Question

A researcher intends to estimate the effect of a drug on the scores of human subjects performing a task of psychomotor coordination. The members of a random sample of 9 subjects were given the drug prior to testing. The mean score in this group was 9.42, and the sample variance was 17.31. An independent random sample of 10 subjects was used as a control group and given a placebo prior to testing. The mean score in this control group was 15.45, and the sample variance was 27.11. Assuming that the population distributions are normal with equal variances, find a 90% confidence interval for the difference between the population mean scores.

Answer 1

x1    =	9.420	x2      =	15.450
s1    =	4.161	s2      =	5.207
n1    =	9	n2     =	10
Pooled Variance Sp2=((n1-1)s21+(n2-1)*s22)/(n1+n2-2)=				22.4982

Point estimate : x1-x2=			-6.0300
standard error se =Sp*√(1/n1+1/n2)=			2.1794

for 90 % CI & 17 df value of t=			1.740
margin of error E=t*std error =			3.7921
lower bound=mean difference-E=			-9.822
Upper bound=mean differnce +E=			-2.238
from above 90% confidence interval for population mean =(-9.822,-2.238)