Question

Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample.

A 99% confidence interval for p given that p^=0.7 (this is p hat) and n=110.

Round your answer for the point estimate to two decimal places, and your answers for the margin of error and the confidence interval to three decimal places.

Answer 1

Solution :

Given that,

n = 110

Point Estimate () = 0.70

1 - = 1 - 0.70 = 0.30

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * [(0.70 * 0.30) / 110]

E = 0.113

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.70 - 0.113 < p < 0.70 + 0.113

0.587 < p < 0.813

(0.587, 0.813)