ChiSquare Analysis Goodness of Fit Analysis A Travel Agency

Chi-Square Analysis

Goodness of Fit Analysis

A Travel Agency decided to run a contest to advertise their business. They asked a marketing research firm to determine what would be the most popular vacation for people in the immediate geographic area, to help them get the maximum interest in the travel agency and in the contest. The research firm asked a random sample of individuals: “Where would you like to take a vacation?” and got the results below. Analyze the data and make a recommendation for the travel agency. Note on this problem you are still doing the CHI, but as you move across from left to right for each row, the steps for the CHI are laid out for you. Item 32 then is the CHI value! Just that easy. Your df are number of rows minus one, (3-1), but because there is only one column, so there isn’t a chance for a score to randomly fall into another column, so we leave the column figure at just 1. So you compute df = (3-1)(1).

Vacation Type

O (observed)

O-E

(O-E)²

(O-E)²/E

Las Vegas

22

22-32

(-10) ²

100/32

Mountains

46

Seashore

28

The top row of this table has been completed

for you. Complete the other two rows and then

Use the Above Data to answer items 30 - 37

30. Total from the last column:

31. What is the Expected value: E?

32. What is the Calculated Chi Square for this analysis?

33. What are the degrees of freedom?

34. N =

35. What is the tabled value for Chi-Square?

36. Is the result significant?

37. What do you conclude?

38. If you have continuous data, and are comparing peoples scores taken before \"pre\" and after \"post\" the independent variable, you should use the statistic.

39. If you have continuous data, and are comparing a group average to the published norms for that data, you should use the statistic.

Solution