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Test for Homogeneity of Proportions Worksheet

Test for Homogeneity of Proportions Worksheet
  • Page 1
 1.  
A library authority wanted to find out, ahead of its bulk purchase of books, about library user's preference for the type of books. Towards this, they conducted a random sample survey among males and females and the data is presented in the table:
RomanceBiographies and Self-helpMystery
Male306394300
Female350250400
At α = 0.05, find the chi-square test value and test the claim that a book preference is independent of the gender.
a.
49.44, dependent on the gender
b.
49.44, independent of the gender
c.
5.99, dependent on the gender
d.
5.99, independent of the gender


Solution:

H0: Proportion of men and women among each group p1, p2, p3 are the same; or in other words book preference is independent of the gender.
H1: At least one proportion is different from others. There is some difference between the genders on the preference for certain types of books.
[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).]


RomanceBiographies and self helpMysteryTotal
Male3063943001000
Female3502504001000
Total6566447002000

[Calculate row and column sums.]

The expected frequencies are computed as shown:
RomanceBiographies and self helpMysteryTotal
Male1000×6562000 = 3281000×6442000 = 3221000×7002000 = 3501000
Female1000×6562000 = 3281000×6442000 = 3221000×7002000 = 3501000
Total6566447002000

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

The completed table is presented below with the expected frequencies, shown within brackets:
RomanceBiographies and self helpMystery
Male306(328)394(322)300(350)
Female350(328)250(322)400(350)


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

c2 = (306 - 328)2328 + (394 - 322)2322 + (300 - 350)2350 + (350 - 328)2328 + (250 - 322)2322 + (400 - 350)2350
[Expand and simplify.]

c2 = 1.48 + 16.1 + 7.14 + 1.48 + 16.1 + 7.14 = 49.44
[Simplify.]

Since 49.44 > 5.99, the decision is to reject the null hypothesis.

Therefore, there is some gender difference on the preference for certain types of books.


Correct answer : (1)
 2.  
The table shows the number of students passed and failed in the tests conducted by three different examiners Mr.X, Mr.Y and Mr.Z. At α = 0.05, find the chi-square test value and test the hypothesis that the proportions of students failed in the three tests are equal.
Mr. XMr. YMr. Z
Passed504756
Failed5148

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


Solution:

If p1, p2, p3 are the proportions of those failed in the three tests using the P value method, let the hypothesis be
H0: p1 = p2 = p3
H1: At least one proportion is different.
[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).]


Mr. XMr. YMr. ZTotal
Passed504756153
Failed514827
Total556164180

[Calculate row and column sums.]

The expected frequencies are computed as shown:
Mr. XMr. YMr. ZTotal
Passed153×55180 = 46.75153×61180 = 51.85153×64180 = 54.4153
Failed27×55180 = 8.2527×61180 = 9.1527×64180 = 9.627
Total556164180

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

The completed table is presented below with the expected frequencies, shown within brackets:
Mr. XMr. YMr. Z
Passed50(46.75)47(51.85)56(54.4)
Failed5(8.25)14(9.15)8(9.6)


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

c2 = (50 - 46.75)246.75 + (47 - 51.85)251.85 + (56 - 54.4)254.4 + (5 - 8.25)28.25 + (14 - 9.15)29.15 + (8 - 9.6)29.6
[Expand and simplify.]

c2 = 0.226 + 0.454 + 0.047 + 1.28 + 2.57 + 0.267 = 4.844
[Simplify.]

Since 4.844 < 5.99 the decision is not to reject the null hypothesis.

Thus the proportions of failures in the 3 tests are the same at 0.05 level.


Correct answer : (1)
 3.  
A random survey of 80 individuals caught for drunk driving from each of the various age groups yielded the results as shown in the table. At level of significance α = 0.05, find the chi-square test value and test the claim that the proportions of those who admitted to the offence are equal among the age groups.

a.
7.81, equal
b.
1.98, equal
c.
1.98, not equal
d.
7.81, not equal


Solution:

Let the proportions of those who admitted to the offence among each of the above age groups be p1, p2, p3, p4 respectively. Using the P value method let the hypothesis be
H0: p1 = p2 = p3 = p4; the proportions of those who admitted to the offence are equal for all the age groups.
H1: At least one proportion is different; the proportions of those who admitted to the offence are not equal for the age groups.
[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:
[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 = (45 - 45.75)245.75 + (49 - 45.75)245.75 + (48 - 45.75)245.75 + (41 - 45.75)245.75 + (35 - 34.25)234.25 + (31 - 34.25)234.25 + (32 - 34.25)234.25 + (39 - 34.25)234.25
[Expand and simplify.]

c2 = 0.01 + 0.23 + 0.11 + 0.49 + 0.01 + 0.32 + 0.15 + 0.66 = 1.98
[Simplify.]

Since 1.98 < 7.81, the decision is not to reject the null hypothesis.

Therefore, the proportions of those who admitted to the offence are equal among the age groups.


