Chapter 11.2, Problem 33E

a.

To determine

To compute: The proportion of eggs in each group that hatched.

Answer to Problem 33E

	Hatched
Cold	0.3721
Neutral	0.4043
Hot	0.6279

The data support that the researchers believed that eggs would not hatch in cold water.

Explanation of Solution

Given:

	Eggs	Hatched
Cold	27	16
Neural	56	38
Hot	104	75
Total	187	129

Formula used:

The chi-square test statistics value:

  $E_i{}_j=\frac{\text{Row sum}\left(i\right)\times \text{Column Sum}\left(j\right)}{\text{Total sum}}$

  ${\chi }^2{}_{statistic}={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}$

  $\begin{array}{l}df=\left(n-1\right)\left(m-1\right)\\ p-value=P\left({\chi }^2>{\chi }^2{}_{test}\right)\end{array}$

Calculation:

The proportion of every group is calculated by dividing the number of infants by the row total.

The proportion of each group, can be calculated as:

	Eggs	Hatched
Cold	$\frac{27}{27+16}=0.6279$	$\frac{16}{27+16}=0.3721$
Neutral	$\frac{56}{56+38}=0.5957$	$\frac{38}{38+56}=0.4043$
Hot	$\frac{104}{104+75}=0.5810$	$\frac{27}{27+16}=0.6279$

Thus, the proportion of eggs in each group that hatch is,

	Hatched
Cold	0.3721
Neutral	0.4043
Hot	0.6279

The proportion (or probability) that the eggs hatched in old water, is 0.3721, which is less than 0.50. Therefore, the data support that the researchers believed that eggs would not hatch in cold water.

b.

To determine

To explain: There is a statistically significant difference between the three groups or not.

Explanation of Solution

Calculation:

To test whether there is a relationship between any of groups or not.

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:\text{There is no difference between the groups}\\ H_1:\text{There is a difference between the groups}\end{array}$

From the given table,

The expectation ( E_i ),

	Eggs	Hatched
Cold	$\frac{43\times 187}{316}=25.4462$	$\frac{43\times 129}{316}=17.5538$
Neural	$\frac{94\times 187}{316}=55.6266$	$\frac{94\times 129}{316}=38.3734$
Hot	$\frac{179\times 187}{316}=105.9272$	$\frac{129\times 179}{316}=73.0729$

The chi-square test statistics value can be calculated as:

  $\begin{array}{c}{\chi }^2{}_{statistic}=\left(\begin{array}{l}\frac{{\left(27-25.4462\right)}^2}{25.4462}+\frac{{\left(16-17.5538\right)}^2}{17.5538}+\frac{{\left(56-55.6266\right)}^2}{55.6266}\\ +\frac{{\left(38-38.3734\right)}^2}{38.3734}+\frac{{\left(104-105.9272\right)}^2}{105.9272}+\frac{{\left(75-73.0729\right)}^2}{73.0729}\end{array}\right)\\ =0.3244\end{array}$

Thus, the test statistic value is 0.3244.

The p-value, at 2 df, can be calculated as:

  $\begin{array}{c}p-value=P\left({\chi }^2{}_{\left(\frac{L}{2},df\right)}>\left|0.3244\right|\right)\\ =0.8502\end{array}$

Reject the null hypothesis when the p-value is less than the less of significance.

Here, the p-value (0.8502) is greater than the less of significance (0.05).

Thus, the null hypothesis is not rejected. Therefore, the researcher has no enough evidence to conclude that there is difference between the three groups.