Probability and Statistics
posted by Kim on .
A research psychologist believes that professional athletes tend to be more confident than the typical person. From previous testing, he knows that the scores for people in the general population are approximately normally distributed with a mean of 72 and standard deviation of 14. A random sample of 15 athletes produces a sample mean of 76.5 with a sample standard deviation of 12.5. Is there enough evidence at he alpha= 0.10 significance level to indicate that professional athletes are more confident?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
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.