Question

A farmer has decided to use a new additive to grow his crops. He divided his farm into 10 plots and kept records of the corn yield (in bushels) before and after using the additive. The before (first row) and after (second row) results from the 10 different plots are shown below.

Corn Yield

Before 9 9 8 7 6 8 5 9 10 11

After 10 9 9 8 7 10 6 10 10 12

You wish to test the following hypothesis at the 1 percent level of significance.

H0= µ=0 against H1: µd > 0

What decision rule would you use?

a.) Reject H0 if test statistic is less than 2.821

b.) Reject H0 if test statistic is greater than -2.821

c.) Reject H0 if test statistic is greater than 2.821

d.) Reject H0 if test statistic is greater than -2.821 or less than 2.821

Answer 1

Solution:
In this question, we will use paired sample t test because population in both are same
So first we will calculate differences of all sample

Before	After	Difference(After-Before)
9	10	1
9	9	0
8	9	1
7	8	1
6	7	1
8	10	2
5	6	1
9	10	1
10	10	0
11	12	1

Sample Mean of difference = (1+0+1+1+1+2+1+1+0+1)/10 = 9/10 = 0.9
Standard deviation of sample difference (S)= sqrt(((1-0.9)^2 + (0-0.9)^2 + (1-0.9)^2 + (1-0.9)^2 + (1-0.9)^2 +(2-0.9)^2 +(1-0.9)^2+(1-0.9)^2+(0-0.9)^2+(1-0.9)^2)/9) = sqrt(2.9/9) = 0.57
Here null hypothesis H0: d = 0
Alternate Hypothesis Ha: d>0
Test statistic value can be calculated as
Test stat = (Sample mean - d)/S/sqrt(n) = (0.9-0)/0.57/sqrt(10) = 5.01
From t table we found critical value at df = n-1 = 10-1 =9 and alpha = 0.01 and this is right tailed test so test critical vaue = 2.821
So if test statistic value is greater than 2.821 than reject H0 null hypothesis. So its correct answer is C. i.e. reject H0 if test statistic is greater than 2.821.
Here we can see that test stat value is greater than 2.481. So we can reject the null hypothesis.