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:

At $\alpha $ = 0.05, find the chi-square test value and test the claim that a book preference is independent of the gender.

Romance | Biographies and Self-help | Mystery | |

Male | 306 | 394 | 300 |

Female | 350 | 250 | 400 |

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 |

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at

[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 | |||

Female | 1000 | |||

Total | 656 | 644 | 700 | 2000 |

[Expected frequency =

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) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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 $\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 |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at

[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 | |||

Failed | 27 | |||

Total | 55 | 61 | 64 | 180 |

[Expected frequency =

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) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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 $\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 |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (4 - 1) = 3. The critical value at

[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:

[Expected frequency =

The completed table is presented below with the expected frequencies, shown within brackets:

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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 $\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 |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (4 - 1) × (4 - 1) = 9. The critical value at

[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 =

c

[O - Observed frequency, E - Expected frequency.]

c

[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 $\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 |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (4 - 1) = 3. The critical value at

[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:

[Expected frequency =

The completed table is presented below with the expected frequencies, shown within brackets:

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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.

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.

Town A | Town B | Town C | |

Will donate | 60 | 73 | 68 |

Will not donate | 40 | 27 | 32 |

a. | 3.88, equal | ||

b. | 3.88, not equal | ||

c. | 9.21, not equal | ||

d. | 9.21, equal |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at

[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 | |||

Will not donate | 99 | |||

Total | 100 | 100 | 100 | 300 |

[Expected frequency =

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) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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.

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.

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 |

a. | 14.32, not equal | ||

b. | 7.82, equal | ||

c. | 7.82, not equal | ||

d. | 14.32, equal |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (4 - 1) = 3. The critical value at

[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:

South Campus | North Campus | East Campus | West Campus | |

Passed | ||||

Failed |

[Expected frequency =

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) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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.

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.

School A | School B | School C | Total | |

On-campus | 28 | 14 | 21 | 63 |

Off-campus | 22 | 36 | 29 | 87 |

Total | 50 | 50 | 50 | 150 |

a. | 9.21, not equal | ||

b. | 8.04, not equal | ||

c. | 9.21, equal | ||

d. | 8.04, equal |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at

[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:

School A | School B | School C | Total | |

On - Campus | 63 | |||

Off - Campus | 87 | |||

Total | 50 | 50 | 50 | 150 |

[Expected frequency =

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) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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.

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.

New York | Atlanta | Total | |

Will like | 310 | 270 | 580 |

Will not like | 190 | 230 | 420 |

Total | 500 | 500 | 1000 |

a. | 6.56, equal proportion | ||

b. | 6.64, proportion not equal | ||

c. | 6.56, proportion not equal | ||

d. | 6.64, equal proportion |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (2 - 1) = 1. The critical value at

[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:

New York | Atlanta | Total | |

Will like | 580 | ||

Will not like | 420 | ||

Total | 500 | 500 | 1000 |

[Expected frequency =

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) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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.

At $\alpha $ = 0.01, find the chi-square test value and test the claim that equal proportion of students will join the new course.

8^{th} | 9^{th} | 10^{th} | Total | |

Will join | 28 | 24 | 32 | 84 |

Will not join | 22 | 26 | 18 | 66 |

Total | 50 | 50 | 50 | 150 |

a. | 9.21, equal proportion | ||

b. | 2.60, proportion not equal | ||

c. | 2.60, equal proportion | ||

d. | 9.21, proportion not equal |

H

H

[Null and alternative hypotheses.]

The degrees of freedom is (2 - 1) × (3 - 1) = 2. The critical value at

[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:

8^{th} | 9^{th} | 10^{th} | Total | |

Will join | 84 | |||

Will not join | 66 | |||

Total | 50 | 50 | 50 | 150 |

[Expected frequency =

The completed table is presented below with the expected frequencies, shown within brackets:

8^{th} | 9^{th} | 10^{th} | |

Will join | 28(28) | 24(28) | 32(28) |

Will not join | 22(22) | 26(22) | 18(22) |

c

[O - Observed frequency, E - Expected frequency.]

c

[Expand and simplify.]

c

[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)