﻿ Test for Homogeneity of Proportions Worksheet - Page 3 | Problems & Solutions

# Test for Homogeneity of Proportions Worksheet - Page 3

Test for Homogeneity of Proportions Worksheet
• Page 3
21.
A research surveys 125 people in each of four neighborhoods to determine the percentage of those whose annual income is greater than $50,000. The results are shown here. At $\alpha$ = 0.10, find the chi-square test value and test the claim that the proportions of those whose income is greater than$50,000 are equal in all four neighborhoods.

 a. 6.25, equal b. 27.24, not equal c. 6.25, not equal d. 27.24, equal

#### Solution:

Let the proportions be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4. The proportions of those whose income is greater than $50,000 are equal in all four neighborhoods. H1: At least one proportion is different from the others. [Null and alternative hypotheses.] The degrees of freedom is (2 - 1) × (4 - 1) = 3. The critical value at α = 0.10 from the table chi-square distribution is 6.251. [degrees of freedom = (row - 1) × (column - 1).] The expected frequencies are computed as shown: [Expected frequency = Row sum × Column sumGrand total .] The completed table is presented below with the expected frequencies, shown within brackets: c2 = (O-E)2E [O - Observed frequency, E - Expected frequency.] c2 = (70 - 65)265 + (85 - 65)265 + (45 - 65)265 + (60 - 65)265 + (55 - 60)260 + (40 - 60)260 + (80 - 60)260 + (65 - 60)260 [Expand and simplify.] c2 = 0.38 + 6.15 + 6.15 + 0.38 + 0.42 + 6.67 + 6.67 + 0.42 = 27.24 [Simplify.] Since 27.24 > 6.251, the decision is to reject the null hypothesis. Therefore, the proportions of those whose income is greater than$50,000 are not equal in all four neighborhoods.

22.
A research surveys 500 people in three different states to determine the proportion of the Republicans. The results are shown here.
 State A State B State C Total Republicans 280 244 226 750 Others 220 256 274 750 Total 500 500 500 1500
At $\alpha$ = 0.10, find the chi-square test value and test the claim that the proportions of Republicans are equal in all three states.
 a. 12.1, equal b. 4.6, not equal c. 12.1, not equal d. 4.6, equal

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of Republicans are equal in all three states.
H1: At least one proportion is different from the others.
[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at α = 0.10 from the table chi-square distribution is 4.605.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
 State A State B State C Total Republicans 750×5001500 = 250 750×5001500 = 250 750×5001500 = 250 750 Others 750×5002000 = 250 750×5001500 = 250 750×5001500 = 250 750 Total 500 500 500 1500

[Expected frequency = Row sum × Column sumGrand total .]

The completed table is presented below with the expected frequencies, shown within brackets:
 State A State B State C Republicans 280(250) 244(250) 226(250) Others 220(250) 256(250) 274(250)

c2 = (O-E)2E
[O - Observed frequency, E - Expected frequency.]

c2 = (280 - 250)2250 + (244 - 250)2250 + (226 - 250)2250 + (220 - 250)2250 + (256 - 250)2250 + (274 - 250)2250
[Expand and simplify.]

c2 = 3.6 + 0.144 + 2.304 + 3.6 + 0.144 + 2.304 = 12.096 » 12.10
[Simplify.]

Since 12.10 > 4.605, the decision is to reject the null hypothesis.

Therefore, the proportions of Republicans are not equal in all three states.

23.
A survey was conducted in 3 towns to determine the proportions of people having pets in their home. The results are shown here.
 Town A Town B Town C Total Have pets 32 45 40 117 Don't have pets 53 40 45 138 Total 85 85 85 255
At $\alpha$ = 0.05, find the chi-square test value and test the claim that the proportions are equal.
 a. 4.07, equal b. 4.07, not equal c. 5.99, not equal d. 5.99, equal

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3.The proportions of people having pets in their home are equal.
H1: At least one proportion is different from the others.
[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at α = 0.05 from the table giving percentile values for the chi-square distribution is 5.99.
[degrees of freedom = (row - 1) x (column - 1).]

