Posted by **Anonymous** on Sunday, April 1, 2012 at 4:02am.

A football coach claims that the average weight of all the opposing teams’ members is 225 pounds. For a test of the claim, a sample of 50 players is taken from all the opposing teams. The mean is found to be 230 pounds, and the standard deviation is 15 pounds. At a = 0.01, test the coach’s claim. Find the P-value and make the decision.

- statistics -
**PsyDAG**, Sunday, April 1, 2012 at 1:04pm
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. Is it larger or smaller than .01?

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