Construct a 90 percent confidence interval for the proportion of all kernels that would not pop. ?

Question

Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 86.

a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.

b) Check the normality assumption.

c) Try the Very Quick Rule. Does it work well here? Why, or why not? d) Why might this sample not be typical?

Answer 1

Binomial problem -- n=773, p=86/773~=.11125, mean=np=86

Standard error of proprtion, s =sqrt(p(1-p)/n) = sqrt(.11125*.88875/773) = .01131

Using the normal distribution as an approximation of the binomial:

Z value for .45 (half of .90) from a standard normal table = 1.645

The 90% confidence interval is p +/- Zs = (.11125-1.645*.01131, .11125+1.645*.01131) = (.09265,.12985)

I don't know what methods you are to use for testing normality, but the binomial is inherently normal except for small values of np and n(p-1), and these are relatively large (>= 5).

I have no idea what the 'Very Quick Rule' might be.