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November 26, 2014

Homework Help: Statistics: 82% Confidence Interval

Posted by melanie on Sunday, June 22, 2008 at 4:54pm.

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!

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