Chapter 15, Problem 30CE

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 0.05 level of significance.

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.

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 30CE

The null hypothesis is:

$H_0:$ Correct response and type of cola are independent.

And the alternative hypothesis is:

$H_1:$ Correct response and type of cola are not independent.

The degrees of freedom for the contingency table is 1.

The critical-value using EXCEL is 3.841.

The chi-square test statistics at 0.05 level of significance is 1.50.

The p-value for the hypothesis test is 0.221.

There is enough evidence to conclude that the correct response and type of cola are independent.

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

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(0.05,1)”
• Output using EXCEL software is given below:

Thus, the critical-value using EXCEL is 3.841.

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 1.50 and the p-value for the hypothesis test is 0.221.

  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, $p\text{-value}\left(=0.221\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

Thus, the data provide sufficient evidence to conclude that the correct response and type of cola are independent.

Take $\alpha =0.01$

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

That is, $p\text{-value}\left(=0.221\right)>\alpha \left(=0.01\right)$.

Therefore, the null hypothesis is not rejected.

Thus, the data provide sufficient evidence to conclude that the correct response and type of cola 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 the 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 “yes” responses to the regular cola is denoted as $p_1$ and is obtained as follows:

$\begin{array}{c}{p}_1=\frac{x_1}{n_1}\\ =\frac{6}{24}\\ =0.25\end{array}$

Where $x_1$ is the number of “yes” responses to the regular cola, $n_1$ is the number of response for regular cola.

The proportion of number of “yes” responses to the diet cola is denoted as $p_2$ and is obtained as follows:

$\begin{array}{c}{p}_2=\frac{x_2}{n_1}\\ =\frac{10}{24}\\ =0.4167\end{array}$

Where $x_2$ is the number of “yes” responses to the diet cola and $n_2$ is the number of response for diet cola.

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

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{6+10}{24+24}\\ =\frac{16}{48}\\ =0.333\end{array}$

Test statistic:

The z-test statistics can be obtained as follows:

$\begin{array}{c}{z}_{\text{calc}}=\frac{p_1-p_2}{\sqrt{\overline{p}\left(1-\overline{p}\right)\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.25-0.4167}{\sqrt{\left(0.333\right)\left(1-0.333\right)\frac{1}{24}+\frac{1}{24}}}\\ =\frac{-0.1667}{\sqrt{\left(0.333\right)\left(0.667\right)\left(0.083\right)}}\\ =\frac{-0.1667}{\sqrt{\left(0.018435213\right)}}\\ =\frac{-0.1667}{0.135776334}\\ =-1.22\end{array}$

Thus, the z-test statistic is –1.22.

The square of the z-test statistic is,

$\begin{array}{c}{z}^2=\left(-1.22\right)\times \left(-1.22\right)\\ =1.4884\\ \simeq 1.5\end{array}$

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(–1.22,1))”
• Output using EXCEL software is given below:

Thus, the p-value using EXCEL is 1.778, which is not same as the p-value obtained in chi-square test. But the square of the z-test statistic is same as the chi-square statistics.

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