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 $76.50.

(a)

Develop a 95% confidence interval estimate of the mean amount spent per day in dollars by a family of four visiting Niagara Falls. (Round your answers to the nearest cent.)

$  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? Explain.

No. The overall average daily vacation expenditure of $215.60 per day is between the upper and lower limits of the confidence interval for the population mean at Niagara Falls. This suggests we cannot determine if the population mean at Niagara Falls is greater than the overall average daily vacation expenditure.

Yes. The lower limit for the confidence interval for the population mean at Niagara Falls is greater than overall average daily vacation expenditure of $215.60 per day. This suggests the population mean at Niagara Falls is greater than the overall average.    

Yes. The upper limit for the confidence interval for the population mean at Niagara Falls is less than overall average daily vacation expenditure of $215.60 per day. This suggests the population mean at Niagara Falls is less than the overall average.

No. The lower limit for the confidence interval for the population mean at Niagara Falls is greater than overall average daily vacation expenditure of $215.60 per day. This suggests we cannot determine if the population mean at Niagara Falls is greater than the overall average daily vacation expenditure.

Yes. The overall average daily vacation expenditure of $215.60 per day is between the upper and lower limits of the confidence interval for the population mean at Niagara Falls. This suggests the population mean at Niagara Falls is less than the overall average.

Answer 1

a) The formula of the confidence interval is given by,

t_{​​​​​​​/2}* (s/ n)

In the above problem,

Sample size (n) =64

Sample mean (​​​​​​​) = 252.45

Standard deviation (s) = 76.50

Confidence level = 95% = 0.95

Now, degree of freedom (df) = n-1 = 64-1 = 63

For determining ‘t_/2’, value from the standard normal distribution of our desired confidence interval, we first need to find the value of ​​​​​​​, i.e., ​​​​​​​ = 1-0.95 = 0.05

Therefore, ​​​​​​​/2 = 0.05/ 2 = 0.025

Now, we will find the value of t_{​​​​​​​/2} in the t-distribution table with respect to degree of freedom (df) = 63 and /2 = 0.025, we will get the result:

t_{​​​​​​​/2} = 1.998

Therefore, margin of error = t_{​​​​​​​/2}* (s/ n)

                                                = 1.998* (76.50/ 64) = 19.1095

Therefore, our confidence interval will become:

              ​​​​​​​ -t_{​​​​​​​/2}* (s/ n) < ​​​​​​​ < ​​​​​​​ + t_{​​​​​​​/2}* (s/ n)

Or, 252.45 – 19.1059 < < 252.45+19.1059

Or, 233.3441 < < 271.5559

b) Yes. The lower limit for the confidence interval for the population mean at Niagara Falls is greater than overall average daily vacation expenditure of $215.60 per day. This suggests the population mean at Niagara Falls is greater than the overall average.