Chapter 11.1, Problem 12E

To determine

The test for the given data at the significance level of the proposed generic model.

Answer to Problem 12E

  $P$ -value is greater than the significance level which means that fail to reject $H_0$ .

Explanation of Solution

Given:

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

Significance level: $5\%=0.05$

Ratio: $R_1:R_2:\text{}{W}_1:\text{}{W}_2=\text{}9:3:3:1$

  $n=4$

Concept Used:

Null hypothesis: $H_0$

The null hypothesis is rejected when the value of $P$ is less or equal to the significance level.

Calculation:

The null hypothesis: $H_0$

The phenotype occur in a ratio of $9:3:3:1$

  $P_{tc}=\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$16$}\right.,P_{tp}=\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$16$}\right.,P_{dc}=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$16$}\right.,P_{dp}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$16$}\right.$

Alternative hypothesis: $H_1$

At least one of the offspring proportion is incorrect.

Now,

Take the sum: $S=9+3+3+1=16$

Draw a table consisting of observed value and expected value:

The expected value is calculated as:

  $\begin{array}{l}{E}_1=T\times \frac{R_1}{R}=1611\times \frac{9}{16}=906.19\\ E_2=T\times \frac{R_2}{R}=1611\times \frac{3}{16}=302.06\\ E_3=T\times \frac{W_1}{R}=1611\times \frac{3}{16}=302.06\\ E_4=T\times \frac{W_2}{R}=1611\times \frac{1}{16}=100.69\end{array}$

Phenotype	Tall cut	Tall potato	Dwarf Cut	Dwarfpotato	Total
Expected value $O_i$	$926$	$288$	$293$	$104$	$1611$
Expected value $E_i$	$906.19$	$302.06$	$302.06$	$100.69$	$1611$

Now, use the test −statistics:

  $\begin{array}{l}{x}^2={\displaystyle \sum _{i=1}^4\frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ x^2=\frac{{\left(926-906.19\right)}^2}{906.19}+\frac{{\left(288-302.06\right)}^2}{302.06}+\frac{{\left(293-302.06\right)}^2}{302.06}+\frac{{\left(104-100.69\right)}^2}{100.69}\\ x^2=1.468\end{array}$

Degree of freedom:

  $df=n-1=4-1=3$

For $x^2=1.468$ and $df=3$ the $P$ -value is $0.6900$ .

So, fail to reject the null hypothesis because $P$ -value is not less or equal to, it is greater than the significance level which means that there is no difference in the distribution of generic traits.

  $P>\propto :\text{}0.6900>\text{}0.05$ : fail to reject $H_0$ .

Hence, there is no enough proof which shows that the proposed generic model of tomato plant does not obey the given standard ratio.

Conclusion:

Therefore, $P>\propto :\text{}0.6900>\text{}0.05$