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A consumer advocate analyzes the nicotine content in two brands of cigarettes. A sample of 20 cigarettes of Brand A resulted in an verage nicotine content of 1.68 milligrams with a standard deviation of 0.22 milligram; 25 cigarettes of Brand B yielded an average nicotine content of 1.95 milligrams with a standard deviation of 0.24 milligram. Brand A Brand B X = 1.68 X2 = 1.95 S1 = 0.22 S2 = 0.24 n, = 20 n2 = 25 %3D Construct the 95% confidence interval for the difference between the two popula- tion means. Nicotine content is assumed to be normally distributed. In addition, the pOnulation yariances are unknown but assumed equal.

A consumer advocate analyzes the nicotine content in two brands of cigarettes. A sample of 20 cigarettes of Brand A resulted in an verage nicotine content of 1.68 milligrams with a standard deviation of 0.22 milligram; 25 cigarettes of Brand B yielded an average nicotine content of 1.95 milligrams with a standard deviation of 0.24 milligram. Brand A Brand B X = 1.68 X2 = 1.95 S1 = 0.22 S2 = 0.24 n, = 20 n2 = 25 %3D Construct the 95% confidence interval for the difference between the two popula- tion means. Nicotine content is assumed to be normally distributed. In addition, the pOnulation yariances are unknown but assumed equal.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Page of 6 s3, where the weights applied are their respective degrees of freedom relative to the total number of degrees of freedom. In the case when of and o are unknown and cannot be assumed equal, we cannot calculate a pooled estimate of the population variance. EXAMPLE A consumer advocate analyzes the nicotine content in two brands of cigarettes. A sample of 20 cigarettes of Brand A resulted in an verage nicotine content of 1.68 milligrams with a standard deviation of 0.22 milligram; 25 cigarettes of Brand B yielded an average nicotine content of 1.95 milligrams with a standard deviation of 0.24 milligram. Brand A Brand B X1 = 1.68 X2 = 1.95 S1 = 0.22 S2 = 0.24 n, = 20 n2 = 25 Construct the 95% confidence interval for the difference between the two popula- tion means. Nicotine content is assumed to be normally distributed. In addition, the population variances are unknown but assumed equal.
Transcribed Image Text:Page of 6 s3, where the weights applied are their respective degrees of freedom relative to the total number of degrees of freedom. In the case when of and o are unknown and cannot be assumed equal, we cannot calculate a pooled estimate of the population variance. EXAMPLE A consumer advocate analyzes the nicotine content in two brands of cigarettes. A sample of 20 cigarettes of Brand A resulted in an verage nicotine content of 1.68 milligrams with a standard deviation of 0.22 milligram; 25 cigarettes of Brand B yielded an average nicotine content of 1.95 milligrams with a standard deviation of 0.24 milligram. Brand A Brand B X1 = 1.68 X2 = 1.95 S1 = 0.22 S2 = 0.24 n, = 20 n2 = 25 Construct the 95% confidence interval for the difference between the two popula- tion means. Nicotine content is assumed to be normally distributed. In addition, the population variances are unknown but assumed equal.
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