Correct answer : (2)
 4.  
A pizza shop conducted a study to find out whether age has anything to do with the type of pizza preferred. A random sample survey yielded the result as shown. At α = 0.05, find the chi-square test value. State whether the pizza ordered is related to the age of the person or not.

a.
16.92, not related
b.
16.92, related
c.
108.1, not related
d.
108.1, related


Solution:

Let the proportions of those who ordered pizza among each of the above age groups be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4; The types of pizza ordered are independent of the age group.
H1: At least one proportion is different. The type of pizza ordered is related to the age group.
[Null and alternative hypotheses.]

The degrees of freedom is (4 - 1) × (4 - 1) = 9. The critical value at α = 0.05 from the table chi-square distribution is 16.919.
[degrees of freedom = (row - 1) × (column - 1).]


The expected frequencies are computed. The completed table is presented below with the expected frequencies, shown within brackets:
[Expected frequency = Row sum × Column sumGrand total .]

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

c2 = 5.3 + 7.5 + 0.4 + 13 + 10 + 2.4 + 0.16 + 0.29 + 0.29 + 1.28 + 1.6 + 2.7 + 49.8 + 0.2 + 6.8 + 6.4 = 108.1
[Expand and simplify.]

Since 108.1 > 16.919, the decision is to reject the null hypothesis.

At least one proportion could be different; in other words the proportions of those who ordered pizza are not equal among the age groups. Thus the type of pizza ordered is related to the age group.


Correct answer : (4)
 5.  
A random sample was collected on the drinking habit of the people in a town. The results are tabulated as shown. At α = 0.01, find the chi-square test value and check if there appears to be a gender difference with respect to the drinking habit.

a.
95.22, dependent on the gender
b.
95.22, independent of the gender
c.
11.3, independent of the gender
d.
11.3, dependent on the gender


Solution:

Let the proportions be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4. The drinking habits are independent of the gender of the person.
H1: At least one proportion is different from the others. The drinking habits are dependent on the gender of the person.
[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 = (150 - 175)2175 + (450 - 525)2525 + (300 - 225)2225 + (100 - 75)275 + (200 - 175)2175 + (600 - 525)2525 + (150 - 225)2225 + (50 - 75)275
[Expand and simplify.]

c2 = 3.57 + 10.71 + 25 + 8.33 + 3.57 + 10.71 + 25 + 8.33 = 95.22
[Simplify.]

Since 95.22 > 11.345, the decision is to reject the null hypothesis.

Therefore, at least one proportion is different from the others. So, the drinking habits are dependent on the gender of the person.


Correct answer : (1)
 6.  
A researcher surveys 100 people in each of the three towns A, B and C to determine the percentages of those who are willing to donate for a relief fund. The results are shown here.
Town ATown BTown C
Will donate607368
Will not donate402732
At level of significance α = 0.01, find the chi-square test value and test the claim that the proportions of those who will donate for the relief fund are equal in all the three towns.
a.
3.88, equal
b.
3.88, not equal
c.
9.21, not equal
d.
9.21, equal


Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of those who will donate for the relief fund are equal in all the three towns.
H1: At least one proportion is different from the others. The proportions of those who will donate for the relief fund are not equal.
[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).]


Town ATown BTown CTotal
Will donate607368201
Will not donate40273299
Total100100100300

[Calculate row and column sums.]

The expected frequencies are computed as shown:
Town ATown BTown CTotal
Will donate201×100300 = 67201×100300 = 67201×100300 = 67201
Will not donate99×100300 = 3399×100300 = 3399×100300 = 3399
Total100100100300

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

The completed table is presented below with the expected frequencies, shown within brackets:
Town ATown BTown C
Will donate60(67)73(67)68(67)
Will not donate40(33)27(33)32(33)


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

c2 = (60 - 67)267 + (73 - 67)267 + (68 - 67)267 + (40 - 33)233 + (27 - 33)233 + (32 - 33)233
[Expand and simplify.]

c2 = 0.73 + 0.54 + 0.01 + 1.48 + 1.09 + 0.03 = 3.88
[Simplify.]

Since 3.88 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportions of those who will donate for the relief fund are equal in all the three towns.


Correct answer : (1)
 7.  
A school has four campuses. The principal of the school wants to know if the percentage of students passing an aptitude test is the same for all the four campuses. He administers an aptitude test to 100 students in each of the four campuses. The results are shown here.
South CampusNorth CampusEast CampusWest CampusTotal
Passed41302346140
Failed59707754260
Total100100100100400
At level of significance α = 0.05, find the chi-square test value and test the claim that the proportions of the students who pass the test are equal.
a.
14.32, not equal
b.
7.82, equal
c.
7.82, not equal
d.
14.32, equal


Solution:

Let the proportions of students be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4. The proportions of the students who pass the test are equal.
H1: At least one proportion is different from the others. The proportions of the students who pass the test are not equal.
[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:
South CampusNorth CampusEast CampusWest Campus
Passed140×100400 = 35140×100400 = 35140×100400 = 35140×100400 = 35
Failed260×100400 = 65260×100400 = 65260×100400 = 65260×100400 = 65

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

The completed table is presented below with the expected frequencies, shown within brackets:
South CampusNorth CampusEast CampusWest Campus
Passed41(35)30(35)23(35)46(35)
Failed59(65)70(65)77(65)54(65)


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

c2 = (41 - 35)235 + (30 - 35)235 + (23 - 35)235 + (46 - 35)235 + (59 - 65)265 + (70 - 65)265 + (77 - 65)265 + (54 - 65)265
[Expand and simplify.]

c2 = 1.03 + 0.71 + 4.11 + 3.46 + 0.55 + 0.38 + 2.22 + 1.86 = 14.32
[Simplify.]

