Question

I'm having some difficulty with this problem:

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 & the standard deviation for all seasons in all leagues is 34.9, find the 82% Confidence Interval & the 93% Confidence Interval for the mean

Here's what I got for 82% CI:

x1=82%
μ=258.5
σ=34.9
n=15

α=.82
1- α=0.18
α/2=0.09
1-0.09=0.91
z-score=1.35

p^=x/n
p^=0.18/15
p^=0.988

q^=1-p^
q^=1-0.988
q^=0.012

E=zα/2*√p^q^/n
1.35*√(0.988)(0.012)/15
1.35*√0.011856/15
1.35*√7.904
1.35*2.811405343
E=3.795397212

p^ - E <p < p^ + E
0.988 – 3.795397212 < p < 0.988 + 3.795397212
-3.696597212 < p < 4.783397212

However, when I use computer software, I get 245.7828 < mean < 271.2172, which I think is correct. Of course, the software doesn't show me how to get to the answer.

If anyone can help figure out where I went wrong, please point me in the right direction.

Thanks!

Answer 1

Here's the formula I would use for this type of problem:

CI82 = mean + or - 1.35(sd / √n)
...where + or - 1.35 represents the 82% confidence interval, sd = standard deviation, √ = square root, and n = sample size.

With your data:

CI82 = 258.5 + or - 1.35 (34.9 / √15)
....... = 258.5 + or - 1.35 (9.01)
....... = 258.5 + or - 12.16

258.5 - 12.16 = 246.34
258.5 + 12.16 = 270.66

Interval is from 246.34 to 270.66. There may be differences due to rounding.

I hope this helps.

Answer 2

Thank you - that helped a lot!