Confidence intervals how to find?

[semidevil]

so if I want to find the 90% confidence interval..how do I do it?

all I know is that given 220 salads, 179 were contaminated.

i'm asked to find the 90% confidence interval, the true proportion of the contimatned salads.

so the formula is 100(1- a)% confidnece =[(y - z(a/2) * sigma/root(n)), ((y + z(a/2) * sigma/root(n)].

so my a is .9 right? since I am looking for 90%? so do I just do Z(.9/2) and look at the table?

what about n and sigma? n is 220, and what is sigma? what about the y's

 

[juvenal]

Are you familiar with the binomial distribution? I think that's what you want to use to get your sigma (= standard deviation).

 

[xanthym]

so if I want to find the 90% confidence interval..how do I do it?

all I know is that given 220 salads, 179 were contaminated.

i'm asked to find the 90% confidence interval, the true proportion of the contimatned salads.

so the formula is 100(1- a)% confidnece =[(y - z(a/2) * sigma/root(n)), ((y + z(a/2) * sigma/root(n)].

so my a is .9 right? since I am looking for 90%? so do I just do Z(.9/2) and look at the table?

what about n and sigma? n is 220, and what is sigma? what about the y's

Consider the sample of 220 salads to be 220 independent events having the Binomial Distribution. The proportion "p" of contaminated salads will then be Binomially Distributed:
{Observed Proportion} = p = (179/220) = (0.8136)
{Estimated Proportion Std Dev} = sqrt{p(1 - p)/N} = sqrt{(0.8136)(1 - 0.8136)/220) = (0.02626)

Because sample size is large, the Binomial Distr of "p" is approximated by the Normal Distr of "p" having the same Mean and Std Dev. For a 2-Tailed 90% (Normal Distr) Confidence Interval on the Population Proportion μ:
Prob{(-1.645) < Z < (+1.645)} = 0.90
Prob{(-1.645) < {(0.8136) - μ}/(0.02626) < (+1.645)} = 0.90
Prob{(0.8568) > μ > (0.7704)} = 0.90

90% Confidence Interval for Population Proportion μ is (0.7704, 0.8568)


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[LMLuis06]

how did you go from Prob{(-1.645) < {(0.8136) - μ}/(0.02626) < (+1.645)} = 0.90 to Prob{(0.8568) > μ > (0.7704)} = 0.90 ?

 

[LMLuis06]

How can i work on this one? I have been stuck for 2 hours: BRCA 1 is a gene that has been linked to breast cancer. Researchers used DNA analysis to search for BRCA 1 mutations in 169 women with family histories of breast cancer. Of the 169 women tested, 27 has BRCA 1 mutations. Let p denote the probability that a woman with a family history of breast cancer will have a BRCA 1 mutation. Find a 95% confidence interval for p.