Question

A U.S. based internet company offers an on-line proficiency course in
basic accounting. Completion of this online course satisfies the "Fundamentals of Accounting"
course requirement in many MBA programs. In the first semester 315 students have enrolled in
the course. The marketing research manager divided the country into seven regions of
approximately equal populations. The course enrollment values in each of the seven regions
are given below. The management wants to know if there is equal interest in the course across
all regions. (Hint: To answer this question you must determine the expected frequency per
region if it is equal)
a. State the null and alternative hypotheses
b. What is the test statistic?
c. Using a .05 significance level, what is the decision rule?
d. Show the test statistic and essential calculations.
e. Interpret you results

Region    1 2    3    4 5 6    7

enrollment 45 60 30 40 50 55 35

Answer 1

a.

Null Hypothesis H0: There is equal number of average interest in the course across all regions.

Alternative Hypothesis H0: At least one of the region differs in average number of interest in the course.

b.

We will use Chi Square test for Homogeneity.

The test statistic is Chi Square test statistic.

c.

Degree of freedom = number of groups - 1 = 7 - 1 = 6

Critical value of Chi Square test statistic at df = 6 and .05 significance level is 12.59

Decision Rule - Reject H0 if Chi Square test statistic , > 12.59

d.

Expected frequncy = np

where n = 315 and p = 1/7

Expected frequncy = 315 / 7 = 45

Chi Square test statistic,

= 15.56

e.

Since the calculated Chi Square test statistic is greater than the critical value, we reject null hypothesis H0 and conclude that there is significant evidence from the sample data that at least one of the region differs in true average number of interest in the course.