4. A researcher provides data on an experiment involving a person’s ability to p

Question

4. A researcher provides data on an experiment involving a person’s ability to perform a mental task before and after taking one of two drugs. The subjects performed a task that involved mental addition. The subjects were randomly divided into two groups. Each group drank a beverage containing one of two drugs, labeled A, B (placebo). After a period of time for the drugs to take effect, each subject repeated the mental addition test. We want to relate the after test score to the before test score and the drug that was taken. The data from the experiment are presented in Ex4a.xls.

a) (3 points) Is there interaction between drug and before test score at significance level 0.1?

b) (3 points) Is drug a useful predictor variable at significance level 0.05? Note if the interaction is not significant, run regression without it to draw conclusion.

c) (4 points) Suppose there are four drugs labeled A, B (placebo), C, D and the subjects were randomly divided into four groups. We want to relate the after test score to the before test score and the drug that was taken. The data from the experiment are presented in Ex4b.xls. Which drugs have an effect on the after mental task score that is different from the placebo at significance level .1?

BEFORE DRUG AFTER 24 A 24 28 A 30 38 A 39 42 A 41 24 A 27 39 A 46 45 A 56 19 A 25 19 A 18 22 A 25 34 A 31 52 A 52 27 A 38 42 A 45 28 B 28 43 B 41 37 B 37 30 B 33 49 B 39 37 B 38 40 B 41 36 B 38 41 B 36 23 B 18 33 B 32 39 B 33 36 B 35 18 B 19

Explanation / Answer

(a) We have to do a two sample t test to find whether there is an interaction between before and aftertest scores:

This can be done by formulating the null hypothesis and alternte hypothesis as under

H0 : U1 -U2 =0

Ha: U1 - U2 != 0

We have to calculate a t statistic called t = U1-U2/SQRT(s1^2/n1 + s2^2 / n2) where s1 and s2 are the std. deviations and n1 and n2 are the sample numbers and U1 and U2 are the means of the two samples

If the t value calculated gives a p value less that 0.01 we can reject the null hypothesis and claim that there is an interaction. p value can be got looking up a std. normal dist table

(b) To find out if drug is a useful predictor variable at 0.05 level we have to first prove that there is useful interaction at the 0.05 sig level using the method stated in part a, if the interaction is significant then regression analyses will give some conclusion regarding the population, however, if it is not significant then the conclusion that the regression analysis will give will only hold true for the sample and not for the population

Regression can be done using the lm() function in R or the Regression set in the Data Analysis toolpack of Excel

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