Question

Pinworm: In a random sample of 830 adults in the U.S.A., it was found that 70 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?


(b) What is the critical value of z (denoted z_α/2) for a 90% confidence interval?
z_α/2 =  

(c) What is the margin of error (E) for a 90% confidence interval?
E =  

(d) Construct the 90% confidence interval for the proportion of all U.S. adults with pinworm.
< p <   

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

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

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

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

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

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

No, 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.     

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

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

Answer 1