Question

To estimate the true proportion of adult citizens who are in favor of new gun control legislations, a random sample of 125 citizens yielded 60 who were in favor. What is a lower limit of a 95% confidence interval for the proportion of citizens who are in favor of the gun control legislation. Enter your answer with two significant places after the decimal. (For example,0.1254 should be entered as 0.13)

Answer 1

Answer: 0.39

Step-by-step explanation:

Given : Sample size : $n=125$

Number of citizens in favor of new gun control legislation= 60

The proportion of citizens in favor of new gun control legislation: $p=\dfrac{60}{125}=0.48$

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The lower limit of confidence interval for population proportion is given by :-

$p-\ z_{\alpha/2}\ \sqrt{\dfrac{p(1-p)}{n}}\\\\=0.48-(1.96)\sqrt{\dfrac{(0.48)(1-0.48)}{125}}\\\\\approx0.48-0.09=0.39$

Hence,a lower limit of a 95% confidence interval for the proportion of citizens who are in favor of the gun control legislation = 0.39