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Chapter 10: Q. 54 (page 656)

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

12.20716.86925.05022.4298.45620.589

The control group data were

11.0749.68612.0649.3518.1826.642

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

Expert verified

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=8.1or more extreme is equal to 2.47%

Step by step solution

01

Part(a) Step 1: Given Information

x炉1=17.6x炉2=9.5s1=6.34s2=1.95n1=6n2=6

02

Part(a) Step 2: Explanation

Determine the hypothesis:

H0:渭1=渭2

Ha:渭1鈮�渭2

Determine the test statistic:

localid="1650519030821" t=x炉1-x炉2s12n1+s22n2=17.6-9.56.3426+1.9526鈮�2.291

Determine the degrees of freedom:

localid="1650519055597" df=minn1-1,n2-1=min(6-1,6-1)=5

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脳0.01<P<2脳0.02=0.04

(In the output we note that the exact P-value is0.0247).

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

P<0.05鈬�RejectH0

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

03

Part(b) Step 1: Given Information

x炉1=17.6x炉2=9.5s1=6.34s2=1.95n1=6n2=6

04

Part(b) Step 2: Explanation

Determine the hypothesis:

H0:渭1=渭2

Ha:渭1鈮�渭2

Determine the test statistic:

localid="1650519139340" t=x炉1-x炉2s12n1+s22n2=17.6-9.56.3426+1.9526鈮�2.29

Determine the degrees of freedom:

localid="1650519154640" df=minn1-1,n2-1=min(6-1,6-1)=5

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脳0.01<P<2脳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=8.1or more extreme is equal to 2.47%, if the population means are equal.

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