posted by marissa .
it is claimed that the average age of employees in a large company is 50. a sample of 36 employees is taken and the average age was 49. the standard deviation of the population was 5. use a level of significance of 0.5. what do you think of the claim?
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.
Then tell what you think.