A tomato farmer with a very large farm of approximately 2200 acres had heard abo
ID: 3224845 • Letter: A
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?
You should start by running the same test he did in which he came to the decision that he could not conclude a difference.
Perform the test as it should have been done and find if you come to a different conclusion.
Explain why the results were different and why your test was a stronger and more reliable test.
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
Explanation / Answer
Solutions:
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.
The farmer is right to test the data as two independent samples. The reason is that the samples from the two fertilizers are not related. By conducting the two tailed test, the results are as follows:
The hypothesis is that the means of two populations are equal hence
By running the test, the outcome is
t-Test: Two-Sample Assuming Unequal Variances
Variable 1
Variable 2
Mean
14.22
13.94
Variance
0.301714
0.278286
Observations
15
15
Hypothesized Mean Difference
0
df
28
t Stat
1.423933
P(T<=t) one-tail
0.082758
t Critical one-tail
1.701131
P(T<=t) two-tail
0.165515
t Critical two-tail
2.048407
Since t Stat is not greater that t Critical two-tail, we do not reject null hypothesis meaning the observed difference in means (14.22-13.94) is not convinving to say the average yield between the two fertilizers differe significantly.
2.Perform the test as it should have been done and find if you come to a different conclusion.
The farmer should have instead conducted a pooled t-test. This means that he should have compared the means using two-sample t-test with the assumption that the variances are equal and the results are as follows:
t-Test: Two-Sample Assuming Equal Variances
Variable 1
Variable 2
Mean
14.22
13.94
Variance
0.301714
0.278286
Observations
15
15
Pooled Variance
0.29
Hypothesized Mean Difference
0
df
28
t Stat
1.423933
P(T<=t) one-tail
0.082758
t Critical one-tail
1.701131
P(T<=t) two-tail
0.165515
t Critical two-tail
2.048407
The conclusion is that p value>0.05 and this means that the null hypothesis is retained meaning that the average yield between the two fertilizers differ significantly.
3.Explain why the results were different and why your test was a stronger and more reliable test.
It is because the variances in the test I conducted were pooled and assumed as equal while the variances in the test conducted by the farmer were not pooled and were assumed as unequal
t-Test: Two-Sample Assuming Unequal Variances
Variable 1
Variable 2
Mean
14.22
13.94
Variance
0.301714
0.278286
Observations
15
15
Hypothesized Mean Difference
0
df
28
t Stat
1.423933
P(T<=t) one-tail
0.082758
t Critical one-tail
1.701131
P(T<=t) two-tail
0.165515
t Critical two-tail
2.048407
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