Test whether the proportions falling into each of the two response categories for males and females are same or not.

Question

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Answer 1

Question

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.

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Answer 2

Question

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.

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Answer 3

Question

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.

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Answer 4

Question

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.

Answer 6

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Answer 7

Chapter 12.2, Problem 21E

a.

To determine

Test whether the proportions falling into each of the two response categories for males and females are same or not.

Answer 8

a.

Expert Solution

Answer to Problem 21E

There is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The given table represents the classification of male and female students according to their response about whether they usually eat three meals a day or rarely eat three meals a day.

In order to test whether the proportions falling into each of the two response categories for males and females are same or not, the appropriate test is χ2 test for homogeneity.

The steps in chi-square test for homogeneity are as follows:

The null and alternative hypotheses are as follows:

Null hypothesis:

H0: The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

Ha: The proportions falling into each of the two response categories for males and females are not the same.

Level of significance:

α=0.05

Assumptions:

It is given that data are resulted from classifying each person in a random sample of male and female students at a particular college.
The expected cell counts are calculated as shown below:

	Usually eat 3 meals a day	Rarely eat 3 meals a day	Total
Male	48×63139=21.755	48×76139=26.244	48
Female	91×63139=41.244	91×76139=49.755	91
Total	63	76	139

From the expected cell counts table, it is observed that all the expected counts are greater than 5.

Test statistic:

χ2=∑(observed count−expected count)expected count2

Chi-square statistic:

χ2=(26−21.755)221.76+(22−26.245)226.24+(37−41.245)241.24+(54−49.755)249.76=0.828+0.687+0.437+0.362=2.314

Degrees of freedom:

Here, the number of rows is 2 and the number of columns is 2. The degrees of freedom is calculated as follows:

df=(r−1)(c−1)=(2−1)(2−1)=(1)(1)=1

From the calculations, χ2=2.314 and df=1.

P-value:

From Appendix A: Table 8-Upper-Tail Areas for Chi-Square Distributions, the right-tailed area is greater than 0.100 for the chi-squared values that are less than 2.70.

Thus, P-value > 0.100.

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, P-value>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Answer 13

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

Answer 14

b.

To determine

Check whether the calculations and conclusions from Part (a) are consistent with the accompanying Minitab output.

Answer 15

b.

Expert Solution

Explanation of Solution

From Part (a), the chi-square value obtained is χ2=2.314. The conclusion is based on the range of the P-value obtained from the table, that is, there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

In the given MINITAB output, the chi-square value is 2.314 and the P-value is 0.128.

Since 0.128>0.05, it must be concluded that there is no convincing evidence that the proportions falling into each of the two response categories for males and females are not the same.

Hence, the calculations and conclusions from Part (a) are consistent with the accompanying MINITAB output.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

Answer 20

c.

To determine

Check whether the two-sample z test would lead to the same conclusion as in Part (a).

Answer 21

c.

Expert Solution

Answer to Problem 21E

The two-sample z test leads to the same conclusion as in Part (a).

Answer 22

c.

Expert Solution

Answer 23

c.

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

From the given output, the test for two proportions is z=1.53 and P-value is 0.127.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is greater than the level of significance.

That is, 0.127>0.05.

Hence, fail to reject the null hypothesis.

Therefore, there is no convincing evidence that the proportions falling in the two response categories are not the same for males and females.

Thus, the two-sample z test leads to the same conclusion as in Part (a).

Answer 26

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.

Answer 27

d.

To determine

Compare the P-values from the tests in Parts (a) and (c).

Answer 28

d.

Expert Solution

Explanation of Solution

The P-value obtained in Part (a) is greater than 0.100, which is not an exact P-value, but a range for the P-value.

The P-value obtained in Part (c) is 0.127.

Both the P-values are greater than 0.100.

This is not surprising as both the chi-square statistic and the z-statistic measure the probability of getting sample proportions that are far from the expected proportions under null hypothesis.