Chapter 10.2, Problem 20E

a.

To determine

Find the value of the ${\chi }^2$ statistic for testing the hypothesis that the opinion on optimum is independent of gender.

Answer to Problem 20E

The value of chi-square statistic is 0.965.

Explanation of Solution

Calculation:

The contingency table with 2 rows and 2 columns are given. The two rows consist of Optimistic and Pessimistic and the two columns consist of male and female.

Contingency table:

A contingency table is obtained as using two qualitative variables. One of the qualitative variable is row variable that has one category for each row of the table another is column variable has one category for each column of the table.

The hypotheses are:

Null Hypothesis:

$H_0:\text{The opinion on optimism and Gender is independent}\text{.}$

Alternate Hypothesis:

$H_1:\text{The opinion on optimism and Gender is not independent}.$

Now, it is obtained that,

Gender	Male	Female	Row Total
Optimistic	1,148	815	1,963
Pessimistic	225	178	403
Column Total	1,373	993	2,366

Expected frequencies:

The expected frequencies in case of contingency table is obtained as,

$E=\frac{\text{Row total}\cdot \text{Coloumn total}}{\text{Grand total}}$.

Now, using the formula of expected frequency it is found that the expected frequency for the optimistic male is obtained as,

$\begin{array}{c}\frac{\left(1,963\right)\left(1,373\right)}{2,366}=\frac{2,695,199}{2,366}\\ =1,139.137\end{array}$.

Hence, in similar way the expected frequencies are obtained as,

Gender	Male	Female
Optimistic	$\frac{\left(1,963\right)\left(1,373\right)}{2,366}=1,139.137$	$\frac{\left(1,963\right)\left(993\right)}{2,366}=823.863$
Pessimistic	$\frac{\left(403\right)\left(1,373\right)}{2,366}=233.863$	$\frac{\left(403\right)\left(993\right)}{2,366}=169.137$

Chi-Square statistic:

The chi-square statistic is obtained as ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where $O_1,O_2,\mathrm{...},O_k$ be the observed frequencies and $E_1,E_2,\mathrm{...},E_k$ be the expected frequencies for k number of categories.

The accept and reject can be rewritten as,

Male	1
Female	2

Test Statistic:

Software procedure:

Step -by-step software procedure to obtain test statistic using MINITAB software is as follows:

• Select Stat > Table > Cross Tabulation and Chi-Square.
• Check the box of Raw data (categorical variables).
• Under For rows enter Row.
• Under For columns enter Gender.
• Check the box of Count under Display.
• Under Chi-Square, click the box of Chi-Square test.
• Select OK.
• Output using MINITAB software is given below:

Thus, the value of chi-square statistic is 0.965.

b.

To determine

Find the proportion of men who were optimistic.

Answer to Problem 20E

The proportion of men who were optimistic is 0.836.

Explanation of Solution

Calculation:

From part (a), it is found that,

Gender	Male	Female	Row Total
Optimistic	1,148	815	1,963
Pessimistic	225	178	403
Column Total	1,373	993	2,366

Hence, the proportion of men who were optimistic is,

$\frac{1,148}{1,373}\approx \underset{\_}{0.836}$.

Thus, the proportion of men who were optimistic is 0.836.

c.

To determine

Find the proportion of women who were optimistic.

Answer to Problem 20E

The proportion of women who were optimistic is 0.821.

Explanation of Solution

Calculation:

From part (a), it is found that,

Gender	Male	Female	Row Total
Optimistic	1,148	815	1,963
Pessimistic	225	178	403
Column Total	1,373	993	2,366

Hence, the proportion of women who were optimistic is,

$\frac{815}{993}\approx \underset{\_}{0.821}$.

Thus, the proportion of women who were optimistic is 0.821.

d.

To determine

Find the test statistic z for testing the null hypothesis that the two proportions are equal versus the alternative that they are not equal.

Answer to Problem 20E

The test statistic z for testing the null hypothesis that the two proportions are equal versus the alternative that they are not equal is 0.9821.

Explanation of Solution

Calculation:

Assume that $p_1$ and $p_2$ are the population proportion of men and proportion of women who were optimistic, respectively.

It is also assumed that ${\widehat{p}}_1$ and ${\widehat{p}}_2$ are the sample proportion of men and proportion of women who were optimistic, respectively.

