Question

Gregor Mendel $(1822-1884)$, an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of $3$smooth peas for every $1$wrinkled pea. In one experiment, he observed $423$smooth and $133$wrinkled peas. Assume that the conditions for inference are met.

a. . State appropriate hypotheses for testing Mendel’s claim about the true proportion of smooth peas.

b. Calculate the standardized test statistic and P-value.

c. Interpret the P-value. What conclusion would you make?

Short answer

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

The null hypothesis: $H_0:p_1=0.75$

The alternate hypothesis: $H_a:p_1\ne 0.75$

Part b. The $p-value$value and standardized test statistics are $0.25,0.3453$respectively.

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

Step-by-step solution

Part a. Step 1. 

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

Observed value: $3$smooth and wrinkled peas

Part a. Step 2. Explanation

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

$$\begin{gathered} H_0:p_1=\frac{3}{3+1}=0.75 \\ {p}_2=1-p_1=1-0.75=0.25 \end{gathered}$$

or $H_0:p_1=0.75$

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

$$\begin{gathered} H_a:atleast{p}_{1}isincorrect \\ {H}_a:p_1\ne 0.75 \end{gathered}$$

Hence, null hypothesis and alternate hypothesis are stated above.

Part b. Step 1. Given information

$$\begin{gathered} p_1=0.75 \\ {p}_2=0.25 \\ {O}_1=423 \\ {O}_2=133 \\ \infty =0.05 \end{gathered}$$

Part b. Step 2. Explanation

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

For the first distribution:

$$\begin{gathered} n=556 \\ {p}_1=0.75 \\ {O}_1=423 \end{gathered}$$

Expected frequency is given by:

$E_1=n{p}_1=556\times 0.75=417$

The chi-square subtotal is given by:

$$\begin{gathered} {X^2}_{sub1}=\frac{{(O_1-E_1)}^2}{E_1} \\ =\frac{{(423-417)}^2}{417} \\ =0.0863 \end{gathered}$$

For the second distribution:

$$\begin{gathered} n=556 \\ {p}_2=0.25 \\ {O}_2=133 \end{gathered}$$

Expected frequency is given by:

$$\begin{gathered} E_2=n{p}_2 \\ =556\times 0.25 \\ =139 \end{gathered}$$

The chi-square subtotal is given by:

$$\begin{gathered} {X^2}_{sub2}=\frac{{(O_2-E_2)}^2}{E_2} \\ =\frac{{(133-139)}^2}{139} \\ =0.259 \end{gathered}$$

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

$$\begin{gathered} X^2={X^2}_{sub1}+{X^2}_{sub2} \\ =0.0863+0.259 \\ =0.3453 \end{gathered}$$

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 $df=1,theP-value$is:

$P>0.25$

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

Part c. Step 1. Explanation

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.

$$\begin{gathered} P>\infty \\ 0.25>0.05 \end{gathered}$$

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.

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