A sample statistic and margin of error are given. Find the confidence interval likely to contain the population parameterof interest and answer the question.
-The CEO of a company claims that $90 \%$ of its employees have very high job satisfaction. A poll of the company's employees revealed that 87 percent had very high job satisfaction with a margin of error of $4.6$ percentage points. Can we conclude that the CEO was lying?
A) $82.4 \%$ to $87 \%$; yes
B) $87 \%$ to $91.6 \%$; no
C) $82.4 \%$ to $91.6 \%$; no
D) $84.7 \%$ to $89.3 \%$; yes

Question

Quizplus · Accepted Answer

The confidence interval is calculated by adding and subtracting the margin of error from the sample statistic. Therefore, the confidence interval is $87\% - 4.6\% = 82.4\%$ to $87\% + 4.6\% = 91.6\%$. Since the CEO's claim of $90\%$ falls within this interval, we cannot conclude that the CEO was lying.

A Sample Statistic and Margin of Error Are Given 90%90 \%90%

A Sample Statistic and Margin of Error Are Given $90 \%$