A market researcher is interested in studying the mean daily cost of meals and l

Question

A market researcher is interested in studying the mean daily cost of meals and lodging for families planning to vacation in Hawaii. He collects data from a random sample of 45 families and leans that the mean daily cost of meals and lodging is $657 with a standard deviation of $40.

a) Using the information above, construct the 99% confidence interval for the mean cost of daily lodging and meals for a family of 4 to vacation in Hawaii.

b) A tourist agency in Hawaii claims that the mean cost of meals and lodging for a family of 4 traveling in Hawaii is at most $650. At a 5% level of significance, do you have enough evidence to reject the tourist agency’s claim?

Explanation / Answer

a) ME = z* s/sqrt(n) = 2.576(40)/sqrt(45) = 15 Interval limits are 657 - 15 and 657 + 15 so interval is (642, 672) b) z = (657-650)/[40/sqrt(45)] = 1.17, so P=value is 0.12 > 0.05 There is no evidence that the mean cost is greater than $650

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