Chapter 11.2, Problem 34E

a.

To determine

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

Answer to Problem 34E

	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:

Data set it can be written as:

	Male	Female	Total
PBM	9	11	20
NLCP	8	11	19
PL-LCP	8	11	19
TG-LCP	10	9	19
Total	35	42	77

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:

	Male	Female
PBM	$\frac{9}{20}=0.45$	$\frac{11}{20}=0.55$
NLCP	$\frac{8}{19}=0.4211$	$\frac{11}{19}=0.5789$
PL-LCP	$\frac{8}{19}=0.4211$	$\frac{11}{19}=0.5789$
TG-LCP	$\frac{10}{19}=0.5260$	$\frac{9}{19}=0.4737$

Thus, the proportion of in each treatment is,

	Female
PBM	0.55
NLCP	0.5789
PL-LCP	0.5789
TG-LCP	0.4737

In most of the categories proportion of females are greater than the categories proportion of male.

Thus, it is unbalanced between the groups by the gender.

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 ),

	Male	Female
PBM	9.090909	10.90909
NLCP	8.636364	10.36364
PL-LCP	8.636364	10.36364
TG-LCP	8.636364	10.36364

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

  $\begin{array}{c}{\chi }^2{}_{statistic}=\left(\begin{array}{l}\frac{{\left(9-9.0909\right)}^2}{9.0909}+\frac{{\left(11-10.9091\right)}^2}{10.9091}+\frac{{\left(8-8.6364\right)}^2}{8.6364}\\ +\frac{{\left(11-10.3636\right)}^2}{10.3636}+\frac{{\left(8-8.6364\right)}^2}{8.6364}+\frac{{\left(11-10.3636\right)}^2}{10.3636}\\ +\frac{{\left(10-8.3636\right)}^2}{8.3636}+\frac{{\left(9-10.3636\right)}^2}{10.3636}\end{array}\right)\\ =0.9036\end{array}$

Thus, the test statistic value is 0.9036.

The p-value, at 3 df, can be calculated as:

  $\begin{array}{c}p-value=P\left({\chi }^2{}_{\left(\frac{L}{2},df\right)}>\left|0.9036\right|\right)\\ =0.1754\end{array}$

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

Here, the p-value (0.1754) 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 a relationship between any of groups.