Question

Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 74 of those had a pinworm infestation. You want to find the 90% 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 90% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to 3 decimal places. < p < (two answers here)

(c) Based on your answer to part (b), are you 90% confident that more than 5% of all U.S. adults have pinworm?

No, because 0.05 is below the lower limit of the confidence interval.

No, because 0.05 is above the lower limit of the confidence interval.

Yes, because 0.05 is below the lower limit of the confidence interval.

Yes, because 0.05 is above the lower limit of the confidence interval.

(d) In Sludge County, the proportion of adults with pinworm is found to be 0.14. Based on your answer to (b), does Sludge County's pinworm infestation rate appear to be above the national average?

No, because 0.14 is below the upper limit of the confidence interval.

No, because 0.14 is above the upper limit of the confidence interval.

Yes, because 0.14 is above the upper limit of the confidence interval.

Yes, because 0.14 is below the upper limit of the confidence interval.

Answer #1

a)

Point estimate for the proportion = 74/830 = 0.089

b)

sample proportion, = 0.089

sample size, n = 830

Standard error, SE = sqrt(pcap * (1 - pcap)/n)

SE = sqrt(0.089 * (1 - 0.089)/830) = 0.0099

Given CI level is 90%, hence α = 1 - 0.9 = 0.1

α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE

ME = 1.64 * 0.0099

ME = 0.0162

CI = (pcap - z*SE, pcap + z*SE)

CI = (0.089 - 1.64 * 0.0099 , 0.089 + 1.64 * 0.0099)

CI = (0.073 , 0.105)

c)

Yes, because 0.05 is below the lower limit of the confidence
interval.

d)

Yes, because 0.14 is above the upper limit of the confidence
interval.

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