Question

A sample of 40 observations is selected from a normal population where the population standard deviation is 25. The sample mean is 75.

a. Determine the standard error of the mean. (Round the final answer to 3 decimal places.)

The standard error of the mean is  .

b. Determine the 90% confidence interval for the population mean. (Round the z-value to 2 decimal places. Round the final answers to 3 decimal places.)

The 90% confidence interval for the population mean is between  and  .

c. If you wanted a wider interval, would you increase or decrease the confidence level?

Answer 1

Solution :

Given that mean x-bar = 75 , standard deviation σ = 25 , n = 40

a. => standard error of the mean is 3.953

=> standard error of the mean = standard deviation/sqrt(n)

= 25/sqrt(40)

= 3.9528

= 3.953 (rounded)

b. => The 90% confidence interval for the population mean is between 68.478 and 81.522

=> for 90% confidence interval , Z = 1.65

=> The 90% confidence interval for the population mean is

=> x-bar +/- Z*σ/sqrt(n)

=> 75 +/- 1.65*25/sqrt(40)

=> (68.478 , 81.522) (rounded)

c.
=> If increase the confidence level , we get a wider interval