A farmer noticed that he had a lot of aphids growing on his lettuce. He randomly
ID: 3221506 • Letter: A
Question
A farmer noticed that he had a lot of aphids growing on his lettuce. He randomly selected 10 farms. Each farm was randomly selected to either receive an insecticide spray or to the control group (no spray). At the conclusion of the experiment, one plot on each farm was selected and the number of aphids per leaf was measured. A one-tailed test of significance was performed to determine if the insecticide reduced the number of aphids. Here are two possible outputs from MINITAB. Only one of these outputs is correct. Some of the output is hidden. Which of the following is the appropriate test statistic and a possible P-value?
Two-sample T for not spray vs spray
Difference = mu (not spray) - mu (spray)
Estimate for difference: 1.04
T-Test of difference = 0 (vs > 0) : T-Value = 1.896 P-Value = **** DF = ****
Paired T for not spray - spray
T-Test of mean difference = 0 (vs > 0) : T-Value = 1.887 P-Value = ****
a. 1.896, 0.013
b. 1.896, 0.065
c. 1.896, 0.131
d. 1.887, 0.059
e. 1.887, 0.118
I know B is the correct answer.
Can you please show and explain to me why C is incorrect?
N Mean StDev SE Mean Not Spray 5 4.09 **** **** Spray 5 3.05 **** ****Explanation / Answer
Solution: Yes, the correct answer is b . The answer is (1.896 , 0.065 )
I will explain the why C is incorret.
In above question only one option is correct.
The given example is paired t test . In this test we take mean of difference. on the basis of that the mean of difference and standard deviation we used in paired t test. and the degrees of freedom also change
The degrees of freedom in this test for the difference is d.f.=n-1. = 5- 1. =4
Why the n is 5 . Because we testing on the basis of difference. Therefore the d.f is 4
To find the p-value from excel we ned to t statistic and degress of freedom
The t statistic is 1.896 and d.f. is 4 . The p value is0.065
Excel command is =T.DIST.RT(1.896,4) = 0.065
The answer is b
Therefore C is incorrect.
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