Question

Question: According to the Bureau of Labor Statistics, the average hourly wage in the United States was $26.84 in April 2018. To confirm this wage, a random sample of 36 hourly workers was selected during the month. The average wage for this sample was $25.35. Assume the standard deviation of wages for the country is $4.50. a. Are the results of this sample consistent with the claim made by the Bureau of Labor Statistics using a 95% confidence interval? b. What is the margin of error for this sample? My Question: According to the answer, we use a T-distribution of 2.03 for the confidence interval of 95% rather than the Z-distribution of 1.96. Why? The sample size is greater than 30 (36) and the population standard deviation is known ($4.50).

Answer 1

Here we have to test that

where

n = sample size = 36

Sample mean =

Assume the standard deviation of wages for the country is $4.50.

That means population standard deviation =

So here we use z interval.

Confidence level = c = 0.95

95% Confidence interval for population mean is

where zc is z critical value for (1+c)/2 = (1+0.95)/2 = 0.975

zc = 1.96

Margin of error (e) :

e = 1.47

Margin of errror = 1.47

95% Confidence interval for population mean is (23.88, 26.82).

Here confidence interval does not contain null value 26.84

So we reject H0.

Conclusion : There is insufficient evidence to conclude that the average hourly wage in the United States was $26.84.