Chapter 11.2, Problem 36E

To determine

To perform chi square test of independence.

Answer to Problem 36E

There is convincing evidence that distribution of opinion is not same for each race.

Explanation of Solution

Given:

	Black	Hispanic	White	Total
Excellent	12	34	22	68
Good	69	55	81	205
Fair	75	61	60	196
Poor	24	24	24	72
Don't know	22	28	14	64
Total	202	202	201	605

Calculation:

Null and alternative hypotheses:

  $H_0$ : The distribution of opinion is same for each race.

  $H_a$ :The distribution of opinion is not same for each race.

Using excel formula,

		Black	Hispanic	White	Total
Excellent	Observed	12	34	22	68
	Expected	22.70	22.70	22.59	68.00
	(O - E)² / E	5.05	5.62	0.02	10.68
Good	Observed	69	55	81	205
	Expected	68.45	68.45	68.11	205.00
	(O - E)² / E	0.00	2.64	2.44	5.09
Fair	Observed	75	61	60	196
	Expected	65.44	65.44	65.12	196.00
	(O - E)² / E	1.40	0.30	0.40	2.10
Poor	Observed	24	24	24	72
	Expected	24.04	24.04	23.92	72.00
	(O - E)² / E	0.00	0.00	0.00	0.00
Don't know	Observed	22	28	14	64
	Expected	21.37	21.37	21.26	64.00
	(O - E)² / E	0.02	2.06	2.48	4.56
Total	Observed	202	202	201	605
	Expected	202.00	202.00	201.00	605.00
	(O - E)² / E	6.47	10.62	5.34	22.43
		22.43	chi-square
		8	Df
		.0042	p-value

Decision: P-value = 0.0042< 0.05, reject H0.

Conclusion: There is convincing evidence that distribution of opinion is not same for each race.