Question

Banking fees have received much attention during the recent economic recession as bankslook for ways to recover from the crisis. A sample of 31 customers paid an average fee of $11.53 permonth on their checking accounts. Assume the population standard deviation is $1.50. Calculatethe margin of error for a 90% confidence interval for the mean banking fee.

Answer 1

Answer:

The margin of error for a 90% confidence interval for the mean banking fee is of $0.44.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Sample of 31:

This means that $n = 31$

Assume the population standard deviation is $1.50.

This means that $\sigma = 1.5$

Calculate the margin of error for a 90% confidence interval for the mean banking fee.

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.645\frac{1.5}{\sqrt{31}}$

$M = 0.44$

The margin of error for a 90% confidence interval for the mean banking fee is of $0.44.