39. the winning team's scores in 13 high school basketball games were recorded.

if the sample mean is 9.5 points, the sample standard deviation is 0.25 points and the variable is approximately normally distributed, find the 98% confidence interval of the true mean.

Step-by-step explanation:

We want to determine a 98% confidence interval for the true mean score of basketball games

Number of sample, n = 13

Mean, u = 9.5 points

Standard deviation, s = 0.25 points

For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

9.5 ± 2.33 × 0.25/√13

= 9.5 ± 2.33 × 0.0693

= 9.5 ± 0.161469

The lower end of the confidence interval is 9.5 - 0.161469 =9.34

The upper end of the confidence interval is 9.5 + 0.161469 =9.66

Therefore, with 98% confidence interval, the true mean score is between 9.34 points and and 9.66 points.

answer: just subtact the units and then add the non negative : happy to

step-by-step explanation: