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. Using a 98% confidence interval estimate, you wish to determine whether the average amount a family of four will spend when vacationing at Niagara Falls per day is different from the average amount for a family of four claimed by the American Automobile Association.

a) If the amount spent by a family of four while vacationing at Niagara Falls is skewed to the right, would you expect the median amount spent to be greater than or less than $252:45? Explain.

b) Is it necessary to assume a normal distribution on the population to estimate the value of the mean spending per day? Explain.

c) What is the sampling distribution of the sample statistic when you construct a 98% confidence interval? Explain in detail.

d) Construct a 98% confidence interval estimate of the population mean amount of a family of four when vacationing at Niagara Falls per day and interpret the interval.

e) Based on result in part d), does it appear that the population mean amount spent per day by families visiting Niagara Fall differs the mean amount per day for a family of four reported by the American Automobile Association? Explain.

Answer 1

a)

If the amount spent by a family of four while vacationing at Niagara Falls is skewed to the right, the median amount spent is expected to be less than $252:45.

This is because if a distribution is right-skewed, the mean is greater than the median.

b)

In order to set up a confidence interval we must estimate the sampling distribution of the mean. The sample, itself, does not provide enough information for us to do this. So, in parametric statistics, we find out information concerning shape by assuming that the sampling distribution of the mean is normal.

The Central Limit Theorem states that, given random and independent samples of N observations each, the distribution of sample means approaches normality as the size of N increases, regardless of the shape of the population distribution. i.e. No matter what distribution we start with, the distribution of sample means becomes normal as the size of the samples increases.

Hence it is necessary to assume a normal distribution on the population to estimate the value of the mean spending per day.

c)

Z distribution.

When n<=30, we use t distribution. When n>30, we use Z distribution.

d)

CI = Mean +- (z*)*Z-alpha

z* = σ/√n

Therefore, 98% confidence interval estimate of the population mean amount of a family of four when vacationing at Niagara Falls per day is:

(230.79, 274.11)

i.e. we are 98% confident the mean value is between 230.79 and 274.11

e)

Yes, because 215.6 falls outside of our calculated CI.