Question

The weekly salaries of sociologists in the United States are normally distributed and have a known population standard deviation of 425 dollars and an unknown population mean. A random sample of 22 sociologists is taken and gives a sample mean of 1520 dollars.

Find the margin of error for the confidence interval for the population mean with a 98% confidence level.
z0.10 z0.05 z0.025 z0.01 z0.005
1.282 1.645 1.960 2.326 2.576
You may use a calculator or the common z values above. Round the final answer to two decimal places.

Answer 1

`ME =z_(\alpha/2) (\sigma)/(√(n))`

The confidence level is 0.98 and the significance is
`\alpha=1-0.98 =0.02` and
`\alpha/2 =0.01` and the critical value using the table is:

`z_(\alpha/2)= 2.326`

And replacing we got:

`ME=2.326 (425)/(√(22))= 210.760\ approx 210.76`

Explanation:

For this case we have the following info given:

`\sigma = 425` represent the population deviation

`n =22` the sample size

`\bar X =1520` represent the sample mean

We want to find the margin of error for the confidence interval for the population mean and we know that is given by:

`ME =z_(\alpha/2) (\sigma)/(√(n))`

The confidence level is 0.98 and the significance is
`\alpha=1-0.98 =0.02` and
`\alpha/2 =0.01` and the critical value using the table is:

`z_(\alpha/2)= 2.326`

And replacing we got:

`ME=2.326 (425)/(√(22))= 210.760\ approx 210.76`