Probability math

Question

There are two kinds of potato, A and B. The question is if the reason if one gets eaten by worm is because the worms prefer one of the kinds. From 200 samples of potatoes from each kind 12 of A and 17 of B gets eaten by worms. How do you determine, by calculating an appropriate confidence interval with a confidence level of 95%, whether the risk of a potato getting eaten by worm depends on which variety it is.

Answer 1

n=200, use binomial approximation to normal distribution

as there's just 2 mutualy exhaustive events, worms or no worms

A=12,pa=12/200,q=188/200, s=sqr(12(188/200)=sqr138.28

B=17,pb=17/200,q=183/200, s=sqr(17(183/200)

but usual approach is to use a pooled or common standard devation

var = pa(1/na+ 1/nb) = (12/200)(1/12 + 1/17)= about.005588

s = sqrvar = about .074

Ho:pb=pa

Ha:pb>/<pa

CI = mean+/-zs for 95% Confidence alpha=5%

= 12/200 +/- zs = ,06 +/-1.96(.074) = (-.06, .18)

pb= 17/200 = .085 is within the CI

fail to reject Ho

no greater risk using A or B

but if you did a one tail test, it's more likely you may reject Ho: pb=/<pa