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Test for Homogeneity of Proportions Worksheet - Page 3

Test for Homogeneity of Proportions Worksheet
  • Page 3
 21.  
A research surveys 125 people in each of four neighborhoods to determine the percentage of those whose annual income is greater than $50,000. The results are shown here. At α = 0.10, find the chi-square test value and test the claim that the proportions of those whose income is greater than $50,000 are equal in all four neighborhoods.

a.
6.25, equal
b.
27.24, not equal
c.
6.25, not equal
d.
27.24, equal


Solution:

Let the proportions be p1, p2, p3, p4 respectively.
H0: p1 = p2 = p3 = p4. The proportions of those whose income is greater than $50,000 are equal in all four neighborhoods.
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:
[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 = (70 - 65)265 + (85 - 65)265 + (45 - 65)265 + (60 - 65)265 + (55 - 60)260 + (40 - 60)260 + (80 - 60)260 + (65 - 60)260
[Expand and simplify.]

c2 = 0.38 + 6.15 + 6.15 + 0.38 + 0.42 + 6.67 + 6.67 + 0.42 = 27.24
[Simplify.]

Since 27.24 > 6.251, the decision is to reject the null hypothesis.

Therefore, the proportions of those whose income is greater than $50,000 are not equal in all four neighborhoods.


Correct answer : (2)
 22.  
A research surveys 500 people in three different states to determine the proportion of the Republicans. The results are shown here.
State AState BState CTotal
Republicans280244226750
Others220256274750
Total5005005001500
At α = 0.10, find the chi-square test value and test the claim that the proportions of Republicans are equal in all three states.
a.
12.1, equal
b.
4.6, not equal
c.
12.1, not equal
d.
4.6, equal


Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of Republicans are equal in all three states.
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 = 0.10 from the table chi-square distribution is 4.605.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
State AState BState CTotal
Republicans750×5001500 = 250750×5001500 = 250750×5001500 = 250750
Others750×5002000 = 250750×5001500 = 250750×5001500 = 250750
Total5005005001500

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

The completed table is presented below with the expected frequencies, shown within brackets:
State AState BState C
Republicans280(250)244(250)226(250)
Others220(250)256(250)274(250)


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

c2 = (280 - 250)2250 + (244 - 250)2250 + (226 - 250)2250 + (220 - 250)2250 + (256 - 250)2250 + (274 - 250)2250
[Expand and simplify.]

c2 = 3.6 + 0.144 + 2.304 + 3.6 + 0.144 + 2.304 = 12.096 » 12.10
[Simplify.]

Since 12.10 > 4.605, the decision is to reject the null hypothesis.

Therefore, the proportions of Republicans are not equal in all three states.


Correct answer : (3)
 23.  
A survey was conducted in 3 towns to determine the proportions of people having pets in their home. The results are shown here.
Town ATown BTown CTotal
Have pets324540117
Don't have pets534045138
Total858585255
At α = 0.05, find the chi-square test value and test the claim that the proportions are equal.
a.
4.07, equal
b.
4.07, not equal
c.
5.99, not equal
d.
5.99, equal


Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3.The proportions of people having pets in their 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 giving percentile values for the chi-square distribution is 5.99.
[degrees of freedom = (row - 1) x (column - 1).]

The expected frequencies are computed as shown:
Town ATown BTown CTotal
Have pets117×85255 = 39117×85255 = 39117×85255 = 39117
Don't have pets138×85255 = 46138×85255 = 46138×85255 = 46138
Total858585255

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

The completed table is presented below with the expected frequencies, shown within brackets:
Town ATown BTown C
Have pets32(39)45(39)40(39)
Don't have pets53(46)40(46)45(46)


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

c2 = (32 - 39)239 + (45 - 39)239 + (40 - 39)239 + (53 - 46)246 + (40 - 46)246 + (45 - 46)246
[Expand and simplify.]

c2 = 1.26 + 0.92 + 0.03 + 1.06 + 0.78 + 0.02 = 4.07
[Simplify.]

Since 4.07 < 5.99, the decision is not to reject the null hypothesis.

Therefore, the proportions of people having pets in their home are equal.


Correct answer : (1)
 24.  
Three batches of 100 animals were inoculated and were exposed to the infection of a disease. The results about the survival of the animals are tabulated as shown.
Batch IBatch IIBatch IIITotal
Survived685470192
Dead324630108
Total100100100300
At α = 0.05, find the chi-square test value and test the claim that the proportions of animals which survived are equal for all the three batches.
a.
6.59, equal
b.
9.21, equal
c.
6.59, not equal
d.
9.21, not equal


Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of animals which survived in the three batches 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:
Batch IBatch IIBatch IIITotal
Survived192×100300 = 64192×100300 = 64192×100300 = 64192
Dead108×100300 = 36108×100300 = 36108×100300 = 36108
Total100100100300

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

The completed table is presented below with the expected frequencies, shown within brackets:
Batch IBatch IIBatch III
Survived68(64)54(64)70(64)
Dead32(36)46(36)30(36)


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

c2 = (68 - 64)264 + (54 - 64)264 + (70 - 64)264 + (32 - 36)236 + (46 - 36)236 + (30 - 36)236
[Expand and simplify.]

c2 = 0.25 + 1.56 + 0.56 + 0.44 + 2.78 + 1 = 6.59
[Simplify.]

Since 6.59 < 9.210, the decision is not to reject the null hypothesis.

Therefore, the proportions of animals which survived in the three batches are equal.


Correct answer : (1)
 25.  
A survey was conducted to find out the proportions of boys who work in the age group 15 through 17. The results are shown.
15 year olds16 year olds17 year oldsTotal
Work514758156
Don't work495342144
Total100100100300
At α = 0.05, find the chi-square test value and test the claim that the proportions of boys who work are equal.
a.
5.99, not equal
b.
2.48, not equal
c.
2.48, equal
d.
5.99, equal


Solution:

Let the proportions be p1, p2, p3 respectively.
H0: p1 = p2 = p3. The proportions of boys who work 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 5.99.
[degrees of freedom = (row - 1) × (column - 1).]

The expected frequencies are computed as shown:
15 year olds16 year olds17 year oldsTotal
Work156×100300 = 52156×100300 = 52156×100300 = 52156
Don't work144×100300 = 48144×100300 = 48144×100300 = 48144
Total100100100300

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

The completed table is presented below with the expected frequencies, shown within brackets:
15 year olds16 year olds17 year olds
Work51(52)47(52)58(52)
Don't work49(48)53(48)42(48)


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

c2 = (51 - 52)252 + (47 - 52)252 + (58 - 52)252 + (49 - 48)248 + (53 - 48)248 + (42 - 48)248
[Expand and simplify.]

c2 = 0.02 + 0.48 + 0.69 + 0.02 + 0.52 + 0.75 = 2.48
[Simplify.]

Since 2.48 < 5.99, the decision is not to reject the null hypothesis.

Therefore, the proportions of boys who work are equal.


Correct answer : (3)

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