﻿ Test for Homogeneity of Proportions Worksheet - Page 2 | Problems & Solutions

# Test for Homogeneity of Proportions Worksheet - Page 2

Test for Homogeneity of Proportions Worksheet
• Page 2
11.
Random samples of teachers were asked if their job was interesting. The results for 3 different years are presented in the table.
 2000 2002 2004 Total Is interesting 280 247 292 819 Is not interesting 220 253 208 681 Total 500 500 500 1500
At $\alpha$ = 0.01, find the chi-square test value and test the claim that proportion of teachers who feel being a teacher is an interesting occupation is the same each year.
 a. 8.77, proportion is the same b. 9.21, proportion is not the same c. 9.21, proportion is the same d. 8.77, proportion is not the same

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3.
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:
 2000 2002 2004 Is interesting 819×5001500 = 273 819×5001500 = 273 819×5001500 = 273 Is not interesting 681×5001500 = 227 681×5001500 = 227 681×5001500 = 227

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

The completed table is presented below with the expected frequencies, shown within brackets:
 2000 2002 2004 Is interesting 280(273) 247(273) 292(273) Is not interesting 220(227) 253(227) 208(227)

The completed table is presented below with the expected frequencies, shown within brackets:
[O - Observed frequency, E - Expected frequency.]

c2 = (280 - 273)2273 + (247 - 273)2273 + (292 - 273)2273 + (220 - 227)2227 + (253 - 227)2227 + (208 - 227)2227
[Expand and simplify.]

c2 = 0.18 + 2.48 + 1.32 + 0.22 + 2.98 + 1.59 = 8.77
[Simplify.]

Since 8.77 < 9.210, the decision is not to reject the null hypothesis.

Therefore, proportion of individuals who feel being a teacher is an interesting occupation is the same each year.

12.
Jail inmates can be classified into one of four crime categories. Suppose a random sample of 1000 female inmates and a random sample of 1000 male inmates were taken. Each inmate was classified into a crime category. The results are tabulated as shown. At $\alpha$ = 0.01, find the chi-square test value and test the claim that the population category proportions are the same in all the populations under study.

 a. 41.18, same b. 11.35, different c. 41.18, different d. 11.35, same

#### Solution:

H0: The population proportions of the categories are the same for male and female inmates.
H1: At least one proportion is different from the others.
[Null and alternative hypotheses.]

The degrees of freedom is (4 - 1) × (2 - 1) = 3. The critical value at α = 0.01 from the table chi-square distribution is 11.345.
[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 = (250 - 225)2225 + (200 - 225)2225 + (300 - 260)2260 + (220 - 260)2260 + (180 - 230)2230 + (280 - 230)2230 + (270 - 285)2285 + (300 - 285)2285
[Expand and simplify.]

c2 = 2.78 + 2.78 + 6.15 + 6.15 + 10.87 + 10.87 + 0.79 + 0.79 = 41.18
[Simplify.]

Since 41.18 > 11.345, the decision is to reject the null hypothesis.

Therefore, the population proportions of the categories are different for male and female inmates.

13.
A survey was conducted in 3 companies to know whether the employees were satisfied with their working conditions. The results of the survey are tabulated as shown.
 Company A Company B Company C Total Satisfied 52 45 36 133 Not satisfied 40 47 56 143 Total 92 92 92 276
At $\alpha$ = 0.01, find the chi-square test value and test the claim that equal proportion of employees are satisfied with the working conditions in their companies.
 a. 9.21, equal proportion b. 5.60, proportion not equal c. 9.21, proportion not equal d. 5.60, equal proportion

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3 . The proportion of employees satisfied with their companies working condition 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:
 Company A Company B Company C Satisfied 133×92276 = 44.3 133×92276 = 44.3 133×92276 = 44.3 Not satisfied 143×92276 = 47.7 143×92276 = 47.7 143×92276 = 47.7

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

The completed table is presented below with the expected frequencies, shown within brackets:
 Company A Company B Company C Satisfied 52(44.3) 45(44.3) 36(44.3) Not satisfied 40(47.7) 47(47.7) 56(47.7)

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

c2 = (52 - 44.3)244.3 + (45 - 44.3)244.3 + (36 - 44.3)244.3 + (40 - 47.7)247.7 + (47 - 47.7)247.7 + (56 - 47.7)247.7
[Expand and simplify.]

c2 = 1.34 + 0.01 + 1.55 + 1.25 + 0.01 + 1.44 = 5.60
[Simplify.]

Since 5.60 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportion of employees satisfied with their companies working condition are equal.

