Chapter 11.2, Problem 30E

a)

To determine

To state the null and alternative hypotheses.

Answer to Problem 30E

  $H_0$ : The distribution of hatching is same for each water temperature.

  $H_a$ : The distribution of hatching is not same for each water temperature.

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

Calculation:

The null and alternative hypotheses:

  $H_0$ : The distribution of hatching is same for each water temperature.

  $H_a$ : The distribution of hatching is not same for each water temperature.

b)

To determine

To calculate expected counts.

Answer to Problem 30E

	Cold	Neutral	Hot
Yes	18.63	38.63	71.74
No	8.37	17.37	32.26

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

Formula:

  $E=\frac{r\times c}{n}$

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

Calculation:

Therefore, expected count is,

  $\begin{array}{l}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{129\times 27}{187}=18.63\\ .\\ .\\ .\\ E_{23}=\frac{r_2\times c_3}{n}=\frac{58\times 104}{187}=32.26\end{array}$

Therefore, table of expected count is,

	Cold	Neutral	Hot
Yes	18.63	38.63	71.74
No	8.37	17.37	32.26

c)

To determine

To calculate value of chi square test statistic.

Answer to Problem 30E

The value of chi square test statistic is 1.7033

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

Expected count is

	Cold	Neutral	Hot
Yes	18.63	38.63	71.74
No	8.37	17.37	32.26

Formula:

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

Calculation:

  $\begin{array}{l}{\chi }^2=\frac{{(16-18.63)}^2}{18.63}+\frac{{(38-38.63)}^2}{38.63}+\frac{{(75-71.74)}^2}{71.74}\\ +\frac{{(11-8.37)}^2}{8.37}+\frac{{(18-17.37)}^2}{17.37}+\frac{{(29-32.26)}^2}{32.26}\\ {\chi }^2=1.7033\end{array}$