Provide an appropriate response.
-A local bank needs information concerning the savings account balances of its customers. A random sample of 15 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.
A) $( \$ 513.17 , \$ 860.33 )$
B) $( \$ 238.23 , \$ 326.41 )$
C) $( \$ 326.21 , \$ 437.90 )$
D) $( \$ 487.31 , \$ 563.80 )$

Question

Quizplus · Accepted Answer

To find a 98% confidence interval for the true mean, we use the formula for the confidence interval for the mean, which is $\bar{x} \pm z\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $z$ is the z-score corresponding to the confidence level. Given $\bar{x} = \$686.75$, $s = \$256.20$, and $n = 15$, and knowing that the z-score for a 98% confidence level is approximately 2.33, we calculate the margin of error and add/subtract it from the sample mean to find the confidence interval, which matches choice A.

Provide an Appropriate Response $686.75\$ 686.75$686.75 With a Standard Deviation Of $256.20\$ 256.20$256.20

Provide an Appropriate Response $\$ 686.75$ With a Standard Deviation Of $\$ 256.20$