Question

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.

Answer 1

Answer: D) 0.438 < p < 0.505

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

$\hat{p}$ = Sample proportion.

z* = critical value.

Given : A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.

i.e. n= 865

$\hat{p}=\dfrac{408}{865}\approx0.4717$

Two-tailed critical avlue for 95% confidence interval : z* = 1.96

Then, the 95 % confidence interval for the true proportion of all voters in the state who favor approval will be :-

$0.4717\pm (1.96)\sqrt{\dfrac{0.4717(1-0.4717)}{865}}\\\\\approx0.4717\pm 0.03327\\\\=(0.4717-0.03327,\ 0.4717+0.03327)=(0.43843,\ 0.50497)\approx(0.438,\ 0.505)$

Thus, the required 95% confidence interval : (0.438, 0.505)

Hence, the correct answer is D) 0.438 < p < 0.505