Question

a cable company is planning to offer an improved cable package to its subscribers . Managers are trying to determine the proportion of customers who are likely to purchase the new package to within 5% of the true proportion and want to present their results in a 95 % confidence interval . Last time they offered an improved package to their customers , a 95 % confidence interval was calculated to be (0.54, 0.64) . What is the minimum sample sized needed ?

Answer 1

Solution,

Given that,

$\hat p$ = 0.54 + 0.64 / 2 = 0.59

1 - $\hat p$ = 1 - 0.59 = 0.41

margin of error = E = 0.5

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96   

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

= (1.96 / 0.5)² * 0.59 * 0.41

= 3.72

sample size = n = 4