Chapter 15, Problem 21CE

To determine

State the null and alternative hypothesis.

Find the degrees of freedom for the contingency table.

Find the critical value of chi-square from Appendix E or from Excel’s function.

Calculate the chi-square test statistics at 0.10 level of significance and interpret the p-value.

Check whether the conclusion is sensitive to the level of significance chosen, identify the cells that contribute to the chi-square test statistic and check for the small expected frequencies.

Calculate the p-value using Excel’s function if necessary.

Perform a two-tailed, two-sample z test for ${\pi }_1={\pi }_2$ and verify that $z^2$ is the same as the chi-square statistics.

Answer to Problem 21CE

The null hypothesis is:

$H_0:$ Grade and order handed in are independent.

And the alternative hypothesis is:

$H_1:$ Grade and order handed in are not independent.

The degrees of freedom for the contingency table is 1.

The critical-value using EXCEL is 2.706.

There is no enough evidence to conclude that grade and order handed in are independent.

The p-value for the hypothesis test is 0.382.

The conclusion is not sensitive to the level of significance chosen.

The cells (1, 1) and (1, 2) contribute the most to the chi-square test statistic.

There is no expected frequencies that are too small.

It is verified that $z^2$ is the same as the chi-square statistics.

Explanation of Solution

Calculation:

The table summarizes the grade and order of papers handed in.

The claim is to test whether the data provide sufficient evidence to conclude that the grade and order handed in are independent. If the claim is rejected, then the grade and order handed in are not independent.

The test hypotheses are given below:

Null hypothesis:

$H_0:$ Grade and order handed in are independent.

Alternative hypothesis:

$H_1:$ Grade and order handed in are not independent.

The degrees of freedom can be obtained as follows:

$\text{d}\text{.f}=\left(r-1\right)\left(c-1\right)$

Substitute 2 for r and 2 for c.

$\begin{array}{c}\text{d}\text{.f }=\left(2-1\right)\left(2-1\right)\\ =\left(1\right)\left(1\right)\\ =1\end{array}$

Thus, the degrees of freedom for the contingency table is 1.

Procedure for critical-value using EXCEL:

Step-by-step software procedure to obtain critical-value using EXCEL software is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=CHISQ.INV.RT(0.10,1)”
• Output using EXCEL software is given below:

Thus, the critical-value using EXCEL is 2.706.

Test statistic:

Software procedure:

Step by step procedure to obtain the chi-square test statistics and p-value using the MINITAB software:

• Choose Stat > Tables >Cross Tabulation and Chi-Square.
• Choose Row data (categorical variables).
• In Rows, choose Grade.
• In Columns, choose Order handed in.
• In Frequencies, choose Count.
• In Display, select Counts.
• In chi-square, select Chi-square test, Expected cell counts and Each cell’s contribution to chi-square.
• Click OK.

Output using the MINITAB software is given below:

Thus, the test statistic is 0.764 and the p-value for the hypothesis test is 0.382.

  Rejection rule:

If the p-value is less than or equal to the significance level, then reject the null hypothesis $H_0$. Otherwise, do not reject $H_0$.

Conclusion:

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

That is, $p\text{-value}\left(=0.382\right)>\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is not rejected.

Thus, the data provide sufficient evidence to conclude that the grade and order handed in are independent.

Take $\alpha =0.05$.

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

That is, $p\text{-value}\left(=0.382\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

Thus, the data provide sufficient evidence to conclude that the grade and order handed in are independent.

Thus, the conclusion is same for both the significance levels.

Hence, the conclusion is not sensitive to the level of significance chosen.

The cells (1, 1) and (1, 2) contribute most to the chi-square test statistic.

Since all $e_j\ge 2$, there is no expected frequencies that are too small.

Two-tailed, two-sample z test:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\pi }_1-{\pi }_2=0$

Alternative hypothesis:

$H_1:{\pi }_1-{\pi }_2\ne 0$

The proportion of students who finish an exam first and get “B or better” on the exam is denoted as $p_1$ and is obtained as follows:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{11}{25}\\ =0.44\end{array}$

Where $x_1$ is the number of students who finish an exam first and get “B or better” on the exam, $n_1$ is the number of students who finish an exam first.

The proportion of students who finish later and get “B or better” on the exam is denoted as $p_2$ and is obtained as follows:

$\begin{array}{c}{p}_2=\frac{x_2}{n_1}\\ =\frac{8}{25}\\ =0.32\end{array}$

Where $x_2$ is the number of students who finish later and get “B or better” on the exam and $n_2$ is the number of students who finish an exam later.

The pooled proportion is denoted as $\overline{p}$ obtained as follows:

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{11+8}{25+25}\\ =\frac{19}{50}\\ =0.38\end{array}$

Test statistic:

The z-test statistics can be obtained as follows:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{\overline{p}\left(1-\overline{p}\right)\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.44-0.32}{\sqrt{\left(0.38\right)\left(1-0.38\right)\frac{1}{25}+\frac{1}{25}}}\\ =\frac{0.12}{\sqrt{\left(0.38\right)\left(0.62\right)\left(0.08\right)}}\\ =\frac{0.12}{\sqrt{\left(0.018848\right)}}\\ =\frac{0.12}{0.137288}\\ =0.8740\end{array}$

Thus, the z-test statistic is 0.87.

The square of the z-test statistic is,

$\begin{array}{c}{z}^2=0.87\times 0.87\\ =0.7569\\ \simeq 0.76\end{array}$

Thus the square of the z-test statistic is same as the chi-square statistics.

Procedure for p-value using EXCEL:

Step-by-step software procedure to obtain p-value using EXCEL software is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=2*(1-NORM.S.DIST(0.8740,1))”
• Output using EXCEL software is given below:

Thus, the p-value using EXCEL is 0.382, which is also same as the p-value obtained in chi-square test.

Thus, it is verified that $z^2$ is the same as the chi-square statistics.