Chapter 11: Q.17 (page 694)

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. The data were produced in such a way that the Random and Independent conditions are met. Carry out a chi-square goodness-of-fit test based on Mendel’s prediction. What do you conclude?

Short Answer

Expert verified

There is insufficient evidence to reject the Mendel's belief.

Step by step solution

Given information

Given the question that, Mendel predicted a ratio of three smooth peas to one wrinkled pea. In one experiment, he discovered 423 smooth and 133 wrinkled peas. The data were generated in such a way that the requirements for Random and Independent were met.

A chi-square goodness-of-fit test based on Mendel's prediction is required.

Explanation

The test statistic's formula will be

$χ^{2} = \sum \frac{(O - E)^{2}}{E}$

These are the proportions:

localid="1650620228150" $p_{2} = 1 - 0.75 = 0.25$

The following are the null and alternative hypotheses:

$H_{0} : p_{1} = p_{2}$

$H_{a}$ : At least one of $p_{i}$ is different

The test statistic is calculated as follows:

Observed value	Expected value	(O-E)	${(O - E)}^{2}$	$\frac{{(O - E)}^{2}}{E}$
$423$	$417$	$6$	$36$	$0.0863$
$133$	$139$	$- 6$	$36$	$0.259$
				$\sum \frac{(O - E)^{2}}{E} = 0.3453$

Test statistic.

The test statistic is:

$χ^{2} = \sum \frac{(O - E)^{2}}{E} = 0.3453$

The degree of freedom is calculated as:

$D e g r e e o f f r e e d o m = N u m b e r o f c a t e g o r i e s - 1 = 2 - 1 = 1$

The $p$ -value using chi-square table at 1 degree of freedom is $0.3453 .$

The significance level has been exceeded by the P-value. The null hypothesis remains viable. As a result, there is little evidence to disprove Mendel's hypothesis.