Question

(2) To obtain an estimate of the proportion p, of New York City residents who fol that the quality of lile has become worse in the past few years, a telephonc poll by Time/CNN on August 2-5, 2015, revealed that 773 out of 1498 residents said that life has become worse. (a) Find a 95% confidence interval for p. (b) We wish to update this survey. Ilow large a sample is required to be 95% confident that the maximum error of the estimate of p is 0.01.

Answer 1

Solution :

Given that,

a) n = 1498

x = 773

Point estimate = sample proportion = $\hat p$ = x / n = 773 / 1498 = 0.516

1 - $\hat p$ = 1 - 0.516 = 0.484

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.960}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.516 * 0.484) / 1498)

= 0.025

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E

= 0.516 ± 0.025

= ( 0.491, 0.541 )

b) margin of error = E = 0.01

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.01)² * 0.516 * 0.484

= 9594.16

sample size = n = 9595