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.

The confidence level is 0.98 and the significance is and and the critical value using the table is:

And replacing we got:

Step-by-step explanation:

For this case we have the following info given:

represent the population deviation

the sample size

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:

The confidence level is 0.98 and the significance is and and the critical value using the table is:

And replacing we got:

i think is g(1)=-1.

hope that i can