14.
A Television channel made a survey in three cities to find out if any changes need be done to improve their viewership. The results of the survey are as shown.
 City A City B City C Total Changes needed 28 14 21 63 Changes not needed 22 36 29 87 Total 50 50 50 150
At $\alpha$ = 0.01, find the chi-square test value and test the claim that the proportions of the population who want the changes to be done are the same for the cities under study.
 a. 9.21, proportions are the same b. 8.04, proportions are the same c. 9.21, proportions are not the same d. 8.04, proportions are not the same

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of the population who want the changes to be done are the same for the cities under study.
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:
 City A City B City C Changes needed 63×50150 = 21 63×50150 = 21 63×50150 = 21 Changes not needed 87×50150 = 29 87×50150 = 29 87×50150 = 29

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

The completed table is presented below with the expected frequencies, shown within brackets:
 City A City B City C Changes needed 28(21) 14(21) 21(21) Changes not needed 22(29) 36(29) 29(29)

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

c2 = (28 - 21)221 + (14 - 21)221 + (21 - 21)221 + (22 - 29)229 + (36 - 29)229 + (29 - 29)229
[Expand and simplify.]

c2 = 2.33 + 2.33 + 0 + 1.69 + 1.69 + 0 = 8.04
[Simplify.]

Since 8.04 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportions of the population who want the changes to be done are the same for the cities under study.

15.
A survey was conducted in 4 schools to determine if the proportions of students who prefer playing in the weekends are equal. The results are shown here. At $\alpha$ = 0.05, find the chi-square test value and test the claim that the proportions are equal.
 Prefer playing in weekends School A School B School C School D Total Yes 54 58 60 64 236 No 32 28 26 22 108 Total 86 86 86 86 344
 a. 2.80, equal b. 7.82, not equal c. 7.82, equal d. 2.80, not equal

#### Solution:

Let the proportions be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4. The proportions of students who prefer playing in the weekends are equal.
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 = 0.05 from the table chi-square distribution is 7.815.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
 Prefer playing in weekends School A School B School C School D Yes 236×86344 = 59 236×86344 = 59 236×86344 = 59 236×86344 = 59 No 108×86344 = 27 108×86344 = 27 108×86344 = 27 108×86344 = 27

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

The completed table is presented below with the expected frequencies, shown within brackets:
 Prefer playing in weekends School A School B School C School D Yes 54(59) 58(59) 60(59) 64(59) No 32(27) 28(27) 26(27) 22(27)

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

c2 = (54 - 59)259 + (58 - 59)259 + (60 - 59)259 + (64 - 59)259 + (32 - 27)227 + (28 - 27)227 + (26 - 27)227 + (22 - 27)227
[Expand and simplify.]

c2 = 0.4 + 0.02 + 0.02 + 0.4 + 0.93 + 0.04 + 0.04 + 0.93 = 2.80
[Simplify.]

Since 2.80 < 7.815, the decision is not to reject the null hypothesis.

Therefore, the proportions of students who prefer playing in the weekends are equal.

16.
A survey was conducted in 3 towns to determine if the proportions of people having access to the internet from home are equal. The results are shown here. At $\alpha$ = 0.05, find the chi-square test value and test the claim that the proportions are equal.
 Access to internet Town A Town B Town C Total Yes 90 75 120 285 No 60 75 30 165 Total 150 150 150 450

 a. 5.99, equal b. 30.14, equal c. 5.99, not equal d. 30.14, not equal

#### Solution:

Let the proportions p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of people having access to the internet from 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 chi-square distribution is 5.991.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
 Access to internet Town A Town B Town C Total Yes 285×150450 = 95 285×150450 = 95 285×150450 = 95 285 No 165×150450 = 55 165×150450 = 55 165×150450 = 55 165 Total 150 150 150 450

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

The completed table is presented below with the expected frequencies, shown within brackets:
 Access to internet Town A Town B Town C Yes 90(95) 75(95) 120(95) No 60(55) 75(55) 30(55)

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

c2 = (90 - 95)295 + (75 - 95)295 + (120 - 95)295 + (60 - 55)255 + (75 - 55)255 + (30 - 55)255
[Expand and simplify.]

c2 = 0.27 + 4.21 + 6.58 + 0.45 + 7.27 + 11.36 = 30.14
[Simplify.]

Since 30.14 > 5.991, the decision is to reject the null hypothesis.

Therefore, the proportions of people having access to the internet from home are not equal.

