Question

An agricultural field trial compares the yield of two varieties of tomatoes for commercial use – one is the common variety used by farmers and the other is a new hybrid. The researchers plant each type of tomato on one half of each of 10 plots. At harvest time the difference in the yield of the hybrid tomato minus the common variety in pounds per plant (ppp) is calculated for each plot. The average difference is 1.05 ppp with a standard deviation of 1.23 ppp. In testing whether the hybrid tomato has a higher yield with a significance level of α = 0.05, what is the p-value? You may assume the conditions to carry out the relevant hypothesis test are satisfied.

Answer 1

Answer:

p-value 0.0124

Step-by-step explanation:

Hello!

In this experiment, 10 plots were randomly sampled and each plot was divided in a half, one half was planted with the common variety of tomato and the other half was planted with the new hybrid variety. At harvest time it was measured the pounds per plant harvested and the variable difference was established. This study variable is defined as the difference between the pounds per plant of the hybrid variety minus the pound per plant of the common variety. Symbolized: Xd: Xnew - Xcommon.

The purpose of this experiment is to test whether there is a difference between the yield of both species. This is a classic example of a paired test, in which you want to put two dependent samples to test. The easiest way to recognize this type of test, if it is not specified in the problem, is that both variables are measured in the same sampling unit. In this case, the sampling unit is "one plot" that was divided and both verities of tomatoes were planted on it.

This test is also called "paired samples T-test" and as a statistic, you have to use the Student t.

The data given is:

sample: 10 plots

sample mean: Xd[bar]: 1.05ppp

sample standard deviation Sd: 1.23ppp

If the hybrid has a higher yield, that would mean that the difference between them would be positive if the difference is positive, its safe to assume that the population mean of the difference will be positive as well. Symbolically: μd>0

With this, you can state the hypothesis as:

H₀:μd≤0

H₁:μd>0

α: 0.05

t= Xd[bar] - μd ~ $t_{n-1}$

       Sd/n

t$t_{obs}$= (1.05 - 0) / (1.23/√10) = 2.69

This is a one-tailed test with critical value $t_{9;0.95} = 1.83$

the p-value for this test is also one tailed.

p($t_{9}$≥2.69) = 1-P($t_{9}$<2.69) = 1 - 0.9876 = 0.0124

The decision is to reject the null hypothesis.

I hope it helps!