# A group of 100 college student

A group of 100 college students and 100 college graduates wasasked to reply to the question, “In general, do you enjoy reading acollege textbook?” The following are the frequency counts for theirresponses on the 3-option item.

 Usually Sometimes Never Students 30 25 45 Graduates 50 20 30

The critical value that you would use for an alpha level of .10in the problem comparing college students’ and graduates’ enjoymentof reading textbooks is?

Solution:

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

Null hypothesis: H0: A frequency of reading a collegetextbook is independent of whether person is student orgraduate.

Alternative hypothesis: Ha: A frequency of reading acollege textbook is not independent of whether person is student orgraduate.

We are given level of significance = α = 0.10

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expectedfrequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 1*2 = 2

α = 0.10

Critical value = 4.60517

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

 Observed Frequencies Column variable Row variable Usually Sometimes Never Total Students 30 25 45 100 Graduates 50 20 30 100 Total 80 45 75 200
 Expected Frequencies Column variable Row variable Usually Sometimes Never Total Students 40 22.5 37.5 100 Graduates 40 22.5 37.5 100 Total 80 45 75 200
 (O – E) -10 2.5 7.5 10 -2.5 -7.5
 (O – E)^2/E 2.5 0.277778 1.5 2.5 0.277778 1.5

Chi square = ∑[(O – E)^2/E] = 8.555556

P-value = 0.013873

(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 a frequency ofreading a college textbook is not independent of whether person isstudent or graduate.

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