Probability statistics and two tailed test

Question

A tomato farmer with a very large farm of approximately 2200 acres had heard about a new type of rather expensive fertilizer which would supposedly significantly increase his production. The frugal farmer wanted to test the new fertilizer before committing the large investment required to fertilize a farm of his size. He therefore selected 15 parcels of land on his property and divided them each into two portions. He bought just enough of the new fertilizer to spread over one half of each parcel and then spread the old fertilizer over the other half of each parcel. His yields in pounds per tomato plant were as follows:

Parcel New Fertilizer Old Fertilizer
1 14.2 14.0
2 14.1 13.9
3 14.5 14.4
4 15.0 14.8
5 13.9 13.6
6 14.5 14.1
7 14.7 14.0
8 13.7 13.7
9 14.0 13.3
10 13.8 13.7
11 14.2 14.1
12 15.4 14.9
13 13.2 12.8
14 13.8 13.8
15 14.3 14.0



The farmer had taken statistics many years ago when in college and consequently made a couple of mistakes when testing to find if the new fertilizer was more effective: (1) He tested the data as two independent samples, and (2) He performed a two-tailed test. He decided that he was unable to conclude that there was a difference between the two fertilizers.



What if you were the fertilizer sales representative and your job was to prove the superiority of the new product to the farmer?



(1) You should start by running the same test he did in which he came to the decision that he could not conclude a difference.



(2) Perform the test as it should have been done and find if you come to a different conclusion.



(3) Explain why the results were different and why your test was a stronger and more reliable test.

Answer 1

(1)

 

If you perform a two independent sample t-test  of 

   H_0f : _{mean yield under the new fertilizer = mean yield under the old fertilizer}

    vs

   H_1f : mean yield under the new fertilizer ≠ mean yield under the old fertilizer

 

you would get a t-statistic of 1.4239. The p-value for the two-sided test is 0.1655, which is greater than the traditional significance level of 0.05. Thus, the farmer's null hypothesis is not rejected.

 

Why the farmer did come to this conclusion?  The reason is that the farmer failed to take into consideration that there is dependency among the yields obtained in each parcel. The effect of ignoring the dependency is that is that the variability (standard error) for the estimate of the difference in the mean yield is inflated. 

 

(2)

 

One way of accounting for the dependency among yields obtained in each parcel is to run a paired t-test (or equivalently, a one-sample t-test on the differences in the yield). 

 

A t-test on the sample of differences in the yields (i.e. yield under new minus yield under old for each parcel) results a t-statistic of  4.8359, which has a p-value of 0.0002641. Thus, H_0f is rejected at 0.05 significance level. The reason for the higher t-statistic is due to the fact the standard error for the estimate of the difference in the mean yield has been correctly estimated. 

 

(3)

 

The effect of fertilizer is a within-parcel effect. In (2), we used a valid estimate of the standard error for the estimator of the within-parcel effect.  

 

 

 

What if you were the fertilizer sales representative and your job was to prove the superiority of the new product to the farmer?

 

For this one, run a one-sided paired t-test but with the  hypothesis test set up as follows

 

H_0s : mean yield under the new fertilizer ≤ mean yield under the old fertilizer
vs
H_1s : mean yield under the new fertilizer > mean yield under the old fertilizer

 

 

I hope this helps. 

 

Thanks,

Kidane