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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.
200020022004Total
Is interesting280247292819
Is not interesting220253208681
Total5005005001500
At α = 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:
200020022004
Is interesting819×5001500 = 273819×5001500 = 273819×5001500 = 273
Is not interesting681×5001500 = 227681×5001500 = 227681×5001500 = 227

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

The completed table is presented below with the expected frequencies, shown within brackets:
200020022004
Is interesting280(273)247(273)292(273)
Is not interesting220(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.


Correct answer : (1)
 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 α = 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.


Correct answer : (3)
 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 ACompany BCompany CTotal
Satisfied524536133
Not satisfied404756143
Total929292276
At α = 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 ACompany BCompany C
Satisfied133×92276 = 44.3133×92276 = 44.3133×92276 = 44.3
Not satisfied143×92276 = 47.7143×92276 = 47.7143×92276 = 47.7

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

The completed table is presented below with the expected frequencies, shown within brackets:
Company ACompany BCompany C
Satisfied52(44.3)45(44.3)36(44.3)
Not satisfied40(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.


Correct answer : (4)
 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 ACity BCity CTotal
Changes needed28142163
Changes not needed22362987
Total505050150
At α = 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 ACity BCity C
Changes needed63×50150 = 2163×50150 = 2163×50150 = 21
Changes not needed87×50150 = 2987×50150 = 2987×50150 = 29

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

The completed table is presented below with the expected frequencies, shown within brackets:
City ACity BCity C
Changes needed28(21)14(21)21(21)
Changes not needed22(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.


Correct answer : (2)
 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 α = 0.05, find the chi-square test value and test the claim that the proportions are equal.
Prefer playing in weekendsSchool ASchool BSchool CSchool DTotal
Yes54586064236
No32282622108
Total86868686344
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 weekendsSchool ASchool BSchool CSchool D
Yes236×86344 = 59236×86344 = 59236×86344 = 59236×86344 = 59
No108×86344 = 27108×86344 = 27108×86344 = 27108×86344 = 27

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

The completed table is presented below with the expected frequencies, shown within brackets:
Prefer playing in weekendsSchool ASchool BSchool CSchool D
Yes54(59)58(59)60(59)64(59)
No32(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.


Correct answer : (1)
 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 α = 0.05, find the chi-square test value and test the claim that the proportions are equal.
Access to internetTown ATown BTown CTotal
Yes9075120285
No607530165
Total150150150450

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 internetTown ATown BTown CTotal
Yes285×150450 = 95285×150450 = 95285×150450 = 95285
No165×150450 = 55165×150450 = 55165×150450 = 55165
Total150150150450

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

The completed table is presented below with the expected frequencies, shown within brackets:
Access to internetTown ATown BTown C
Yes90(95)75(95)120(95)
No60(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.


Correct answer : (4)
 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.
EastWestNorthSouthTotal
Play basketball55704261228
Don't play basketball45305839172
Total100100100100400
At α = 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:
EastWestNorthSouth
Play basketball228×100400 = 57228×100400 = 57228×100400 = 57228×100400 = 57
Don't play basketball172×100400 = 43172×100400 = 43172×100400 = 43172×100400 = 43

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

The completed table is presented below with the expected frequencies, shown within brackets:
EastWestNorthSouth
Play basketball55(57)70(57)42(57)61(57)
Don't play basketball45(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.


Correct answer : (1)
 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.
MarchAprilMayTotal
Male9082104276
Female606846174
Total150150150450
At α = 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:
MarchAprilMayTotal
Male276×150450 = 92276×150450 = 92276×150450 = 92276
Female174×150450 = 58174×150450 = 58174×150450 = 58174
Total150150150450

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

The completed table is presented below with the expected frequencies, shown within brackets:
MarchAprilMay
Male90(92)82(92)104(92)
Female60(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.


Correct answer : (1)
 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 α = 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.


Correct answer : (4)
 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 α = 0.01, find the chi-square test value and test the claim that the proportions are equal.
Admitted to the hospitalCity ACity BCity CTotal
Yes283647111
No726453189
Total100100100300
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 hospitalCity ACity BCity CTotal
Yes111×100300 = 37111×100300 = 37111×100300 = 37111
No189×100300 = 63189×100300 = 63189×100300 = 63189
Total100100100300

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

The completed table is presented below with the expected frequencies, shown within brackets:
Admitted to the hospitalCity ACity BCity C
Yes28(37)36(37)47(37)
No72(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.


Correct answer : (4)

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