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.

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.

2000 | 2002 | 2004 | Total | |

Is interesting | 280 | 247 | 292 | 819 |

Is not interesting | 220 | 253 | 208 | 681 |

Total | 500 | 500 | 500 | 1500 |

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 |

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:

2000 | 2002 | 2004 | |

Is interesting | |||

Is not interesting |

[Expected frequency =

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.]

c

[Expand and simplify.]

c

[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 $\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 |

H

[Null and alternative hypotheses.]

The degrees of freedom is (4 - 1) × (2 - 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 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.

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.

Company A | Company B | Company C | Total | |

Satisfied | 52 | 45 | 36 | 133 |

Not satisfied | 40 | 47 | 56 | 143 |

Total | 92 | 92 | 92 | 276 |

a. | 9.21, equal proportion | ||

b. | 5.60, proportion not equal | ||

c. | 9.21, proportion not equal | ||

d. | 5.60, equal proportion |

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:

Company A | Company B | Company C | |

Satisfied | |||

Not satisfied |

[Expected frequency =

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) |

c

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

c

[Expand and simplify.]

c

[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.

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.

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 |

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 |

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:

City A | City B | City C | |

Changes needed | |||

Changes not needed |

[Expected frequency =

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) |

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 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 $\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 |

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:

Prefer playing in weekends | School A | School B | School C | School D |

Yes | ||||

No |

[Expected frequency =

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) |

c

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

c

[Expand and simplify.]

c

[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 $\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 |

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:

Access to internet | Town A | Town B | Town C | Total |

Yes | 285 | |||

No | 165 | |||

Total | 150 | 150 | 150 | 450 |

[Expected frequency =

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) |

c

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

c

[Expand and simplify.]

c

[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.

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?

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 |

a. | 16.89, yes | ||

b. | 6.25, yes | ||

c. | 6.25, no | ||

d. | 16.89, no |

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:

East | West | North | South | |

Play basketball | ||||

Don't play basketball |

[Expected frequency =

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) |

c

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

c

[Expand and simplify.]

c

[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.

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?

March | April | May | Total | |

Male | 90 | 82 | 104 | 276 |

Female | 60 | 68 | 46 | 174 |

Total | 150 | 150 | 150 | 450 |

a. | 6.97, no | ||

b. | 6.97, yes | ||

c. | 9.21, no | ||

d. | 9.21, yes |

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:

March | April | May | Total | |

Male | 276 | |||

Female | 174 | |||

Total | 150 | 150 | 150 | 450 |

[Expected frequency =

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) |

c

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

c

[Expand and simplify.]

c

[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 $\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 |

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 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 $\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 |

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:

Admitted to the hospital | City A | City B | City C | Total |

Yes | 111 | |||

No | 189 | |||

Total | 100 | 100 | 100 | 300 |

[Expected frequency =

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) |

c

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

c

[Expand and simplify.]

c

[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)