A farmer wants to determine how much better this new brand of fertilizer (Brand A) is compared to the old one (Brand B). He decides to divide his field into 1000 plots

Question

 He randomly selects 30 plots to receive Brand A and randomly selects 40 plots to receive Brand B. After this process, if the same plot comes up in both samples he throws it out of each sample and reselects a new plot for each brand. This continues until he has 30 plots that will receive Brand A, 40 plots that will receive Brand B, and 30 plots that will not be a part of his study. At the end of the season he compares the yields between the brands. He finds that Brand A produced yields with mean¯x_a= 20 bushels per plot with s_a= 3.5, while Brand B produced yields with mean¯x_b = 18.5 bushels per plot with s_b= 2.1.

1. Determine if there is a significant (use =.01) difference in yield between plots fertilized with Brands A and B .

Answer 1

n_a = 30, a-bar = 20, s_a = 3.5

n_b = 40, b-bar = 18.5, s_b = 2.1

Here the null hypothesis is that a-bar - b-bar = d = 0 vs. the alternative that a-bar and b-bar ^= 0

a-bar - b-bar = d =20 - 18.5 = 1.5

Find the pooled standard deviation s_p as sqrt (s_a²/n_a + s_b²/n_b)

= sqrt(3.5²/30 + 2.1²/40)

= sqrt(0.51858)

=0.720

The t-statistic is d/s_p

= 1,5/0.720

= 2.0833...

At the 0.01 level of statistical significance, the critical value for t with 68 degrees of freedom, calculated as (n_a-1) + (n_b-1), is 2.390.

The calculated t, 2.0833... < 2.390, so there is no statistically significant difference in yields.