Question

1. A survey conducted by the American Automobile Association (AAA) showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 28 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $70.54 a.

a. Develop a 98% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls.

b. From your results, explain why or why not the average amount spend quoted by the AAA will lie within or outside the confidence interv

1. For studies with the following characteristics, determine which of the following methods would be most appropriate when calculating the confidence interval estimates for the population. (ie using z, t or a more advanced technique. Write out the formula if it’s a z or t test.
   1. σ is unknown; n = 14; the population is normally distributed.
   2. σ is known; n=57; the distribution is approximately normal.   
   3. σ is unknown; n = 85; distribution is not normally distributed.

Answer 1

1) a) DF = 28 - 1 = 27

At 98% confidence level, the critical value is t^* = 2.473

The 98% confidence interval is

$\bar x$ +/- t^* * s/$\sqrt n$

= 252.45 +/- 2.473 * 70.54/$\sqrt {28}$

= 252.45 +/- 32.967

= 219.483, 285.417

b) The interval does not contain 215.60, so the average amount spend quoted by the AAA will lie outside the confidence interval.

2) a) We will use t-test.

- s/n

b) We will use z-test

urlo

c) We will use z-test.

z = E-u s/n