Question

Poisoning by the pesticide DDT causes convulsions in humans and other mammals. Researchers seek to understand how the convulsions are caused. In a randomized comparative experiment, they compared 6 white rats poisoned with DDT with a control group of 6 unpoisoned rats. Electrical measurements of nerve activity are the main clue to the nature of DDT poisoning. When a nerve is stimulated, its electrical response shows a sharp spike followed by a much smaller second spike. The researchers measured the height of the second spike as a percent of the first spike when a nerve in the rat’s leg was stimulated.33 For the poisoned rats, the results were 12.207 16.869 25.050 22.429 8.456 20.589 The control group data were 11.074 9.686 12.064 9.351 8.182 6.642 (a) Do these data provide convincing evidence that DDT affects the mean relative height of the second spike’s electrical response?
(b) Interpret the P-value from part
(a) in the context of this study.

Answer 1

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 )

$t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$

$Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}$

$Sa^2= \frac{(5*40.20)+(5*3.80)}{6+6-2} = 22$

Sa= 4.69

$t_{H_0}= \frac{17.60-9.50-0}{4.69\sqrt{\frac{1}{6}+\frac{1}{6} } }= 2.99$

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!