Question

Provide an appropriate response. A local bank needs information concerning the checking 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. ($487.31, $563.80) ($326.21, $437.90) ($238.23, $326.41) ($513.17, $860.33)

Answer 1

Answer:

[$489.83, $883.67]

Step-by-step explanation:

The 98% confidence interval is given by the interval

$\bf [\bar x-t\frac{s}{\sqrt n}, \bar x+t\frac{s}{\sqrt n}]$

where

$\bf \bar x$=686.75 is the sample mean  

s = 256.20 is the sample standard deviation  

n = 15 is the sample size

t is the 2% critical value for the Student's t-distribution with 14 degrees of freedom (sample size -1), this is a value such that the area under the Student's t curve outside the interval [-t, t] is 2%=0.02.

We are using the t-distribution for it is the approximation to the Normal distribution for small samples (n<30).

Either by using a table or the computer, we find  

t = 2.9768

In Excel we use the function

TINV(0.01,14)

In OpenOffice Calc

TINV(0.01;14)

and our 98% confidence interval is

$\bf [686.75-2.9768*\frac{256.20}{\sqrt{15}}, 686.75+2.9768*\frac{256.20}{\sqrt{15}}]=\boxed{[\ 489.83,\ 883.67]}$