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 95% 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 95% 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 95% 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.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.Yes, because 0.05 is below the lower limit of the confidence interval.

(d) In Sludge County, the proportion of adults with pinworm is found to be 0.13. Based on your answer to (b), does Sludge County's pinworm infestation rate appear to be above the national average?

Yes, because 0.13 is below the upper limit of the confidence interval.No, because 0.13 is below the upper limit of the confidence interval.    Yes, because 0.13 is above the upper limit of the confidence interval.No, because 0.13 is above the upper limit of the confidence interval.

Answer 1

Solution :

Given that,

n = 830

x = 70

Point estimate = sample proportion = = x / n = 70/830=0.084

1 -   = 1- 0.084 =0.916

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z / 2    * ((( * (1 - )) / n)

= 1.96 (((0.084*0.916) / 830)

= 0.019

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.084-0.019 < p < 0.084+0.019

0.065< p < 0.103

The 95% confidence interval for the population proportion p is : 0.065< p < 0.103