Chapter 11.1, Problem 18E

To determine

To check: Whether there is sufficient evidence to conclude that phenotypes is occurring in ratio of 9:3:3:1.

Answer to Problem 18E

Yes, there is sufficient evidence to conclude that phenotypes are occurring in ratio of 9:3:3:1.

Explanation of Solution

Given:

Table is:

Phenotype	Frequency
Tall cut	926
Tall potato	288
Dwarf cut	293
Dwarf potato	104

Ratio is 9:3:3:1.

Formula used:

The formula to compute the test statistic is

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

Calculation:

The proportions are:

  $\begin{array}{l}{p}_1=\frac{9}{9+3+3+1}=0.5625\\ p_2=\frac{3}{16}\\ p_3=\frac{3}{16}\\ p_4=\frac{5}{16}\end{array}$

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2=p_3=p_4=p_5\\ H_a:\text{ At least one of }{p}_i\text{ is different}\end{array}$

The calculation for test statistic is done as:

Observed value	Expected value	$\left(O-E\right)$	${\left(O-E\right)}^2$	${\left(O-E\right)}^2/E$
926	906.1875	19.8125	392.5352	0.4332
288	302.0625	14.0625	197.7539	0.6547
293	302.0625	-9.0625	82.1289	0.2719
104	100.6875	3.3125	10.9727	0.109
				${\displaystyle \sum {\frac{\left(O-E\right)}{E}}^2=1.4687}$

The test statistic is:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum {\frac{\left(O-E\right)}{E}}^2}\\ =1.4687\end{array}$

The degree of freedom is calculated as:

  $\begin{array}{c}\text{Degree of freedom =Number of categories - 1}\\ =2-1\\ =1\end{array}$

The p-value using chi-square table at 1 degree of freedom is 0.6895

The p -value is above significance level. The null hypothesis does not get rejected. Thus, there is sufficient evidence to conclude that phenotypes are occurring in ratio of 9:3:3:1.