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 $74.50.

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?

Explanation / Answer

Confidence interval = x ± z* * standard error

x = $252.45

s = $74.50

n = 64

Standard error = s/n = 74.50/8 = 9.3125

z* is a appropriate z*-value from the standard normal distribution

= 0.05, 1-/2 = 1- 0.05/2 = 1- 0.025 = 0.975.

the z value proportional to 0.975 is 1.96

95% confidence interval = 252.45 ± (1.96*9.3125)

= (270.7025, 234.1975)

The mean reported by the American Automobile Association differs from the confidence interval estimate. We can be 95% confident that the population mean amount spend per day by families visiting Niagara Falls lies between $270.7025 and $234.1975.

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