SOLUTION: 3) Students from school A and B are compared on the basis of their scores on an aptitude test. Two random samples of 90 and 100 students are selected from school A and B respective

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Question 1067747: 3) Students from school A and B are compared on the basis of their scores on an aptitude test. Two random samples of 90 and 100 students are selected from school A and B respectively. The sample means are 76.4 and 81.2, where as the sample standard deviation is 8.2 and 7.6 respectively. Establish a 98% confidence interval for the difference in population mean scores between students A and B.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
t=(mean1-mean2)/ sqrt {(s1^2/n1)+(s2^2/n2)}
=(76.4-81.2)/sqrt{8.2^2/90 + 7.6^2/100}
=-4.8/1.15=-4.17

the SE is sqrt {(s1^2/n1)+(s2^2/n2)}
8.2^2/90 + 7.6^2/100=1.325
sqrt of that is 1.15
98% CI for t df=188 is 2.34(1.15)=+/-2.69
This is added or subtracted to the mean
(-7.49, -2.11)