Q.18 - The Practice of Statistics for AP

Question

The paper “Linkage Studies of the Tomato” (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio $9:3:3:1$.

The data were produced in such a way that the Random and Independent conditions are met. Carry out a chi-square goodness-of-fit test using these data. What do you conclude? 

Step-by-Step Solution

Answer

There is sufficient evidence to conclude that phenotypes are occurring in ratio of $9: 3: 3: 1$

Step 1: Given information

Given the nature of the query, The following data on phenotypes arising from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes was published in the publication "Linkage Studies of the Tomato" (Transactions of the Canadian Institute, 1931). We want to see if the following frequencies are in accordance with genetic principles, which specify that phenotypes should occur in a $9:3:3:1$ ratio.

Using these data, we must do a chi-square goodness-of-fit test.

Step 2: Explanation

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

Ratio is $9:3:3:1.$

The test statistic will be calculated using the formula

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

The proportions are:

$$\begin{gathered} p_1=\frac{9}{9+3+3+1} \\ =0.5625 \end{gathered}$$

$p_2=\frac{3}{16}$

$p_3=\frac{3}{16}$

$p_4=\frac{5}{16}$

The null and alternative hypotheses is given below:

$H_0:p_1$

$H_0:p_2$

$H_0:p_3$

$H_0:p_4$

$H_0:p_5$

$H_a:$ At least one of $p_i$is different

Step 3: Calculation for test statistic

The test statistic is calculated as follows:

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

The test statistic is:

$$\begin{gathered} {\chi }^2=\sum \frac{(O-E{)}^2}{E} \\ =1.4687 \end{gathered}$$

The degree of freedom is calculated as:

$$\begin{gathered} Degree of freedom=Number of categories-1 \\ =2-1 \\ =1 \end{gathered}$$

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 $9: 3: 3: 1$