Question

Problem Set 1: Chi Square Test of Goodness of Fit Research Scenario: A political psychologist is curious about the effects of a town hall meeting on people’s intentions to support a state proposition that would legalize gambling. He interviews people as they leave and asks them whether their opinion about the proposition has changed as a result of the meeting. He records these frequencies in the table below. Using this table, enter the data into a new SPSS data file and run a Chi Square Test of Goodness of Fit to test whether the frequencies are equal across the categories. Create a bar chart to show the relationship between the variables.

Less likely to support	No change	More likely to support
25	12	9

1. Paste SPSS output. (7 pts)

2. Write an APA-style Results section based on your analysis. Include your bar chart as an APA-style figure as demonstrated in the APA writing presentation. (Results = 8 pts; Graph = 5 pts)

Answer 1

A political psychologist is curious about the effects of a town hall meeting on people’s intentions to support a state proposition that would legalize gambling. He interviews people as they leave and asks them whether their opinion about the proposition has changed as a result of the meeting. He records these frequencies in the table below.

Here we have given three categories less likely to support, no change and more likely to support.

Hypothesis for the test is,

H : p1 = p2 = p3  

H1 : Atleast one population proportion differ.

Assume alpha = level of significance = 0.05

We have given observed frequencies. Now we need to find expected frequencies.

The formula for expected frequencies is,

E = N * p

where p is probability.

Assume three categories are equally likely.

So probability for each category will be 1/3

N = sample size = 46

E1 = E2 = E3 = (1/3)*46 = 15.33

The test statistic is,

O	E	(O-E)^2	(O-E)^2/E
25	15.33333	93.44444	6.094203
12	15.33333	11.11111	0.724638
9	15.33333	40.11111	2.615942
46			9.434783

Test statistic = 9.43

Now we have to find P-value for taking decision.

P-value we can find in excel.

syntax :

=CHIDIST(x, deg_freedom)

where x is test statistic.

deg_freedom = k-1 = 3-1 = 2

P-value = 0.0089

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : Atleast one population proportion differs.

Bar chart :