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 $70.00. 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).

Answer 1

Answer:

The 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls is ($234.96, $269.94).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 64 - 1 = 63

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 63 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 1.9983

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9983\frac{70}{\sqrt{64}} = 17.49$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 252.45 - 17.49 = $234.96.

The upper end of the interval is the sample mean added to M. So it is 252.45 + 17.49 = $269.94.

The 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls is ($234.96, $269.94).