Chapter 11.1, Problem 18P

(i)

To determine

To graph: A cluster bar graph showing the percentages of Congress members from each party who spent each designated amount in their respective home districts.

Explanation of Solution

Calculation: To find the percentage table:

Party	Less than 5 Billion	5 to 10 Billion	More than 10 Billion
Democratic	$\frac{8}{45}\times 100\approx 18$	$\frac{15}{45}\times 100\approx 33$	$\frac{22}{45}\times 100\approx 49$
Republican	$\frac{12}{47}\times 100\approx 26$	$\frac{19}{47}\times 100\approx 40$	$\frac{16}{47}\times 100\approx 34$

Graph: To create cluster bar graph by using Excel is as follows:

Step 1: Enter the percentage data table in Excel worksheet.

Step 2: Select table and go to Insert > Charts > Column.

The cluster bar graph is obtained as:

(ii)

(a)

To determine

The level of significance and state the null and alternative hypotheses.

Answer to Problem 18P

Solution: The level of significance is 0.01. $H_0$: Same proportion of Democrats and Republicans within each spending level, $H_1$: Different proportion of Democrats and Republicans within each spending level.

Explanation of Solution

The level of significance, $\alpha =0.01$.

The null hypothesis for testing is defined as,

$H_0$: Same proportion of Democrats and Republicans within each spending level.

The alternative hypothesis is defined as,

$H_1$: Different proportion of Democrats and Republicans within each spending level.

(b)

To determine

To test: Whether all the expected frequencies are greater than 5. Also determine the value of chi-square statistic for the sample, the sampling distribution that should be used and degrees of freedom for the test.

Answer to Problem 18P

Solution: The value of chi-square statistic for the sample, ${\chi }^2=2.176$, expected frequencies are greater than 5, we should use chi-square distribution and degrees of freedom are 2.

Explanation of Solution

Calculation: The find the ${\chi }^2$ statistic for the sample by using MINITAB software is as:

Step 1: Go to Stat >Tables> Chi-square Test For Association.

Step 2: Select ‘Summarized data in two-way table’ and select <5B, 5-10B, >10Bin ‘Columns containing the table’ box.

Step 3: Select ‘Partyin ‘Rows’ and write any name in ‘Columns’ and click on ‘Statistics’ tick on Chi-square test and Expected cell counts. Then click on OK.

The Minitab output is:

Chi-Square Test for Association: Party, spent

Rows: Party   Columns: spent

	<5B	5-10B	>10B
Democratic	9.78	16.63	18.59
Republican	10.22	17.37	19.41

Cell Contents

      Expected count

Chi-Square Test

	Chi-Square	DF	P-Value
Pearson	2.176	2	0.337
Likelihood Ratio	2.185	2	0.335

Therefore, the obtained Chi-square test statistics is 2.176.

The obtained expected frequencies are:

Party	<5B	5-10B	>10B
Democratic	9.78	16.63	18.59
Republican	10.22	17.37	19.41

So, all expected frequencies are greater than 5.

The chi-square distribution should be used in this study and the obtained degrees of freedom are 2.

(c)

To determine

The P-value of the sample statistic.

Answer to Problem 18P

Solution: The P-value of sample statistic is 0.337.

Explanation of Solution

Calculation: The Minitab output obtained in above part (b) also gives the P-value. So, the P-value for the sample test statistic is 0.337.

(d)

To determine

To explain: Whether we will reject or fails to reject the null hypothesis.

Answer to Problem 18P

Solution: We failed to reject the null hypothesis at significance level 0.05.

Explanation of Solution

The obtained results in part (a), (b) and (c) are,

$\alpha =0.01$

${\chi }^2\text{statistic}=2.176$ with 2 degrees of freedom.

Since the P-value (0.337) is greater than 0.01, hence we failed to reject the null hypothesis of independence at $\alpha =0.01$. Therefore it is concluded that the data are not statistically significant at 0.01 level of significance.

(e)

To determine

To explain: The conclusion in the context of application.

Answer to Problem 18P

Solution: It is concluded that the Stone tools construction material and sitearenot independent.

Explanation of Solution

From above part, it can be seen that we failed to reject the null hypothesis at $\alpha =0.01$ and we can conclude that the data are not statistically significant.

Therefore, at the 1% level of significance, there is sufficient evidence to conclude that congressional members of each political party spent designated amounts in the same proportions.