Suppose the mean income of 35-year-olds in the U.S. is $25,000. A random sample of 150 35-year-olds in California results in a sample mean income of $26,600 and a sample standard deviation of $3800. At the 5% significance level, can we conclude that the true mean income of 35-year-old Californians is greater than that of the nation, in general?

Question

Suppose the mean income of 35-year-olds in the U.S. is $25,000. A random sample of 150 35-year-olds in California results in a sample mean income of $26,600 and a sample standard deviation of $3800. At the 5% significance level, can we conclude that the true mean income of 35-year-old Californians is greater than that of the nation, in general?

Answer 1

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.