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) $.5625 \pm .0129$
B) $.5625 \pm .4048$
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) $.5625 \pm .0129$
B) $.5625 \pm .4048$
C) $.4375 \pm .4048$
D) $.4375 \pm .0129$

Quizplus · Accepted Answer

The sample proportion is 2250/4000 = 0.5625. The standard error for the proportion is calculated as $\sqrt{(0.5625 \times 0.4375) / 4000}$. Using a Z-score of 1.645 for a 90% confidence interval, the margin of error is approximately 0.0129. Thus, the confidence interval is $0.5625 \pm 0.0129$.

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

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