Question

No Smoking: The general social survey conducted a poll of 685 adults in which the subjects were asked whether they agree that the government should prohibit smoking in public places. In addition, each was asked how many people lived in his or her household. The results are summarized in the following contingency table.

Household Size

1 2 3 4 5

Agree 66 109 37 28 30

no opinion 26 56 29 19 30

Disagree 51 79 46 48 31

(a) compute the expected frequencies.

(b) compute the value of the test statistics.

(c) how many degrees of freedom are there?

(d) Test the hypothesis of independence. Use the a=0.05 level of significance. What do you conclude?

Answer 1

Observed Frequencies
	S1	S2	S3	S4	S5	Total
Agree	66	109	37	28	30	270
No opinion	26	56	29	19	30	160
Disagree	51	79	46	48	31	255
Total	143	244	112	95	91	685
Expected Frequencies
	S1	S2	S3	S4	S5	Total
Agree	143 * 270 / 685 = 56.365	244 * 270 / 685 = 96.1752	112 * 270 / 685 = 44.146	95 * 270 / 685 = 37.4453	91 * 270 / 685 = 35.8686	270
No opinion	143 * 160 / 685 = 33.4015	244 * 160 / 685 = 56.9927	112 * 160 / 685 = 26.1606	95 * 160 / 685 = 22.1898	91 * 160 / 685 = 21.2555	160
Disagree	143 * 255 / 685 = 53.2336	244 * 255 / 685 = 90.8321	112 * 255 / 685 = 41.6934	95 * 255 / 685 = 35.365	91 * 255 / 685 = 33.8759	255
Total	143	244	112	95	91	685
	(fo-fe)²/fe
Agree	(66 - 56.365)²/56.365 = 1.647	(109 - 96.1752)²/96.1752 = 1.7102	(37 - 44.146)²/44.146 = 1.1567	(28 - 37.4453)²/37.4453 = 2.3825	(30 - 35.8686)²/35.8686 = 0.9602
No opinion	(26 - 33.4015)²/33.4015 = 1.6401	(56 - 56.9927)²/56.9927 = 0.0173	(29 - 26.1606)²/26.1606 = 0.3082	(19 - 22.1898)²/22.1898 = 0.4585	(30 - 21.2555)²/21.2555 = 3.5975
Disagree	(51 - 53.2336)²/53.2336 = 0.0937	(79 - 90.8321)²/90.8321 = 1.5413	(46 - 41.6934)²/41.6934 = 0.4448	(48 - 35.365)²/35.365 = 4.5142	(31 - 33.8759)²/33.8759 = 0.2442

b) Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 20.7164

c) df = (r-1)(c-1) = 8

d) Critical value:

χ²α = CHISQ.INV.RT(0.05, 8) = 15.5073

p-value:

p-value = CHISQ.DIST.RT(20.7164, 8) = 0.0079

Decision:

p-value < α, Reject the null hypothesis.

There is enough evidence to conclude that there is an association between factors at 0.05 significance level.