A survey conducted by the American Automobile Association showed that a family o

Question

A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $71.50.

a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals).

b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? (YES/NO)

Explanation / Answer

Confidence interval for population mean (Mu) is
= sample mean +/- z value *standard error of mean
= 252.45 +/- 1.96* ( 71.50/sqrt 64)
= 252.45 +/- 17.52
lower boundry is 252.45 - 17.52 = 234.93
upper boundry is 252.45 + 17.52 = 269.97
The CI is ($234.93, $269.97)

b) Since the mean reported by the AMA ($215.60) is NOT falling within the confidence interval, it can be inferred that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the AMA

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