A random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a
90% confidence interval.
A) $.5625 \pm .4048$
B) $.5625 \pm .0129$
C) $.4375 \pm .4048$
D) $.4375 \pm .0129$

Question

A random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a
90% confidence interval. 
A) $.5625 \pm .4048$
B) $.5625 \pm .0129$
C) $.4375 \pm .4048$
D) $.4375 \pm .0129$

Quizplus · Accepted Answer

The proportion of the sample in favor of gun control legislation is $2250 / 4000 = 0.5625$. To calculate the margin of error for a 90% confidence interval, we use the formula for the standard error of a proportion, which is $\sqrt{p(1-p)/n}$, where $p$ is the sample proportion and $n$ is the sample size. Then, we find the z-score for a 90% confidence interval, which is approximately 1.645. The margin of error is $1.645 \times \sqrt{0.5625(1-0.5625)/4000} \approx 0.0129$. Therefore, the 90% confidence interval is $0.5625 \pm 0.0129$.

A Random Sample of 4000 U .5625±.4048.5625 \pm .4048.5625±.4048 B) .5625±.0129.5625 \pm .0129.5625±.0129

A Random Sample of 4000 U $.5625 \pm .4048$ B) $.5625 \pm .0129$