Question

As a part of its annual review of accounts, a brokerage firm selects a random sample of 15 customers. The average account value for these customers was $32,000 with a standard deviation of $8,200.

Determine the 95% confidence interval for the account value.

Answer 1

µ = 32000

s = 8200

n = 15

T score for 95% confidence interval = t_0.025,14 = 2.145

confidence interval $=(\bar{X} \pm t_{0.025,14}*\frac{s}{\sqrt{n}})$

                               $=(32000 \pm 2.145*\frac{8200}{\sqrt{15}})$

                               $=(32000 \pm 4541.46)$

                               = (27458.54 , 36541.46)