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.

THIS IS THE COMPLETE QUESTION BELOW;

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

Margin error =210.8

Given:

standard deviation of 425

sample mean x=1520 dollars.

random sample n= 22

From the question We need to to calculate the margin of error for the confidence interval for the population mean .

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

answer: 457.14

step-by-step explanation:

205 - 45 = 160

160 / .35 = 457.142857

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step-by-step explanation: