Question

n 1987, the General Social Survey asked, "Have you ever been active in a veteran's group? " For this question, 52 people said that they definitely did out of 98 randomly selected people. What is the 95% confidence interval for the proportion of all Americans who have been active in a veteran's organization?A. (0.432, 0.629)B. (0.525, 0.535)C. (0.0448, 0.614)D. Assumptions are not met. Can not make confidence interval.

Answer 1

Answer: A. (0.432, 0.629)

Step-by-step explanation:

Let $\hat{p}$ be the sample proportion.

As peer given, we have

n= 98

$\hat{p}=\dfrac{52}{98}=\dfrac{26}{49}$

Critical value for 95% confidence interval : $z_{\alpha/2}=1.960$

Formula for confidence interval :-

$\dfrac{26}{49}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $\dfrac{26}{49}\pm (1.96)\sqrt{\dfrac{\dfrac{26}{49}(1-\dfrac{26}{49})}{98}}$

$\approx\dfrac{26}{49}\pm 0.0988092369474= (\dfrac{26}{49}+0.0988092369474,\ 0.531+0.0988092369474)\\\\=(0.431803007951,\ 0.629421481845)\\\\\approx(0.432,\ 0.629)$

Hence, the correct  95% confidence interval : A. (0.432, 0.629)