Question

Pinworm: In a random sample of 810 adults in the U.S.A., it was found that 84 of those had a pinworm infestation. You want to find the 99% confidence interval for the proportion of all U.S. adults with pinworm.

(a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places.


(b) Construct the 99% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to 3 decimal places.
< p <

(c) Based on your answer to part (b), are you 99% confident that more than 6% of all U.S. adults have pinworm?

Yes, because 0.06 is above the lower limit of the confidence interval.
No, because 0.06 is below the lower limit of the confidence interval.   
Yes, because 0.06 is below the lower limit of the confidence interval.
No, because 0.06 is above the lower limit of the confidence interval.

(d) In Sludge County, the proportion of adults with pinworm is found to be 0.16. Based on your answer to (b), does Sludge County's pinworm infestation rate appear to be above the national average?

No, because 0.16 is below the upper limit of the confidence interval.
Yes, because 0.16 is above the upper limit of the confidence interval.    
Yes, because 0.16 is below the upper limit of the confidence interval.
No, because 0.16 is above the upper limit of the confidence interval.

Answer 1

a) For i from 1 to 810, le Xi = 1 if the ith person has pinworm and 0 otherwise. These are Bernoulli(p) independent samples.

The point estimate of mean is the sample mean= = 84/810 = 0.1037

b) The estimatedstandard deviation of the mean is

The 99% confidence interval is bounded by points where is the point over which in N(0,1)

the probability would be 0.005. This point is 2.575829. Thus the confidence interval is (0.0761071, 0.1312929) on substituting these values. Rounding off,

0.076 < p < 0.131

c) Yes, because 0.06 is below the lower limit.

d) Yes, because 0.16 is above the upper limit.