Question

Recent revenue shortfalls in a Midwestern state led to a reduction in the state budget for higher education. To offset the reduction, the largest state university proposed a 25% tuition increase. It was determined that such an increase was needed to simply compensate for the lost support from the state. Random samples of 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors from the university were asked whether they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university’s budget at current levels. The results are given in the following table.

Is this a test of independence or not? Conduct a chi-squared test for this situation. Please report the expected counts as a matrix Make the 4 steps cleared as outlined on page 325 of your textbook.

	Year
Strongly Opposed	Freshmen	Sophomore	Junior	Senior
Yes	39	36	29	18
No	11	14	21	32

Answer 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: The year of student and whether they are strongly opposed are independent.

Alternative hypothesis: H_a: The year of student and whether they are strongly opposed are not independent.

We assume level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 4

Degrees of freedom = df = (r – 1)*(c – 1) = 1*3 = 3

α = 0.05

Critical value = 7.814727764

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Year
Strongly Opposed	Freshmen	Sophomore	Junior	Senior	Total
Yes	39	36	29	18	122
No	11	14	21	32	78
Total	50	50	50	50	200

Expected Frequencies
	Year
Strongly Opposed	Freshmen	Sophomore	Junior	Senior	Total
Yes	30.5	30.5	30.5	30.5	122
No	19.5	19.5	19.5	19.5	78
Total	50	50	50	50	200

Calculations
(O - E)
8.5	5.5	-1.5	-12.5
-8.5	-5.5	1.5	12.5

(O - E)^2/E
2.368852	0.991803	0.07377	5.122951
3.705128	1.551282	0.115385	8.012821

Chi square = ∑[(O – E)^2/E] = 21.94199243

P-value = 0.000067

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the year of student and whether they are strongly opposed are not independent.