Question

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) ($532.86, $840.64) C) (S513.14, $860.36) B) (S544.87, $828.63) D) (S326.21, $437.90)

Answer 1

Answer:

(513.14, 860.36) is a 98% confidence interval for the true mean

Step-by-step explanation:

We have a small sample size of n = 15, the mean balance was $\bar{x} = 686.75$ with a standard deviation of s = 256.20. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the 100$(\alpha/2)$th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a $100(1-\alpha)$% = 98% confidence interval, we have that $\alpha = 0.02$ and the confidence interval is $686.75\pm t_{0.01}(\frac{256.20}{\sqrt{15}})$ where $t_{0.01}$ is the 1st quantile of the t distribution with 14 df, i.e., $t_{0.01} = -2.6245$. Then, we have $686.75\pm (-2.6245)(\frac{256.20}{\sqrt{15}})$ and the 98% confidence interval is given by (513.1379, 860.3621)