Question

 

A random sample of student is asked their opinions on a proposed core curriculum change. The results are as follows:

Class	Opinion
	Favoring	Opposing
Freshman	120	80
Sophomore	70	130
Junior	60	70
Senior	40	60

A)Test the hypothesis (using ? = 0.05) using the Chi-Square B)Test that opinion on the change is independent of class standing. C)Find the p-value for this test and comment on the data.

Answer 1

Steps to be followed for doing Chi Square test in Minitab 17:

1. Enter a table in Minitab with the levels of one variable as columns and the levels of the second variable as rows. Note: It does not matter which variable, X or Y, is the column and which is the row. Enter the counts of the XY combinations into the table as shown:

2. After entering the data, choose Stat > Tables > Chi-Square Test for Association...

3. In the window of Chi-Square Test for Association, select 'Summarized data in two-way table' from dropdown menu. In Columns containing table select Favouring and Opposing.

4. Click Statistics button on the window of Chi-Square Test for Association and ensure 3 options are checked as shown in screenshot below. Click Ok

4. Click Ok button on the window of Chi-Square Test for Association

Following is the screenshot of output obtained using Minitab:

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H₀?: The two variables are independent

H_a?: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is ?=0.05.

If p-value > 0.05, we do not reject null hypothesis

If p-value < 0.05, we reject null hypothesis

Decision about the null hypothesis

Since it is observed that p-value (0.000) < 0.05, it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables i.e. class and opinion are dependent, at the 0.05 significance level.