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Supplementary Exercises 385 A survey conducted by the American Automobile Association (AAA) showed that a fam- ily 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 45. day and a sample standard deviation of $74.50 Develop a 95% confidence interval estimate of the mean amount spent per day by a a. family of four visiting Niagara Falls. 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 b. by the American Automobile Association? Explain.
math   Statistics-And-Probability   
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a. The sample of n = 64 families of four vacationing at Niagara Falls resulted in a sample mean of x 252.45 per day and sample standard deviation s 74.5 The confidence interval estimate of the mean amount spent per day be a family of four visiting Niagara Falls would be Using NORM.S.INV function we can get the z value Now 95% confidence level mean middle 95% area 95% Area to the left of z-2.5% + 95%-97.5% -NORM.S.INV(0.975) gives z -1.959963985 ~ 1.9600 So, 95% confidence interval would be 74.5 V64 74.5 74.5 V64 74.5 + 1.965 252.45-1.96 252.45 1.96 252.45 -1.96 -,252.4 (252.45 - 1.96 9.3125, 252.45+1.96 9.3125) (252.45 - 18.25216461,252.4518.25216461) (234.1978354, 270.7021646) So, we 95% confidence, the mean amount spent per day be a family of four visiting Niagara Falls would be between $234.20 and $270.70 b. According to survey conducted by the American Automobile Association family of four spends an average of $215.60 per day while on vacation, which does not lie in an interval $234.20 and $270.70, hence it appears mean amount spent per day be a family of four visiting Niagara Falls differs from the mean reported by the American Automobile Association

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