Part One
The farmer tested the hypothesis
H0: μ1 - μ2 = 0
H1: μ1 - μ2 ≠ 0
First, the farmer must have tested for the equality of variances to determine whether the t-test that will be used will assume equal variances or not before proceeding to the t-test. In this case, the highlighted value P (F<=f) one-tail is greater than 0.05. As such, the t-test should assume equality of variances.
Undertaking the analysis in Excel, the following output is obtained. The excel working apply the formulae used in manual computation to generate the outcomes as follows:
Examining the t statistic computed, the value that is obtained is within the critical region at 28 degrees of freedom. As such, the farmer cannot conclude that the new fertilizer is better.
Part 2
However, what the farmer did not consider is that the data is paired. As such, each plot had two corresponding portions where one portion was put under old fertilizers while the other was put under the new fertilizer. Once the paired Two Sample for Means test is undertaken, the following output is obtained
Part Three
The paired mean test shows that the alternate hypothesis is accepted since the t value computed is outside the critical t value at 14 degrees of freedom. As such, it shows that the new fertilizer is better than the old fertilizer since the means have been found to be statistically different while the new fertilizer has a higher mean. In other words, the rejection of the null hypothesis means that the two means are statistically different and since the new fertilizer has a higher value, it means the new fertilizer has a better yield.
The method is superior to the farmer's approach since the pairs have to be matched since every plot is divided into two sections with each half being put under one type of fertilizer. Each plot outcome has to be considered as a pair.
