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 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 ${\chi }^2$ test for homogeneity.

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

The null and alternative hypotheses are as follows:

Null hypothesis:

$H_0:$ The proportions falling into each of the two response categories for males and females are the same.

Alternative hypothesis:

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

Level of significance:

$\alpha =0.05$

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	$\frac{48\times 63}{139}=21.755$	$\frac{48\times 76}{139}=26.244$	48
Female	$\frac{91\times 63}{139}=41.244$	$\frac{91\times 76}{139}=49.755$	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:

${\chi }^2={{\displaystyle \sum \frac{(\text{observed count}-\text{expected count})}{\text{expected count}}}}^2$

Chi-square statistic:

$\begin{array}{c}{\chi }^2=\frac{{\left(26-21.755\right)}^2}{21.76}+\frac{{\left(22-26.245\right)}^2}{26.24}+\frac{{\left(37-41.245\right)}^2}{41.24}+\frac{{\left(54-49.755\right)}^2}{49.76}\\ =0.828+0.687+0.437+0.362\\ =2.314\end{array}$

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:

$\begin{array}{c}df=(r-1)(c-1)\\ =(2-1)(2-1)\\ =(1)(1)\\ =1\end{array}$

From the calculations, ${\chi }^2=2.314$ 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, $P\text{-value}>0.05$.

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.

Explanation of Solution

From Part (a), the chi-square value obtained is ${\chi }^2=2.314$. 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).

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).

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.