Question

In a random sample of 810 adults in the U.S.A., it was found that 76 of those had a pinworm infestation. You want to find the 99% 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 for a 99% confidence interval?

(c) What is the margin of error (E) for the 99% confidence interval?

(d) What is the lower bound of the interval?

(e) What is the upper bound of the interval?

Answer 1

Solution :

Given that,

n = 810

x = 76

A ) = x / n = 76 / 810 = 0.094

1 - = 1 - 0.094 = 0.916

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

B ) Z_/2 = Z_0.005 =2.576

C ) Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.094 * 0.916) / 810)

                           = 0.027

A 99*% confidence interval for population proportion p is ,

- E < P < + E

0.094 - 0.027 < p < 0.094 + 0.027

0.067 < p < 0.121

D ) The lower bound of the interval IS 0.067

E ) The upper bound of the interval is 0.121