Arandomized experiment was performed to determine whether two fertilizers

To test for the hypothesis is that there is a difference in the yield between fertilizer A and fertilizer B which is at 5 percent significance level.
From the MINIT AB output, the t-test statistics is -2.47
The p-value for the test is from the MINIT AB output the p-value of the test is 0.019
To conclude I can say that p-value in this context is less than 0.05 which is 0.019. Therefore the alternative or null hypothesis is rejected at five percent level of significance.
There is a sufficient evidence to indicate that there is a difference in the yield between fertilizer A and B. The result is statically significant.

Ithink sanforcisco california (i am not 100% sure) if you feel the opposite so that lol