In: Statistics and Probability
Pinworm: In a random sample of 820 adults in the U.S.A., it was found that 76 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) What is the critical value of z (denoted
zα/2) for a 90% confidence interval?
Use the value from the table or, if using software, round
to 2 decimal places.
zα/2 =
(c) What is the margin of error (E) for a 90% confidence
interval? Round your answer to 3 decimal
places.
E =
(d) Construct the 90% confidence interval for the proportion of all
U.S. adults with pinworm. Round your answers to 3 decimal
places.
<p<
(e) Based on your answer to part (d), are you 90% confident that
more than 6% of all U.S. adults have pinworm?
No, 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.
Yes, because 0.06 is above the lower limit of the confidence interval.
Yes, because 0.06 is below the lower limit of the confidence interval.
(f) In Sludge County, the proportion of adults with pinworm is
found to be 0.15. Based on your answer to (d), does Sludge County's
pinworm infestation rate appear to be greater than the national
average?
No, because 0.15 is above the upper limit of the confidence interval.
Yes, because 0.15 is above the upper limit of the confidence interval.
No, because 0.15 is below the upper limit of the confidence interval.
Yes, because 0.15 is below the upper limit of the confidence interval.