Question

1. A poll found that out of 600 people polled, 72% of them owned and regularly used data on their smartphone.

a. Determine a 99% CI.

b. Determine the margin of error for the CI in part (a). Round to three decimal places.

c. For a 99% CI and a margin of error of 0.01, determine the necessary sample size for estimating ?.

Answer 1

Solution :

Given that,

n = 600

Point estimate = sample proportion = = 0.72

1 - = 0.28

a)

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 * (((0.72 * 0.28) / 600)

= 0.047

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.72 - 0.047 < p < 0.72 + 0.047

0.673 < p < 0.767

( 0.673 , 0.767 )

b)

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

= 2.576 * (((0.72 * 0.28) / 600)

= 0.047

c)

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576 / 0.01)² * 0.72 * 0.28

= 13377.72

sample size = 13378