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I have a few questions I'm working on for Statistics hw. Here's one of them - please help me, I'm stuck.

It is known that 2% of the population has a general injection allergy. Suppose the Health Dept. randomly selected 250 individuals from the population and there were 7 people known to have the allergy .

a. construct a 99% confidence interval for the population proportion p.

I can't figure this out because i don't have the numbers or the standard deviation. i was thinking p(1-p)/n, which gives .0089 when I find the root. then is it P(Z=.15) = P(z=14.61?) The problem is that I can't find this number on my table of z scores.

b. what is the probability that the proportion in the sample that has the injection allergy will be as great as 4%?

Is this the same as above but P(Z=.04)?

Thank you for you help!


CI99 = p + or - (2.58)(sqrt of pq/n)
...where sqrt = square root; p = x/n; q = 1 - p; and + or - 2.58 represents the 99% confidence interval using a z-table.

With your sample data:
CI99 = 7/250 + or - (2.58)[sqrt of (7/250)(243/250)/250]

Convert fractions to decimals.
I'll let you take it from there.

Perhaps someone else can help you with part b.

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