ = population mean = 93 s = standard deviation = 9 n = sample size = 49

Substituting these values into the formula, we get:

t = (95 - 93) / (9/√49) = 2 / (9/7) = 2 / 1.28571429 = 1.55431223

So, the value of the test statistic is approximately 1.56. Therefore, the correct answer is t48 = 1.56.

This problem has been solved

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