Question

In a random sample of 820 adults in the U.S.A., it was found that 86 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 z_α/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
z_α/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 7% of all U.S. adults have pinworm?

No, because 0.07 is above the lower limit of the confidence interval.

Yes, because 0.07 is above the lower limit of the confidence interval.    

Yes, because 0.07 is below the lower limit of the confidence interval.

No, because 0.07 is below the lower limit of the confidence interval.

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

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

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

Yes, 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.

Answer 1

(A) we know that point estimate for the proportion of all U.S. adults with pinworm = favourable/total outcome = 86/820 = 0.105

(B) Using z distribution table for 90% confidence interval, we get z critical value = 1.64 {check z table for alpha= 1-0.90= 0.10}

(c) Margin of error =

where z = 1.64, n = 820 and

Margin of error =

(D) Confidence interval =

we have ME = 0.018 and

Confidence interval =

Confidence interval is (0.087, 0.123) or 8.7% to 12.3%

(E) Null value is 7% and we know that when the null value is within confidence interval, then we dont reject the null hypothesis. In this case, confidence interval is from 8.7% to 12%, this means null value is not included in the confidence interval. Therefore, we can say that the result is significant

option C is correct

(F) Yes, the average is looking greater than national average because the sludge county's average is outside the 90% confidence interval for the mean national value.

Option D is correct