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 $75.50.

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?
Select Yes or No

Answer 1

a) we have sample size n = 64

Sample mean Xbar = 252.45

Sample standard deviation s = 75.50

The 95% confidence interval for population mean is

xbar - E < < xbar + E

Where E = t​​​​​​a/2*( s/√n)

For a = 0.05 , degrees of freedom = n -1 = 63

t​​​​​a/2,n-1 = t​​​​​​0.025, 63 = 1.998

E = 1.998*(75.50/√64) = 18.86

252.45 - 18.86 < < 252.45 + 18.86

233.59 <   < 271.31

So the 95% confidence interval is ( 233.59 , 271.31)

b) 215.60 does not involve in the confidence interval

So answer is Yes