The random variables $x_1$ and $x_2$ are the number of men and women who were optimistic, respectively and $n_1$ and $n_2$ are the samples sizes of men and women who were optimistic, respectively.

The assumptions for performing a Hypothesis Test for the difference between two population proportions are defined as,

• The two random samples are independent to each other.
• Each population size is at least 20 times of the sample size.
• The individuals in the each sample are divided into two categories.
• The minimum sample size in each category is 10.

A random sample of 1,373 men and another random sample of 993 women are asked in the General Survey that whether they were optimistic about the future. There is no statistical relationship between these two samples. Hence, the two random samples are independent to each other.

The number of men and women in United States are much larger than the drawn samples. Hence, population size is more than 20 times of the sample size

The individuals in the each sample are classified in two categories. One is optimistic and another is pessimistic.

The sample size of men is 1,148 and the sample size of women is 815.

Now, as all the assumptions for performing a Hypothesis Test for the difference between two populations proportions are satisfied, then one can proceed to perform a Hypothesis Test for the difference between two population proportions.

The hypotheses are:

Null Hypothesis:

$H_0:p_1=p_2$.

That is, the population proportions of men and proportion of women who were optimistic are same.

Alternative Hypothesis:

$H_1:p_1\ne p_2$.

That is, the population proportions of men and proportion of women who were optimistic are not same.

The test statistic z is defend as $z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$.

The pooled proportion is defined as $\widehat{p}=\frac{x_1+x_2}{n_1+n_2}$.

From part (a), it is found that,

Gender	Male	Female	Row Total
Optimistic	1,148	815	1,963
Pessimistic	225	178	403
Column Total	1,373	993	2,366

Thus, $x_1=1,148$, $x_2=815$, $n_1=1,373$ and $n_2=993$.

From part (b), it is found that,

Hence, the proportion of men who were optimistic is 0.836.

Therefore, ${\widehat{p}}_1=0.836$.

From part (c) it is found that proportion of women who were optimistic is 0.821.

Therefore, ${\widehat{p}}_2=0.821$.

Hence,

$\begin{array}{c}\widehat{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{1,148+815}{1,373+993}\\ =\frac{1,963}{2,366}\\ =0.8297\\ \approx 0.830\end{array}$

Thus, the test statistic value is,

 $\begin{array}{c}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{\widehat{p}\left(1-\widehat{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{0.836-0.821}{\sqrt{0.830\left(1-0.830\right)\left(\frac{1}{1,373}+\frac{1}{993}\right)}}\\ =\frac{0.015}{\sqrt{\left(0.830\right)\left(0.170\right)\left(0.00165\right)}}\\ =\frac{0.015}{\sqrt{0.0002332}}\end{array}$

    $\begin{array}{c}=\frac{0.015}{0.01527}\\ =\underset{\_}{0.9821}.\end{array}$

e.

To determine

Prove that ${\chi }^2=z^2$.

Explanation of Solution

Calculation:

From part (a), it is found that $z=0.9821$.

Hence,

$\begin{array}{c}{z}^2={\left(0.9821\right)}^2\\ =0.9645\\ \approx 0.965\end{array}$

From part (a), it is found that value of chi-square statistic is 0.965.

Hence, it is proved that ${\chi }^2=z^2$.

f.

To determine

Find the P-values for each of these tests using technology.

Prove that P-values are equal.

Answer to Problem 20E

The P-values for each of these tests is 0.326.

Explanation of Solution

Calculation:

From part (a), it is found that the P-value for the chi square test is 0.326.

Using z test:

Software procedure:

Step by step procedure to obtain the P-values using the MINITAB software is given below:

• Choose Stat > Basic Statistics > 2 Proportions.
• Choose Summarized data.
• In First sample, enter Trials as 1,373 and Events as 1,148.
• In Second sample, enter Trials as 993 and Events as 815.
• Check Perform hypothesis test.
• Under Test Method choose Use the pooled estimate of the proportion.
• Click OK.

The output is MINITAB software is given below:

Therefore, the P-value is 0.326.

Hence, it is proved that the P-values for each of these tests are same.

g.

To determine

Conclude that when a contingency table has two rows and two columns, the chi square tests equivalent to the test for the difference between proportions.

Explanation of Solution

It is found that the P-values for chi-square test and z test are same. Moreover, the square of z test statistic is same as the chi-square test statistic.

Hence, it can be concluded that when a contingency table has two rows and two columns, the chi square tests equivalent to the test for the difference between proportions.