A survey conducted by the American Automobile Association (AAA) showed that a family of four spends an average of per day while on vacation. Suppose a sample of families of four vacationing at Niagara Falls resulted in a sample mean of per day and a sample standard deviation of . a. Develop a confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to decimals). $ to $ 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

Comparing the confidence interval value and the mean value reported by the American Automobile Association ;it can be concluded that the mean reported by the American Automobile Association differ from the mean spent at Niagara Fall as 215.60 falls below the confidence interval.

swapnalimalwadeVT swapnalimalwadeVT · Answer 2 · 2022-04-12T11:34:15+02:00

The confidence interval is (233.8436 ; 271.0564).

We have given that the,

Sample mean, x bar = 252.45

Sample standard deviation, s = 74.50

Sample size, n = 64

α = 0.05

What is the confidence interval ?

The confidance interval is Mean ± margin of error.

[tex]Margin of Error = t\alpha /2 \times s/\sqrt{n}[/tex]

[tex]df = n - 1 \\df= 64 - 1 \\df= 63[/tex]

[tex]t(\alpha/2, df = 63) = 1.998[/tex]

[tex]Margin of Error = 1.998 \times 74.50/\sqrt64[/tex]

Margin of Error = 18.6064

The confidence interval is given by,

252.45 ± 18.6064

Therefore we get the confidence interval is (233.8436 ; 271.0564)

Comparing the confidence interval value and the mean value reported by the American Automobile Association ;it can be concluded that the mean reported by the American Automobile Association differ from the mean spent at Niagara Fall as 215.60 falls below the confidence interval.

To learn more about the confidence interval visit:

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