Chapter 9.2, Problem 49E

a.

To determine

The appropriate hypotheses for the true proportion of the smooth peas described by the Mendel are to be explained.

Answer to Problem 49E

The appropriate hypotheses for the true proportion of the smooth peas described by the Mendel are:

The null hypothesis: $H_0:\text{}{p}_1=0.75$

The alternate hypothesis: $H_a:\text{}{p}_1\ne 0.75$

Explanation of Solution

Given:

Predicted value: there is ratio of $3$ smooth peas for every $3$ wrinkled pea

Observed value: $3$ smooth and $3$ wrinkled peas

The proportions are equal to the mentioned probabilities by the Mendel is stated by the null hypothesis:

  $\begin{array}{l}{H}_0:\text{}{p}_1=\frac{3}{3+1}=0.75\\ p_2=1-p_1=1-0.75=0.25\end{array}$

Or

  $H_0:\text{}{p}_1=0.75$

Exactly opposite to the null hypothesis is stated by the alternate hypothesis:

  $H_a$ : at least $p_1$ is incorrect.

Or

  $H_a:\text{}{p}_1\ne 0.75$

Conclusion:

Hence, null hypothesis and alternate hypothesis are stated above.

b.

To determine

The $P-$ value and standardized test statistics are to be estimated.

Answer to Problem 49E

The $P-$ value and standardized test statistics are $0.25,0.3453$ respectively.

Explanation of Solution

Given:

  $\begin{array}{l}{p}_1=0.75\\ p_2=0.25\\ O_1=423\\ O_2=133\\ \propto =0.05\end{array}$

Concept Used:

The expected frequencies are calculated as:

  $E=np$

Where,

Sample size: $n$

Probability: $p$

The chi-square subtotal is calculated as:

  $X^2{}_{sub}=\frac{{\left(O-E\right)}^2}{E}$

Where,

Observed frequency: $O$

Expected frequency: $E$

Calculation:

The sample size two categories $\left(c\right)$ is: $n=O_1+O_2=423+139=556$

For the first distribution:

  $\begin{array}{l}n=556\\ p_1=0.75\\ O_1=423\end{array}$

Expected frequency is given by:

  $E_1=n{p}_1=556\times 0.75=417$

The chi-square subtotal is given by:

  $X^2{}_{sub1}=\frac{{\left(O_1-E_1\right)}^2}{E_1}=\frac{{\left(423-417\right)}^2}{417}=0.0863$

For the second distribution:

  $\begin{array}{l}n=556\\ p_2=0.25\\ O_2=133\end{array}$

Expected frequency is given by:

  $E_2=n{p}_2=556\times 0.25=139$

The chi-square subtotal is given by:

  $X^2{}_{sub2}=\frac{{\left(O_2-E_2\right)}^2}{E_2}=\frac{{\left(133-139\right)}^2}{139}=0.259$

Distribution	Observed frequency $\left(O\right)$	Expected frequency $E=np$	$X^2{}_{sub}=\frac{{\left(O-E\right)}^2}{E}$
$0.75$	$423$	$417$	$0.0863$
$0.25$	$133$	$139$	$0.259$
$n$	$556$	SUM	$0.3453$

The value of the test-statistics is sum of all chi-square subtotals:

  $X^2=X^2{}_{sub2}+X^2{}_{sub2}=0.0863+0.259=0.3453$

Now the degree of the freedom of the experiment is the number of categories decreased by $1$ :

  $df=c-1=2-1=1$

From the table containing the $X^2$ -value in the row $df=1$ , the $P-$ value is:

  $P>0.25$

Conclusion:

By using the chi-square subtotals, value of the test-statistics is estimated.

c.

To determine

The meaning of the $P-$ value and conclusion of the claim of the Mendel’s belief is to be interpreted.

Answer to Problem 49E

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

Explanation of Solution

From the answer of above part:

  $P>0.25$

If the $P-$ value is less than or equal to the significance level then the null hypothesis is rejected.

  $\begin{array}{l}P>\propto \\ 0.25>0.05\end{array}$

And here the $P-$ value is greater than the significance level that means fail to reject null hypothesis.

As fail to reject the null hypothesis $H_0$ which clearly represents that there are no sufficient evidences are provided to reject the null hypothesis or to reject the claim described by the Mendel.

Conclusion:

Hence, the null hypothesis $H_0$ is not rejected.