Question

A university official wishes to determine whether the degree of the instructor is related to the students’ opinion of the quality of instruction received. A sample of students’ evaluations of various instructors is selected, and the data in the table below are obtained. At a = 0.10, can the officials conclude that the degree of the instructor is related to the opinions of the students about the instructor’s effectiveness in the class?

	Degree of instructor
Opinion Rating	B.S.	M.S.	Ph.D.
Excellent Average Poor	14 16 3	9 5 12	4 7 16

a) Identify the appropriate test. Why do you consider this test to be the most appropriate for analyzing this study?

b) State the hypotheses.

Answer 1

a) We will use Chi-square test for independence of 2 variables.

The 2 variables are the degree of the instructor and the opinion rating. We want to test if the degree of the instructor is related to the opinions of the students about the instructor’s effectiveness in the class. That is we want ot test if the2 variables are independent. Chi-square test is suitable for testing the euqality of more than 2 proportions.

b) The following are the hypotheses

The tables given is the observed frequency . We find the column and row total as below

Degree of instructor
Opinion Rating	B.S.	M.S.	Ph.D.	Total
Excellent	14	9	4	27
Average	16	5	7	28
Poor	3	12	16	31
Total	33	26	27	86

Nexct we want to find the frequency expected () if the 2 variables are independent, that is the degree of the instructor is not related to the opinions of the students about the instructor’s effectiveness in the class

Let us consider the first cell where 14 have rated an instructor with B.S. degree as excellent.

Let A be the event that the degree of instructor is B.S.

B be the event that the opinion rating is excellent

The probability of the first cell is

The expected frequency () of the first cell assuming independence is

We can generalize this for other cells as

where

CT - is the column total of corresponding to the cell

RT is the row total corresponding to the cell

N = 86 is the total number of observations

Finaly we get the test statistics using

The following excel table shows the calculations

The values are

We note that the expected frequency of no cell is less than 5, as a prerequisit to applying the test.

The chi-square statistics is 19.51

The degrees of freedom is (number of rows -1) x (number of columns -1 ) = (3-1) x (3-1) = 4

The critical values for alpha = 0.1 is given by the chi-square value corresponding to the area under the right tail = 0.1 for df=4

Using chi-square tables we get the critical value as 7.779

The test statistics of 19.61 is greater than the critical value 7.779. Hence we reject the null hypothesis.

We conclude that there is sufficient evidence to support the claim that the degree of the instructor is related to the opinions of the students about the instructor’s effectiveness in the class.