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!

- Statistics: 82% Confidence Interval -
**MathGuru**, Sunday, June 22, 2008 at 8:05pm
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**, Sunday, June 22, 2008 at 9:33pm
Thank you - that helped a lot!

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