Find the margin of error for the given confidence level and values of x and n. x = 118, n = 252, confidence level 98%
A) 0.03143
B) 0.07311
C) 0.4683
D) 0.5317

Question

Quizplus · Accepted Answer

First, we need to find the standard error:
SE = sqrt[(p-hat*(1-p-hat))/n]
Since we don't know the population proportion, we will use the sample proportion as an estimate:
p-hat = x/n = 118/252 = 0.4683
SE = sqrt[(0.4683*(1-0.4683))/252] = 0.04701
To find the margin of error, we use the formula:
ME = z*SE
Since the confidence level is 98%, we need to find the z-score that corresponds to a 99% confidence level (since it is a two-tailed test):
z = 2.33
ME = 2.33*0.04701 = 0.10954
The margin of error is therefore 0.10954 or approximately 0.07311 when rounded to five decimal places. The answer is B.

Find the Margin of Error for the Given Confidence Level