Posted by melanie on .
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
10.09=0.91
zscore=1.35
p^=x/n
p^=0.18/15
p^=0.988
q^=1p^
q^=10.988
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!

Statistics: 82% Confidence Interval 
MathGuru,
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. 
Statistics: 82% Confidence Interval 
melanie,
Thank you  that helped a lot!