Math
posted by Alison .
Let x be a random variable representing the monthly cost of joining a gym. We may assume that x has a normal distribution and that the population standard deviation is $5.20. A fitness magazine advertises that the mean monthly cost of joining a gym is $35. You work for a consumer advocacy group and are asked to test this claim. You find that a random sample of 40 health club monthly costs has a mean of $37.30. If you assume that the population mean is $35, find the standardized sample test statistic.

Are you talking about a Z score for the difference between means?
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/probability related to the Z score.
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