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 $77.50. 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?

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 intervalThe first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. Sodf = 64 - 1 = 6395% confidence intervalNow, 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 [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 1.9983The margin of error is:[tex]M = T\frac{s}{\sqrt{n}} = 1.9983\frac{70}{\sqrt{64}} = 17.49[/tex]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).