Question

An agronomist hopes that a new fertilizer she has developed will enable grape growers to increase the yield of each grapevine by more than 5 pounds. To test this fertilizer she applied it to 44 vines and used the traditional growing strategies on 47 other vines. The fertilized vines produced a mean of 58.4 pounds of grapes with standard deviation 3.7 pounds, while the unfertilized vines yielded an average of 52.1 pounds with standard deviation 3.4 pounds of grapes. Do these experimental results confirm the agronomist’s expectations?

Answer 1

Answer:

Yes. Fertilization increases grape yields by more than 5 pounds.

Step-by-step explanation:

Let f-fertilized and o-old(unfertilized)

#First, we use our data to calculate the standard error:

$SE(\bar y_f-\bar y_o)=\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}\\\\=\sqrt{\frac{3.7^2}{44}+\frac{3.4^2}{47}}\\\\=0.7464$

#State both null and alternative hypothesis:

$H_o:\mu_f-\mu_o=0\\\\H_A=\mu_f-\mu_o>0$

#We determine our degrees of freedom as 87.02(using R), we now compute the t-value as:

$t=\frac{\bar y_f-\bar y_o}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{53.4-52.1}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{1.3}{0.7464}\\\\t=1.7417\\\\P=P(t_{87.0>1.7417}=1-0.9575=0.0.0425$

Since, the p-value is low, we Reject the null hypothesis. The is enough evidence suggesting that grape yields increase by more than 5 pounds than mean yields of  unfertilized grapes.