Answer:
Step-by-step explanation:
Hello!
The researchers made a comparative experiment to test the convulsive effects of DDT. They had two groups of 6 rats each, one of them was given the pesticide, the other wasn't treated with any kind of substance (control group) Then the nerve activity was measured.
Given the samples:
X₁: Electrical response of a rat's leg after the rat has eaten poison. (height of the second spike)
12.207, 16.869, 25.050, 22.429, 8.456, 20.589
n₁= 6
X[bar]₁= 17.60
S₁= 6.34
X₂: Electrical response of a rat's leg. (height of the second spike)
11.074, 9.686, 12.064, 9.351, 8.182, 6.642
n₂= 6
X[bar]₂= 9.50
S₂= 1.95
The objective is to compare both samples electrical response to determine if the relative height of the second spike is affected by DDT (if it is the mean height of the spike of rats that are poisoned will be different to the mean height of the spike of the rats that aren't poisoned)
H₀: μ₁ = μ₂
H₁: μ₁ ≠ μ₂
α: 0.05 (the significance level is not determined so I've choosen the most common one.)
I've performed a normality test for both samples, with p-value 0.7078 X₁ has a normal distribution and p-value 0.9367 X₂ has a normal distribution.
To compare both means, considering that both samples are small, and there is no information regarding the population variances, the most accurate statistic to use is the student t.
The p-value for variance homogenecy test is 0.1108 so the population variances are unknown but equal, so the test is a pooled t with pooled sample variance:
(Note I've compared all p-values to α: 0.05 )
Sa= 4.69
This test is two tailed with n₁+n₂-2 degrees of freedom, the p-value is:
p- value: 0.013574
The decision at 5% significance level is to reject the null hypothesis.
b. The p-value in a. indicates that 1.3574% of all samples taken will result that the mean height of the spike of rats poisoned with DDT will be different from the mean height of the spike of rats that were not poisoned.
I hope it helps!