Question

(1 point) (For the following question, use the methods described in lecture - do not use R prop.test(). You must use at least four decimal places for all intermediate calculations and your final answer) A poll is taken in which 373 out of 500 randomly selected voters indicated their preference for a certain candidate. (a) Find the "Large-Sample" 95% confidence interval for p. <p> (b) Find the margin of error for this 95% confidence interval for p. (c) Without doing any calculations, indicate whether the margin of error is larger or smaller or the same for an 80% confidence interval. A. larger B. smaller C. same

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = $\hat p$ = x / n = 373 / 500 = 0.7460

1 - $\hat p$ = 1 - 0.7460 = 0.2540

Z$\alpha$/2 = Z_0.025 = 1.96

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

= 1.96 ($\sqrt$((0.7460 * 0.2540) / 500)

= 0.0382

A 95% confidence interval for population proportion p is ,

$\hat p$ - E $\leq$ p $\leq$ $\hat p$ + E

0.7460 - 0.0382  $\leq$ p $\leq$ 0.7460 + 0.0382

( 0.7078 $\leq$ p $\leq$ 0.7842 )

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

= 1.96 ($\sqrt$((0.7460 * 0.2540) / 500)

= 0.0382

c) correct option is = B. smaller