A survey conducted by the American Automobile Association showed that a family of four spends an averageof $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Fallsresulted in a sample mean of $252.45 per day and a sample standard deviation of $73.50.a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visitingNiagara Falls (to 2 decimals)b. Based on the confidence interval from part (a), does it appear that the population mean amount spent perday by families visiting Niagara Falls differs from the mean reported by the American Automobile Association?Select

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Asked Nov 16, 2019
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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 $73.50.
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
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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 $73.50. 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

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Expert Answer

Step 1

a,

Calculation of confidence interval:

Here, the sample mean, x-bar is $252.45.

Population mean, µ is 215.60.

Population standard deviation, s = $73.50.

Sample size, n is 64.

The 95% confidenc...

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X-Z0.025 J+Z0.025 Vn =0.95; Vn n where Z025 P(z > zo025)= 0.025;P(z<-Zo025= 0.025 P(-Z0025 <z<Z0025)= 0.95 Thus, P(z Z0025)= 0.975 Using EXCEL formula =NORM.S.INV(0.975),zo975 1.96. Calculation 73.50 73.50 -,252.45 1.96 64 CI 252.45 1.96 64 = (252.45 -18.01, 252.45 18.01) (234.44,270.46). Lower limit 234.44 dollars; Upper limit 270.46 dollars.

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