Over the last 15 Major League Baseball seasons, the mean # of strikeouts by the American League leader is 258.5. Assuming that the # of strikeouts by the league leader is normally distributed and the st. dev. for all seasons in all leagues is 34.9, find the 82% Confidence Interval and the 93% Confidence Interval for the mean.

Question

Over the last 15 Major League Baseball seasons, the mean # of strikeouts by the American League leader is 258.5. Assuming that the # of strikeouts by the league leader is normally distributed and the st. dev. for all seasons in all leagues is 34.9, find the 82% Confidence Interval and the 93% Confidence Interval for the mean.

I get that the 82% would be 1-.82=.18 & 93% would be 1-.93=.07 and that n=15, x-bar=258.5, & st. dev=34.9. But I'm confused about the formula. Would I be using a t-distribution?

Answer 1

A t-distribution is sensitive to the size of the sample when considering confidence intervals. If a z-distribution is used instead of a t-distribution with small sample sizes, the confidence interval may be too narrow and the ability to estimate the true population mean may be diminished. So yes, try a confidence interval formula using a t-distribution.