Question

The winning team's score in 5 high school basketball games was recorded. If the sample mean is 62.3 points and the sample standard deviation is 11.0 points, find the

98% confidence interval of the true mean.

A) 57.4 < µ < 67.2 B) 50.8 < µ < 73.8
C) 25.4 < µ < 99.2 D) 43.9 < µ < 80.7

Answer 1

To find the 98% confidence interval of the true mean, we can use the formula: CI = x ± Z * (σ / √n). For the given sample mean and standard deviation, the 98% confidence interval is (57.4, 67.2).

Step-by-step explanation:

To find the 98% confidence interval of the true mean, we can use the formula:

CI = x ± Z * (σ / √n)

Where CI is the confidence interval, x is the sample mean, Z is the Z-score corresponding to the desired confidence level, σ is the sample standard deviation, and n is the sample size.

For a 98% confidence level, the Z-score is approximately 2.33.

Plugging in the values from the question, the calculation becomes:

CI = 62.3 ± 2.33 * (11.0 / √5)

Calculating the above expression gives us a confidence interval of (57.4, 67.2).

Answer 2

This is 3.747*11/sqrt(5)=18.4

the interval is (49.9,86.7), the SE added to and subtracted from the mean.

It is B

Thank You