Question

Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A previous random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. How many citizens would need to be sampled if a 90% confidence interval was desired to estimate the true proportion to within 4%?

Answer 1

Answer:

416

Step-by-step explanation:

Given that:

A previous random sample = 4000

Sample mean = 2250

$\hat p = \dfrac{x}{n}$

$\hat p = \dfrac{2250}{4000}$

$\hat p = 0.5625$

The confidence interval level = 90%

Desired Margin of error E = 0.04

The level of significance at 90% C.I  = 1 - 0.90 = 0.10

The critical value $Z_{\alpha/2} = Z_{0.10/2}$

$\implies Z_{0.05} = 1.645$   ( From the  z tables)

$n = \hat p \times (1- \hat p ) \bigg ( \dfrac{z_{0.05}}{E} \bigg)^2$

$n = 0.5625 \times (1- 0.5625) \bigg ( \dfrac{1.645}{0.04} \bigg)^2$

$n = 0.5625 \times (0.4375) \bigg ( 41.125 \bigg)^2$

n = 416.21

Thus, the number of citizens required to be sampled is $\simeq$ 416