# 82% Confidence Interval

#### ridley1013

##### New Member
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've got for the 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!

#### bbkaran

##### New Member
1) calculate standard error of the sample mean SEM = sigma/sqrt(n) = 9.011
2) calculate z-score. Your value of 1.35 for 82% looks correct.
3) now multiply the values obtained in steps (1) and (2) above.
This is the limiting value.
4) Add (subtract) the value obtained in step (3) to the mean value 258.5 to get upper (lower) confidence limits.

See: http://www.global-i-tech.co.uk/205.html

where a numerical example is worked out and all the 4 steps given above are
explained in detail.

Hope this helps

bbkaran
www.global-i-tech.co.uk/9.html

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've got for the 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!

#### ridley1013

##### New Member
1) calculate standard error of the sample mean SEM = sigma/sqrt(n) = 9.011
2) calculate z-score. Your value of 1.35 for 82% looks correct.
3) now multiply the values obtained in steps (1) and (2) above.
This is the limiting value.
4) Add (subtract) the value obtained in step (3) to the mean value 258.5 to get upper (lower) confidence limits.

See: http://www.global-i-tech.co.uk/205.html

where a numerical example is worked out and all the 4 steps given above are
explained in detail.

Hope this helps

bbkaran
www.global-i-tech.co.uk/9.html
Thanks - that was very helpful & I can see now I was overthinking it before...