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)

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