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:
t = 
Step-by-step explanation:
Given
D = M + Mrt ( subtract M from both sides )
D - M = Mrt ( divide both sides by Mr )
= t
Answer:
-1/8x-22
Step-by-step explanation:
Insert 3 in the RHS bracket
Insert - in the RHS bracket
Simplify
Answer:
6
(x−7)(x−1) this is the correct answer