Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. $$95 \% \text { confidence; } n = 2388 , x = 1672$$
A) 0.0206
B) 0.0184
C) 0.0156
D) 0.0248

Question

Quizplus · Accepted Answer

First, we need to find the sample proportion: 
$$\hat{p}=\frac{x}{n}=\frac{1672}{2388}=0.7003$$
Next, we need to find the standard error: 
$$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.7003(1-0.7003)}{2388}}=0.0132$$
To find the margin of error, we need to multiply the standard error by the appropriate z-score for the given confidence level. For 95% confidence, the z-score is 1.96. 
$$E=1.96	imes SE=1.96	imes 0.0132=0.0259\approx 0.026$$
Therefore, the margin of error is approximately 0.026. The closest answer choice is B, which is 0.0184.

Assume That a Sample Is Used to Estimate a Population 95% confidence; n=2388,x=167295 \% \text { confidence; } n = 2388 , x = 167295% confidence; n=2388,x=1672

Assume That a Sample Is Used to Estimate a Population $95 \% \text { confidence; } n = 2388 , x = 1672$