The expected frequencies are computed as shown:
 Town A Town B Town C Total Have pets 117×85255 = 39 117×85255 = 39 117×85255 = 39 117 Don't have pets 138×85255 = 46 138×85255 = 46 138×85255 = 46 138 Total 85 85 85 255

[Expected frequency = Row sum × Column sumGrand total.]

The completed table is presented below with the expected frequencies, shown within brackets:
 Town A Town B Town C Have pets 32(39) 45(39) 40(39) Don't have pets 53(46) 40(46) 45(46)

c2 = (O-E)2E
[O - Observed frequency, E - Expected frequency.]

c2 = (32 - 39)239 + (45 - 39)239 + (40 - 39)239 + (53 - 46)246 + (40 - 46)246 + (45 - 46)246
[Expand and simplify.]

c2 = 1.26 + 0.92 + 0.03 + 1.06 + 0.78 + 0.02 = 4.07
[Simplify.]

Since 4.07 < 5.99, the decision is not to reject the null hypothesis.

Therefore, the proportions of people having pets in their home are equal.

24.
Three batches of 100 animals were inoculated and were exposed to the infection of a disease. The results about the survival of the animals are tabulated as shown.
 Batch I Batch II Batch III Total Survived 68 54 70 192 Dead 32 46 30 108 Total 100 100 100 300
At $\alpha$ = 0.05, find the chi-square test value and test the claim that the proportions of animals which survived are equal for all the three batches.
 a. 6.59, equal b. 9.21, equal c. 6.59, not equal d. 9.21, not equal

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of animals which survived in the three batches are equal.
H1: At least one proportion is different from the others.
[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at α = 0.01 from the table chi-square distribution is 9.210.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
 Batch I Batch II Batch III Total Survived 192×100300 = 64 192×100300 = 64 192×100300 = 64 192 Dead 108×100300 = 36 108×100300 = 36 108×100300 = 36 108 Total 100 100 100 300

[Expected frequency = Row sum × Column sumGrand total .]

The completed table is presented below with the expected frequencies, shown within brackets:
 Batch I Batch II Batch III Survived 68(64) 54(64) 70(64) Dead 32(36) 46(36) 30(36)

c2 = (O-E)2E
[O - Observed frequency, E - Expected frequency.]

c2 = (68 - 64)264 + (54 - 64)264 + (70 - 64)264 + (32 - 36)236 + (46 - 36)236 + (30 - 36)236
[Expand and simplify.]

c2 = 0.25 + 1.56 + 0.56 + 0.44 + 2.78 + 1 = 6.59
[Simplify.]

Since 6.59 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportions of animals which survived in the three batches are equal.

25.
A survey was conducted to find out the proportions of boys who work in the age group 15 through 17. The results are shown.
 15 year olds 16 year olds 17 year olds Total Work 51 47 58 156 Don't work 49 53 42 144 Total 100 100 100 300
At $\alpha$ = 0.05, find the chi-square test value and test the claim that the proportions of boys who work are equal.
 a. 5.99, not equal b. 2.48, not equal c. 2.48, equal d. 5.99, equal

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of boys who work are equal.
H1: At least one proportion is different from the others.
[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at α = 0.05 from the table chi-square distribution is 5.99.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
 15 year olds 16 year olds 17 year olds Total Work 156×100300 = 52 156×100300 = 52 156×100300 = 52 156 Don't work 144×100300 = 48 144×100300 = 48 144×100300 = 48 144 Total 100 100 100 300

[Expected frequency = Row sum × Column sumGrand total.]

The completed table is presented below with the expected frequencies, shown within brackets:
 15 year olds 16 year olds 17 year olds Work 51(52) 47(52) 58(52) Don't work 49(48) 53(48) 42(48)

c2 = (O-E)2E
[O - Observed frequency, E - Expected frequency.]

c2 = (51 - 52)252 + (47 - 52)252 + (58 - 52)252 + (49 - 48)248 + (53 - 48)248 + (42 - 48)248
[Expand and simplify.]

c2 = 0.02 + 0.48 + 0.69 + 0.02 + 0.52 + 0.75 = 2.48
[Simplify.]

Since 2.48 < 5.99, the decision is not to reject the null hypothesis.

Therefore, the proportions of boys who work are equal.