A local bank needs information concerning the savings account balances of its customers. A random sample of15 accounts was checked. The mean balance was $686.75 with a standard deviation of $256.20. Find a 98%confidence interval for the true mean. Assume that the account balances are normally distributed. Round to thenearest cent.
A) ($513.17, $860.33)
B) ($238.23, $326.41)
C) ($326.21, $437.90)
D) ($487.31, $563.80)

Question

Quizplus · Accepted Answer

Using the formula for a confidence interval for the mean, which is $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size. For a 98% confidence level, the z-score is approximately 2.33. Plugging in the given values: $686.75 \pm 2.33 \times \frac{256.20}{\sqrt{15}}$ gives an interval of approximately ($513.17, $860.33).

A Local Bank Needs Information Concerning the Savings Account Balances