17.
A researcher surveyed 100 randomly selected persons in each of four regions of the country and asked if they play basketball or not. The results are shown here.
 East West North South Total Play basketball 55 70 42 61 228 Don't play basketball 45 30 58 39 172 Total 100 100 100 100 400
At $\alpha$ = 0.10, find the chi-square test value. Is there enough evidence to reject the claim that the proportions of persons playing basketball are the same in each area?
 a. 16.89, yes b. 6.25, yes c. 6.25, no d. 16.89, no

#### Solution:

Let the proportions be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4. The proportions of persons playing basketball are the same in each area.
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:
 East West North South Play basketball 228×100400 = 57 228×100400 = 57 228×100400 = 57 228×100400 = 57 Don't play basketball 172×100400 = 43 172×100400 = 43 172×100400 = 43 172×100400 = 43

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

The completed table is presented below with the expected frequencies, shown within brackets:
 East West North South Play basketball 55(57) 70(57) 42(57) 61(57) Don't play basketball 45(43) 30(43) 58(43) 39(43)

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

c2 = (55 - 57)257 + (70 - 57)257 + (42 - 57)257 + (61 - 57)257 + (45 - 43)243 + (30 - 43)243 + (58 - 43)243 + (39 - 43)243
[Expand and simplify.]

c2 = 0.07 + 2.96 + 3.96 + 0.28 + 0.09 + 3.93 + 5.23 + 0.37 = 16.89
[Simplify.]

Since 16.89 > 6.251, the decision is to reject the null hypothesis.

Therefore, the proportions of persons playing basketball are not the same in each area.

18.
A hospital emergency room supervisor wishes to determine if the proportions of injuries to males in his hospital are the same for three months, March, April and May. He surveys 150 injuries treated in his emergency room for the three months. The results are tabulated in the table.
 March April May Total Male 90 82 104 276 Female 60 68 46 174 Total 150 150 150 450
At $\alpha$ = 0.01, find the chi-square test value. Is there enough evidence to reject the claim that the proportions of injuries for males are equal for each of the three months?
 a. 6.97, no b. 6.97, yes c. 9.21, no d. 9.21, yes

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of injuries for males are equal for each of the three months.
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:
 March April May Total Male 276×150450 = 92 276×150450 = 92 276×150450 = 92 276 Female 174×150450 = 58 174×150450 = 58 174×150450 = 58 174 Total 150 150 150 450

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

The completed table is presented below with the expected frequencies, shown within brackets:
 March April May Male 90(92) 82(92) 104(92) Female 60(58) 68(58) 46(58)

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

c2 = (90 - 92)292 + (82 - 92)292 + (104 - 92)292 + (60 - 58)258 + (68 - 58)258 + (46 - 58)258
[Expand and simplify.]

c2 = 0.04 + 1.09 + 1.57 + 0.07 + 1.72 + 2.48 = 6.97
[Simplify.]

Since 6.97 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportions of injuries for males are equal for each of the three months.

19.
A researcher surveyed 200 randomly selected persons in four different areas of the city and asked whether they had seen at least 3 wonders of the world. The results are shown here. At $\alpha$ = 0.01, find the chi-square test value. Can it be concluded that the proportions of persons who had seen at least 3 wonders of the world are the same in each area?

 a. 11.35, no b. 11.35, yes c. 15.17, yes d. 15.17, no

#### Solution:

Let the proportions be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3. = p4. The proportions of persons seeing at least 3 wonders of the world are the same in each area.
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 = 0.01 from the table chi-square distribution is 11.345.
[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 = (67 - 82)282 + (90 - 82)282 + (100 - 82)282 + (71 - 82)282 + (133 - 118)2118 + (110 - 118)2118 + (100 - 118)2118 + (129 - 118)2118
[Expand and simplify.]

c2 = 2.74 + 0.78 + 3.95 + 1.48 + 1.91 + 0.54 + 2.75 + 1.02 = 15.17
[Simplify.]

Since 15.17 > 11.345, the decision is to reject the null hypothesis.

Therefore, the proportions of persons seeing at least 3 wonders of the world are not the same in each area.

20.
In a survey, to determine the proportion of people getting admitted in the hospital in the last month from three cities, a researcher sampled 100 persons in each of the three cities and asked if they were admitted to the hospital in the last month. The results are tabulated as shown. At $\alpha$ = 0.01, find the chi-square test value and test the claim that the proportions are equal.
 Admitted to the hospital City A City B City C Total Yes 28 36 47 111 No 72 64 53 189 Total 100 100 100 300
 a. 7.81, the proportions are not equal b. 9.21, the proportions are equal c. 9.21, the proportions are not equal d. 7.81, the proportions are equal

#### Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportion of people getting admitted in the hospital in the last month for three cities 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:
 Admitted to the hospital City A City B City C Total Yes 111×100300 = 37 111×100300 = 37 111×100300 = 37 111 No 189×100300 = 63 189×100300 = 63 189×100300 = 63 189 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:
 Admitted to the hospital City A City B City C Yes 28(37) 36(37) 47(37) No 72(63) 64(63) 53(63)

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

c2 = (28 - 37)237 + (36 - 37)237 + (47 - 37)237 + (72 - 63)263 + (64 - 63)263 + (53 - 63)263
[Expand and simplify.]

c2 = 2.19 + 0.03 + 2.7 + 1.28 + 0.02 + 1.59 = 7.81
[Simplify.]

Since 7.81 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportion of people getting admitted in the hospital in the last month for three cities are equal.