Question

DDT in rats Poisoning by the pesticide DDT causes convulsions in humans and other mammals. Researchers seek to understand how 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. 78 For the poisoned rats the results were

$\begin{array}{llllll}12.207& 16.869& 25.050& 22.429& 8.456& 20.589\end{array}$

The control group data were

$\begin{array}{llllll}11.074& 9.686& 12.064& 9.351& 8.182& 6.642\end{array}$

Computer output for a two-sample t-test on these data from software is shown below. (Note that SAS provides two-sided P-values.)

(a) Do these data provide convincing evidence that DDT affects the mean height of the second spike's electrical response? Carry out a significance test to help answer this question.

(b) Interpret the P-value from part (a) in the context

Short answer

a) Yes

b) If the population means are equal, then the probability of obtaining a sample with a mean difference of $17.6-9.5=8.1$or more extreme is equal to $2.47\%$

Step-by-step solution

Part(a) Step 1: Given Information

$$\begin{gathered} {\overline{x}}_1=17.6 \\ {\overline{x}}_2=9.5 \\ {s}_1=6.34 \\ {s}_2=1.95 \\ {n}_1=6 \\ {n}_2=6 \end{gathered}$$

Part(a) Step 2: Explanation

Determine the hypothesis:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

Determine the test statistic:

localid="1650519030821" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{17.6-9.5}{\sqrt{\frac{6.{34}^2}{6}+\frac{1.{95}^2}{6}}} \\ \approx 2.291 \end{gathered}$$

Determine the degrees of freedom:

localid="1650519055597" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (6-1,6-1) \\ =5 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row $df=5$:

$0.02=2\times 0.01<P<2\times 0.02=0.04$

(In the output we note that the exact P-value is$0.0247$).

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P<0.05\Rightarrow \text{Reject}{H}_0$

There is sufficient evidence to support the claim of a difference.

Part(b) Step 1: Given Information

$$\begin{gathered} {\overline{x}}_1=17.6 \\ {\overline{x}}_2=9.5 \\ {s}_1=6.34 \\ {s}_2=1.95 \\ {n}_1=6 \\ {n}_2=6 \end{gathered}$$

Part(b) Step 2: Explanation

Determine the hypothesis:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

Determine the test statistic:

localid="1650519139340" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{17.6-9.5}{\sqrt{\frac{6.{34}^2}{6}+\frac{1.{95}^2}{6}}} \\ \approx 2.29 \end{gathered}$$

Determine the degrees of freedom:

localid="1650519154640" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (6-1,6-1) \\ =5 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row $df=20$:

$0.02=2\times 0.01<P<2\times 0.02=0.04$

In the output, we note that the exact P-value is localid="1651483737374" $0.0247=2.47\%.$

This means that the probability of obtaining a sample with a mean difference of $17.6-9.5=8.1$or more extreme is equal to $2.47\%$, if the population means are equal.