MATH SOLVE

4 months ago

Q:
# 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

Accepted Solution

A:

Answer: D) 0.438 < p < 0.505Step-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* = 1.96Then, 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