Since 14.32 > 7.815, the decision is to reject the null hypothesis.

Therefore, the proportions of the students who pass the test are not equal.


Correct answer : (1)
 8.  
A survey was conducted across 3 schools in a city to find out whether the students like to stay on-campus or off-campus. The results of the survey are tabulated as shown.
School ASchool BSchool CTotal
On-campus28142163
Off-campus22362987
Total505050150
At level of significance α = 0.01, find the chi-square test value and test the claim that the proportions of the students staying on-campus are equal.
a.
9.21, not equal
b.
8.04, not equal
c.
9.21, equal
d.
8.04, equal


Solution:

Let the proportions of the students be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of the students staying on-campus are equal.
H1: At least one proportion is different from the other. The proportions of the students staying on-campus are not equal.
[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:
School ASchool BSchool CTotal
On - Campus63×50150 = 2163×50150 = 2163×50150 = 2163
Off - Campus87×50150 = 2987×50150 = 2987×50150 = 2987
Total505050150

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

The completed table is presented below with the expected frequencies, shown within brackets:
School ASchool BSchool C
On - Campus28(21)14(21)21(21)
Off - Campus22(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 students staying on-campus are equal.


Correct answer : (4)
 9.  
A company has its head office in New York and a branch in Atlanta. The CEO of the company wanted to know if the workers at the two places would like the introduction of a new rule and a survey was conducted for this purpose. The results of the survey are as shown.
New YorkAtlantaTotal
Will like310270580
Will not like190230420
Total5005001000
At α = 0.01, find the chi-square test value and test the claim that equal proportion of workers will like the introduction of the new rule.
a.
6.56, equal proportion
b.
6.64, proportion not equal
c.
6.56, proportion not equal
d.
6.64, equal proportion


Solution:

Let the proportions be p1, p2 respectively.
H0: p1 = p2. Equal proportion of workers will like the introduction of new plan.
H1: Proportion of workers who will like the introduction of new plan is not equal.
[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (2 - 1) = 1. The critical value at α = 0.01 from the table chi-square distribution is 6.635.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
New YorkAtlantaTotal
Will like580×5001000 = 290580×5001000 = 290580
Will not like420×5001000 = 210420×5001000 = 210420
Total5005001000

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

The completed table is presented below with the expected frequencies, shown within brackets:
New YorkAtlanta
Will like310(290)270(290)
Will not like190(210)230(210)


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

c2 = (310 - 290)2290 + (270 - 290)2290 + (190 - 210)2210 + (230 - 210)2210
[Expand and simplify.]

c2 = 1.38 + 1.38 + 1.9 + 1.9 = 6.56
[Simplify.]

Since 6.56 < 6.635, the decision is not to reject the null hypothesis.

Therefore, proportion of workers who will like the introduction of new plan is equal.


Correct answer : (1)
 10.  
A professor wanted to know whether the proportion of the students joining the new course is same for Grade 8, 9 and 10. He surveys 50 students from each grade. The results are as shown.
8th9th10thTotal
Will join28243284
Will not join22261866
Total505050150
At α = 0.01, find the chi-square test value and test the claim that equal proportion of students will join the new course.
a.
9.21, equal proportion
b.
2.60, proportion not equal
c.
2.60, equal proportion
d.
9.21, proportion not equal


Solution:

Let the proportions of students be p1, p2, p3 respectively.
H0: p1 = p2 = p3. Equal proportion of students will join the new course.
H1: At least one proportion is different from the others. The proportion of students who will join the new course is not equal.
[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:
8th9th10thTotal
Will join84×50150 = 2884×50150 = 2884×50150 = 2884
Will not join66×50150 = 2266×50150 = 2266×50150 = 2266
Total505050150

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

The completed table is presented below with the expected frequencies, shown within brackets:
8th9th10th
Will join28(28)24(28)32(28)
Will not join22(22)26(22)18(22)


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

c2 = (28 - 28)228 + (24 - 28)228 + (32 - 28)228 + (22 - 22)222 + (26 - 22)222 + (18 - 22)222
[Expand and simplify.]

c2 = 0 + 0.57 + 0.57 + 0 + 0.73 + 0.73 = 2.60
[Simplify.]

Since 2.60 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportion of students who will join the new course is equal.


Correct answer : (3)

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