Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of?

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

These and many more . . .

http://answers.yahoo.com/search/search_r...

Answer 2

Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.

For large sample confidence intervals about the population proportion you have:

pHat ± z * sqrt(phat * (1- phat) / n)

where phat is the sample proportion

z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α

n is the sample size

The sample proportion, phat = 0.1112549

The sample size n = 773

The z score for a 0.9 confidence interval is the z score such that 0.05 is in each tail.

z = 1.644854

The confidence interval is:

( phat ± z * sqrt(phat * (1 - phat) / n)

( 0.09265174 , 0.1298580 )



We are 0.9 confident the true value of the population proportion lies in this interval.

Normality should be fine, large enough sample and the proportion is ~11%

c and d are up to you, I've never heard of the "very quick rule."