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.
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.

Answer 1

Answer:

a) The 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls is ($103.58, $401.32).

b) The population mean is inside the 95% confidence interval of the mean. So, it does not 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:

The first step to solve this problem is solving question a), that is, building the confidence interval. Then question b), if the average spent by the families is inside the confidence interval, it does not appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association. Otherwise, it does

a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls.

The sample size is 64.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 64-1 = 63$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 63 and 0.025 in the t-distribution table, we have $T = 1.9983$.

Now, we use the standard deviation of the sample. That is $s = 74.50$

Now, we multiply T and s

$M = T*s = 74.50*1.9983 = 148.87$

For the lower end of the interval, we subtract the mean by M. So 252.45 - 148.87 = $103.58.

For the upper end of the interval, we add the mean to M. So 252.45 + 148.87 = $401.32.

The 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls is ($103.58, $401.32).

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?

The population mean is inside the 95% confidence interval of the mean. So, yes, it does not appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association.