Chapter 21, Problem 21.22E

(a)

To determine

To explain:

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

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:

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:

Thenull and alternative hypothesis fir this test of the given law.

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:

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.

Answer to Problem 21.22E

  $798$ , $266$

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:

Since the sample size are large.

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

Applying chi-square test of independence

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

E (Tall)

  $\begin{array}{l}=\frac{3}{4}\times 1064\\ =798\end{array}$

E (short)

  $\begin{array}{l}=\frac{1}{4}\times 1064\\ =266\end{array}$

(d)

To determine

To find:

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

Answer to Problem 21.22E

The P-value $=0.4361$

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 $=0.6065$

Degree of freedom

  $\begin{array}{l}=n-1\\ =\left(2-1\right)\\ =1\end{array}$

P-value $=0.4361$

This result is not significant.

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