Answers: Chi-Square

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1. In real estate, location is very important. Is it as important in politics? We can ask that by using Chi-Square analysis to look at the relationship of presidential winner and state location. There are 26 states east of the Mississippi River and 24 west of it. If we do a cross-tabulation of the location of states and presidential winner in the 2000 election, we get the following tables.

WINNER * WHERE Crosstabulation
Count	WHERE::
WINNER:	east	west	Total
bush	13	17	30
gore	13	7	20
Total	26	24	50

Chi-Square Test

	Value	df	Asymp. Sig. (2-sided)
Pearson Chi-Square	2.257	1	.133
N of Valid Cases	50

Computed only for a 2x2 table
0 cells (.0%) have expected count less than 5. The minimum expected count is 9.60.

a) If we compute the expected number of eastern states that Bush should have won if geography has no influence, how many do we get?

b) Are these results statistically significant? Explain carefully in a paragraph that someone who has not had statistics would understand.

a) 15.6, found by multiplying 30 by 26 and dividing by 50. (Bush won 60% of the states (30/50). If geography makes no difference, he would have won 60% of the eastern states. 60% of 26 is 15.6.)
b) Not at usual levels. If location does not matter, we would get results as different from the expected values over 13% of the time. So random chance is a plausible explanation for the differences from the expected breakdown.

2. A researcher has surveyed a large number of high schools trying to determine their attitudes toward college. One question is whether or not they have ever heard of Saint Joseph's College. The researcher suspects that as students advance through high school, they learn more about colleges, and hence seniors should be more likely to have heard of Saint Joseph's College than freshmen. She decides to test this hypothesis using Chi-Square. Below are the results she gets:

QUES7 * YEAR Crosstabulation

			YEAR				Total
			2003	2004	2005	2006
QUES7	1	Count	143	87	42	10	282
		Expected Count	132.7	83.7	49.4	16.1	282.0
	2	Count	120	79	56	22	277
		Expected Count	130.3	82.3	48.6	15.9	277.0
Total		Count	263	166	98	32	559
		Expected Count	263.0	166.0	98.0	32.0	559.0

Chi-Square Tests

	Value	df	Asymp. Sig. (2-sided)
Pearson Chi-Square	8.853	____	____
N of Valid Cases	559

0 cells (.0%) have expected count less than 5. The minimum expected count is 15.86.

(For this question, an answer of 1 indicates that the student has heard of SJC, while a 2 indicates they the student has not heard of SJC. Year represents year of graduation, so 2003 is a senior and 2006 is a freshman.)

a) The first cell of the table indicates that 143 seniors have heard of SJC. Where does the expected count of 132.7 come from?
b) Just eyeballing these data, do they tend to support the researcher's hypothesis? Explain.
c) To formally test the hypothesis, what does the researcher establish as the null hypothesis? What is the alternative?
d) The results indicate a Chi-Square statistic of 8.853. Set up the equation that yielded this result. (Put in the numbers that you would begin with--no need to do any of the calculation.)
e) How many degrees of freedom would this test have? Explain.
f) If you were to test the null hypothesis with alpha = .05, what would you decide? Explain.
g) If you were to test the null hypothesis with alpha = .01, what would you decide? Explain.

The degrees of freedom are 3. The level of significance is .031
a) 263*282/559
b) The percentage of students who have heard of the college declines as they get further away from graduation. So, yes, eyeballing the data suggests the researcher's suspicion may be correct. However, it is not clear if the trend is strong enough to be more than random.
c) The claim or null hypothesis, which we want to show is incorrect, is that year in school has no impact on whether students have heard of the college. The alternative is that the year in school matters.
d) (143-137.2)²/137.2 + (120 - 130.3)²/130.3 + etc.
e) 3. (number of row -1)*(number of columns - 1) = 1*3 = 3
f) You would reject the claim that the results are random and say that the results show that year in school influences how aware the students are of the college.
g) You would say that the evidence is not strong enough to reject the claim that the year of students matters in their awareness of the college.

3. A political candidate has a survey done to determine how popular he is with various groups. He finds the following:

	Age:
Preference:	over 65	under 65	Total
support him	18	12	30
oppose him	22	48	70
Total	40	60	100

a) Complete a contingency table showing expected frequencies under the null hypothesis that there is no difference in support across age groups.

	Age:
Preference:	over 65	under 65	Total
support him	_____	_____	30
oppose him	_____	_____	70
Total	40	60	100

b) How many degrees of freedom will we have in the Chi-square test? Explain how you find out.
c) If he wants to test the hypothesis that he is equally supported in different age groups and he sets his alpha level to .05, what is the critical value of the chi-square statistic?
d) If he sets alpha to .01, what is the critical value of the chi-square statistic? What does this value mean?
e) Compute the chi-square statistic in this case.
f) Do you accept or reject the null hypothesis at the .01 level? What does your decision mean?

a) Missing numbers in the first row: 12, 18. In the second row: 28, 42.
b) 1. (2-1)*)2-1) = 1.
c) 3.84, found in a table.
d) 6.64.
e) 7.143
f) Reject the claim that age does not matter. The probability that we get a result like this by random chance if there are no differences by age is only .00752.

4. A statistics professor believes that students in his morning classes do better than students in his afternoon classes. His department chairman says the differences are random and that students do equally well regardless of time. The professor finds his grades for the past semesters and finds the following:

	Letter Grade:
Time the Classes Met:	A	B	C	D	F	Total
Morning	12	18	13	11	6	60
Afternoon	3	7	17	4	9	40
Total	15	25	30	15	15	100

Using the Chi-square test, do you conclude that the morning classes and afternoon classes were somehow different?

The Chi-square statistic is 11.083 with 4 degrees of freedom. The p-value or level of significance is .02564. If we want just strong evidence, setting our cut-off level of significance at .05, then the evidence is enough to conclude that the morning classes are different from the afternoon classes. However, if we wanted overwhelming evidence, setting the cut-off level of significance at .01, then this evidence is still not enough to reject the claim that there is no real difference in the classes. By random chance we would get a result this far from the expected values about 2.5% of the time.

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