A random sample of 16 ATM transactions at the Last National Bank of Flat Rock revealed a mean transaction time of 2.8 minutes with a standard deviation of 1.2 minutes. The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is:
A)± 0.639
B)± 0.588
C)± 0.300
D)± 2.131

Question

Quizplus · Accepted Answer

The width of the 95% confidence interval can be calculated using the formula for the margin of error: $ME = z \times \frac{\sigma}{\sqrt{n}}$, where $z$ is the z-score corresponding to the desired confidence level (for 95%, $z = 1.96$), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the given values: $ME = 1.96 \times \frac{1.2}{\sqrt{16}} = 1.96 \times 0.3 = 0.588$. The width of the confidence interval is twice the margin of error, so $2 \times ME = 2 \times 0.588 = 1.176$. However, the question asks for the width in the form of ± value, so the correct answer is ± 0.588.

A Random Sample of 16 ATM Transactions at the Last