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

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

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)((s)/(√(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)((256.20)/(√(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)((256.20)/(√(15)))` and the 98% confidence interval is given by (513.1379, 860.3621)