For the given case, I would set null hypothesis and alternative hypothesis as following:
Null Hypothesis
H0= New Fertilizer has same effect on the tomato Yield as the Old Fertilizer; μ1 = μ2.
H1 = Tomato yield from New Fertilizer is higher than the old fertilizer; μ1 > μ2
It is assumed 5% level of significance. i.e. α = .05
First one tail F-test is performed which gave result as below
The test shows that there remains homogeneity in the variances of the two populations as F(14,14) =1.08 & p is equal to 0.44 which is greater than 5% indicating that assumption of different hypothesis for the same test used by farmer gave the same result.
Now, as different approach,
Assuming the null hypothesis,
H0: The effect on tomato yield from the both fertilizers is same, i.e. μD = 0
H1: The effect of the new fertilizer is more effective than the old fertilizer, i.e. μD > 0,
Where, μD = Difference between the mean yield of tomato due to new fertilizer and old fertilizer.
Assuming level of significance 5%, i.e. α = .05.
Performing two tailed T-test for paired two samples for means,
Only now the result is valid as p = 0.0001 < 5% and there is enough support to prove that use of the new fertilizer is better than the old fertilizer as t(14) = 4.83
This is because as the land used in checking the effectiveness of the fertilizer was homogeneous for each parcel and hence, the observation of tomato yield could not be counted as an independent. Thus the assumption that the samples are dependent needs test of significant mean difference, which is different than the previous one. Also, this test suits more than the previous one and is relevant for this situation and hence the result obtained is more reliable and accurate in identifying the difference than that of the earlier test.
