Question

Charles Darwin, author of On the Origin of Species (1859), designed an experiment to compare the effects of cross-fertilization and self fertilization on the size of plants. He planted pairs of very similar seedling plants, one self-fertilized and one cross-fertilized, in each of pot$15$at the same time. After a period of time, Darwin measured the heights (in inches) of all the plants. Here are the data:

(a) Explain why it is not appropriate to perform a paired t test in this setting.

(b) A hasty student generates the Minitab output shown below. What conclusion should he draw at the $\alpha =0.05$significance level? Explain

Short answer

(a)Both variables are taken as independent variable. Thus, to test the comparison, will utilize independent t-test.

(b)There is no sufficient evidence to infer that there is any difference among cross and self.

Step-by-step solution

part (a) Step 1: Given information

Given in the question that, Charles Darwin, author of On the Origin of Species (1859), designed an experiment to compare the effects of cross-fertilization and self fertilization on the size of plants. He planted pairs of very similar seedling plants, one self-fertilized and one cross-fertilized, in each of 15 pots at the same time. After a period of time, Darwin measured the heights (in inches) of all the plants. Here are the data

Part (a) Step 2: Explanation

It is given that the designed experiment is going to comparison between the effects of cross-fertilization and self-fertilization on the size of plants.

As per the given data set, there are two variables cross and self.

Since, a paired t-test is valuable for dependent variable (or both experiments has been done on similar land or same items).

Since, both variables are taken as independent variable. Thus, to test the comparison, will utilize independent t-test.

Part (b) Step 1: Given information

Charles Darwin, author of On the Origin of Species (1859), designed an experiment to compare the effects of cross-fertilization and self-fertilization on the size of plants. He planted pairs of very similar seedling plants, one self-fertilized and one cross-fertilized, in each of pots at the same time. After a period of time, Darwin measured the heights (in inches) of all the plants. Here are the data

Part (b) Step 2: Explanation

As per the given table,

The null and alternative hypotheses are:

$H_0:\mu =0$

$H_1:\mu \ne 0$

Variable	$\mathrm{N}$	Mean	St Dev	SE Mean	$\mathrm{T}$	$\mathrm{P}$
Diff	$15$	$2.61$	$4.71$	$1.22$	$2.14$	$0.050$

When the $p$-value does not equal the level of significance, the null hypothesis is rejected. The $p$-value $(0.050)$ isn't quite at the $0.050$ threshold of significance. The null hypothesis is not refuted in this way. As a result, there is insufficient evidence to conclude that there is a distinction between cross and self.