# 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:
 Romance Biographies and Self-help Mystery Male 306 394 300 Female 350 250 400
At $\alpha$ = 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).]

 Romance Biographies and self help Mystery Total Male 306 394 300 1000 Female 350 250 400 1000 Total 656 644 700 2000

[Calculate row and column sums.]

The expected frequencies are computed as shown:
 Romance Biographies and self help Mystery Total Male 1000×6562000 = 328 1000×6442000 = 322 1000×7002000 = 350 1000 Female 1000×6562000 = 328 1000×6442000 = 322 1000×7002000 = 350 1000 Total 656 644 700 2000

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

The completed table is presented below with the expected frequencies, shown within brackets:
 Romance Biographies and self help Mystery Male 306(328) 394(322) 300(350) Female 350(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.

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 $\alpha$ = 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. X Mr. Y Mr. Z Passed 50 47 56 Failed 5 14 8

 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. X Mr. Y Mr. Z Total Passed 50 47 56 153 Failed 5 14 8 27 Total 55 61 64 180

[Calculate row and column sums.]

The expected frequencies are computed as shown:
 Mr. X Mr. Y Mr. Z Total Passed 153×55180 = 46.75 153×61180 = 51.85 153×64180 = 54.4 153 Failed 27×55180 = 8.25 27×61180 = 9.15 27×64180 = 9.6 27 Total 55 61 64 180

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

The completed table is presented below with the expected frequencies, shown within brackets:
 Mr. X Mr. Y Mr. Z Passed 50(46.75) 47(51.85) 56(54.4) Failed 5(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.

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 $\alpha$ = 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.

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 $\alpha$ = 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.

5.
A random sample was collected on the drinking habit of the people in a town. The results are tabulated as shown. At $\alpha$ = 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.

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 A Town B Town C Will donate 60 73 68 Will not donate 40 27 32
At level of significance $\alpha$ = 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 A Town B Town C Total Will donate 60 73 68 201 Will not donate 40 27 32 99 Total 100 100 100 300

[Calculate row and column sums.]

The expected frequencies are computed as shown:
 Town A Town B Town C Total Will donate 201×100300 = 67 201×100300 = 67 201×100300 = 67 201 Will not donate 99×100300 = 33 99×100300 = 33 99×100300 = 33 99 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:
 Town A Town B Town C Will donate 60(67) 73(67) 68(67) Will not donate 40(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.

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 Campus North Campus East Campus West Campus Total Passed 41 30 23 46 140 Failed 59 70 77 54 260 Total 100 100 100 100 400
At level of significance $\alpha$ = 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 Campus North Campus East Campus West Campus Passed 140×100400 = 35 140×100400 = 35 140×100400 = 35 140×100400 = 35 Failed 260×100400 = 65 260×100400 = 65 260×100400 = 65 260×100400 = 65

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

The completed table is presented below with the expected frequencies, shown within brackets:
 South Campus North Campus East Campus West Campus Passed 41(35) 30(35) 23(35) 46(35) Failed 59(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.

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 A School B School C Total On-campus 28 14 21 63 Off-campus 22 36 29 87 Total 50 50 50 150
At level of significance $\alpha$ = 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 A School B School C Total On - Campus 63×50150 = 21 63×50150 = 21 63×50150 = 21 63 Off - Campus 87×50150 = 29 87×50150 = 29 87×50150 = 29 87 Total 50 50 50 150

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

The completed table is presented below with the expected frequencies, shown within brackets:
 School A School B School C On - Campus 28(21) 14(21) 21(21) Off - Campus 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 students staying on-campus are equal.

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 York Atlanta Total Will like 310 270 580 Will not like 190 230 420 Total 500 500 1000
At $\alpha$ = 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 York Atlanta Total Will like 580×5001000 = 290 580×5001000 = 290 580 Will not like 420×5001000 = 210 420×5001000 = 210 420 Total 500 500 1000

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

The completed table is presented below with the expected frequencies, shown within brackets:
 New York Atlanta Will like 310(290) 270(290) Will not like 190(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.

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.
 8th 9th 10th Total Will join 28 24 32 84 Will not join 22 26 18 66 Total 50 50 50 150
At $\alpha$ = 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:
 8th 9th 10th Total Will join 84×50150 = 28 84×50150 = 28 84×50150 = 28 84 Will not join 66×50150 = 22 66×50150 = 22 66×50150 = 22 66 Total 50 50 50 150

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

The completed table is presented below with the expected frequencies, shown within brackets:
 8th 9th 10th Will join 28(28) 24(28) 32(28) Will not join 22(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.