Chapter 21, Problem 21.23E

(a)

To determine

To explain:

Theconditions for a chi-square goodness of fit test of the given law.

Explanation of Solution

Given:

Phenotype	Yellow round	Yellow wrinkled	Green round	Green wrinkled
Count	$315$	$101$	$108$	$32$

Concept used:

Formula

Degree of freedom

  $=n-1$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

The sample is random.

The variable under steady is tall or dawn $f$ .

The expected value of number of observations in each level of the variable is at least $5$ .

All the conditions for chi-square goodness of fit test satisfied.

Since the sample size are large.

The condition for the two samples $z$ test are met.

(b)

To determine

To explain:

The null and alternative hypothesis fir this test of the given law.

Explanation of Solution

Given:

Phenotype	Yellow round	Yellow wrinkled	Green round	Green wrinkled
Count	$315$	$101$	$108$	$32$

Concept used:

Formula

Degree of freedom

  $=n-1$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Since the sample size are large.

The condition for the two samples $z$ test are met.

Null hypothesis $H_0$ :-No relationship between expected ratio and observed ratio.

Alternative hypothesis $H_1$ :-A significant relationship between expected ratio and observed ratio Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\\ \alpha =0.05\end{array}$

Since $p-$ Value is greater than $0.05$ significance level.

So the hypothesis is fail to reject null hypothesis $H_0$

(c)

To determine

To find:

The expected counts under the null hypothesis test of the given law.

Explanation of Solution

Given:

Phenotype	Yellow round	Yellow wrinkled	Green round	Green wrinkled
Count	$315$	$101$	$108$	$32$

Concept used:

Formula

Degree of freedom

  $=n-1$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Since the sample size are large.

The condition for the two samples $z$ test are met.

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

  $\begin{array}{l}{H}_0:pt=\frac{1}{16}\\ pv=\frac{3}{16}\\ p=\frac{9}{16}\\ H_a:H_0\text{is}\text{not}\text{true}\end{array}$

Since the sample size are large.

The condition for the two samples $z$ test are met.

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

Draw the table

	Category	Frequency	Observed counts	$E$
$1$	$\frac{1}{16}$	$315$	$5.31$	$-0.14$
$2$	$\frac{3}{16}$	$101$	$15.94$	$-0.23$
$3$	$\frac{3}{16}$	$108$	$15.94$	$-0.23$
$4$	$\frac{9}{16}$	$32$	$63.75$	$0.16$

(d)

To determine

To find:

The chi-square statistic and the p-valueof the given law.

Answer to Problem 21.23E

p-value is greater than $0.05$ .

Explanation of Solution

Given:

Tall plants $=787$

Dwarf plants $=277$

Concept used:

Formula

Degree of freedom

  $=n-1$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

  $H_0:{\pi }_1={\pi }_2={\pi }_3$

  $H_1:\text{not}\text{all}\text{of}\text{the}{\text{π}}_{\text{j}}\text{are}\text{equal}$

Since the sample size are large.

The condition for the two samples $z$ test are met.

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ {\chi }^2=\frac{{\left(787-798\right)}^2}{798}+\frac{{\left(277-266\right)}^2}{266}\\ {\chi }^2=0.6065\end{array}$

Chi-square $=0.47$

Degree of freedom

  $\begin{array}{l}=n-1\\ =\left(4-1\right)\\ =3\end{array}$

P-value $=0.4361$

The p-value of the test is $0.4361$

Since p-value is greater than $0.05$

So, the null hypothesis is fail to reject.

This result is not significant.

So, the data is not support the Mendel’s second law.