Chapter 11.2, Problem 29E

a)

To determine

To state the null and alternative hypotheses.

Answer to Problem 29E

  $H_0$ : The distribution of candy is same for each survey type.

  $H_a$ : The distribution of candy is not same for each survey type.

Explanation of Solution

Given:

		Survey type
		Red	Blue	Control	Total
Color of candy	Red	13	5	8	26
	Blue	7	15	12	34
	Total	20	20	20	60

Calculation:

The null and alternative hypotheses:

  $H_0$ : The distribution of candy is same for each survey type.

  $H_a$ : The distribution of candy is not same for each survey type.

b)

To determine

To calculate expected counts.

Answer to Problem 29E

	Red	Blue	Control
Red	8.6667	8.6667	8.6667
Blue	11.3333	11.3333	11.3333

Explanation of Solution

Given:

		Survey type
		Red	Blue	Control	Total
Color of candy	Red	13	5	8	26
	Blue	7	15	12	34
	Total	20	20	20	60

Formula:

  $E=\frac{r\times c}{n}$

Where, r = row total, c = column total and n = grand total.

Calculation:

Therefore, expected count is,

  $\begin{array}{l}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{26\times 20}{60}=8.6667\\ .\\ .\\ .\\ E_{23}=\frac{r_2\times c_3}{n}=\frac{34\times 20}{60}=11.3333\end{array}$

Therefore, table of expected count is,

	Red	Blue	Control
Red	8.6667	8.6667	8.6667
Blue	11.3333	11.3333	11.3333

c)

To determine

To calculate value of chi square test statistic.

Answer to Problem 29E

The value of chi square test statistic is 6.6516

Explanation of Solution

Given:

		Survey type
		Red	Blue	Control	Total
Color of candy	Red	13	5	8	26
	Blue	7	15	12	34
	Total	20	20	20	60

Expected count is

	Red	Blue	Control
Red	8.6667	8.6667	8.6667
Blue	11.3333	11.3333	11.3333

Formula:

  ${\chi }^2={\displaystyle \sum \frac{{(O-E)}^2}{E}}$

Calculation:

  $\begin{array}{l}{\chi }^2=\frac{{(13-8.6667)}^2}{8.6667}+\frac{{(5-8.6667)}^2}{8.6667}+\frac{{(8-8.6667)}^2}{8.6667}\\ +\frac{{(7-11.3333)}^2}{11.3333}+\frac{{(12-11.3333)}^2}{11.3333}+\frac{{(12-11.3333)}^2}{11.3333}\\ {\chi }^2=6.6516\end{array}$