Statistics
posted by Erinn on .
Suppose the mean income of 35yearolds in the U.S. is $25,000. A random sample of 150 35yearolds 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 35yearold Californians is greater than that of the nation, in general?

Z = (mean1  mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n1)
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.