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

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

Explanation:

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

`\alpha = (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(\sigma)/(√(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(\sigma)/(√(n))`

`M = 1.645(1.5)/(√(31))`

`M = 0.44`

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