Question

1. For the following data: (a) Compute the value of the chi-square test statistic. (6) Test the hypothesis that X and Y are independent at the a=0.05 level of significance. X3 Y 87 74 34 Y 12 18 X2 Poor 2. Are health and happiness related? Use a=0.05 level of significance. Health Excellent Fair Very Happy 271 261 Happiness Pretty Happy 247 Not Too Happy 33 567 103

Answer 1

Question 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: X and Y are independent.

Alternative hypothesis: H_a: X and Y are not independent.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 1*2 = 2

α = 0.05

Critical value = 5.991465

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	X1	X2	X3	Total
Y1	87	74	34	195
Y2	12	32	18	62
Total	99	106	52	257

Expected Frequencies
	Column variable
Row variable	X1	X2	X3	Total
Y1	75.11673	80.42802	39.45525	195
Y2	23.88327	25.57198	12.54475	62
Total	99	106	52	257

Calculations
(O - E)
11.88327	-6.42802	-5.45525
-11.8833	6.428016	5.455253

(O - E)^2/E
1.879902	0.513744	0.754267
5.912594	1.615807	2.372291

Chi square = ∑[(O – E)^2/E] = 13.0486

P-value = 0.001467

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence to conclude that X and Y are independent.

Question 2

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: Health and happiness are not related.

Alternative hypothesis: H_a: Health and happiness are related.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 3

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4

α = 0.05

Critical value = 9.487729

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	Excellent	Fair	Poor	Total
Very Happy	271	261	20	552
Pretty Happy	247	567	53	867
Not Too Happy	33	103	36	172
Total	551	931	109	1591

Expected Frequencies
	Column variable
Row variable	Excellent	Fair	Poor	Total
Very Happy	191.1703	323.0119	37.81772	552
Pretty Happy	300.2621	507.3394	59.39849	867
Not Too Happy	59.56757	100.6486	11.78378	172
Total	551	931	109	1591

Calculations
(O - E)
79.82967	-62.0119	-17.8177
-53.2621	59.66059	-6.39849
-26.5676	2.351351	24.21622

(O - E)^2/E
33.33559	11.90507	8.394776
9.447916	7.015789	0.689255
11.84933	0.054932	49.76544

Chi square = ∑[(O – E)^2/E] = 132.4581

P-value = 0.0000

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that Health and happiness are related.