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A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.
Construct the 95 % confidence interval for the true proportion of all voters in the state who favor approval.

A) 0.435 < p < 0.508

B) 0.444 < p < 0.500

C) 0.471 < p < 0.472

D) 0.438 < p < 0.505

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JeanaShupp

Answer: D) 0.438 < p < 0.505

Step-by-step explanation:

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

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size

[tex]\hat{p}[/tex] = 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

[tex]\hat{p}=\dfrac{408}{865}\approx0.4717[/tex]

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 :-

[tex]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)[/tex]

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

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

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