Question

A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $279.29. Find a 99% confidence interval for the true mean checking account balance for local customers.

Answer 1

Answer:

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 13

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 3.0123

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.0123\frac{279.29}{\sqrt{14}} = 224.85$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 664.14 - 224.85 = $439.29

The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).