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The manager of a vacation resort has claimed that, on average, a guest spends at least $1600 at the resort during a one week stay, including meals and entertainment. A member of the accounting staff does not believe the amount is that high. They authorize you to settle their dispute. You take a sample of 16 guests that had stayed at the resort over the last several weeks. Tracking the spending of these guests at the resort you determine that the guests spent a mean of $1478. Assume that the population standard deviation is $160. Test the accountant's claim that the mean is actually less than $1600 versus the manager's claim that the guests spend a mean of $1600 or more Show all problem steps. You may assume alpha = .05 .

Answer 1

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.

Answer 2

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