This time let us consider the following scenario: Assume that the population distributions for the maximum sustainable clock speed is approximately normally distributed but the population variances are unknown. Assume that the population variances are equal. A sample of 41 CPUs of brand 1 are found to have a sample mean of 550 MHz and a sample variance of 200. A sample of 21 CPUs of brand 2 are found to have a sample mean of 530 MHz and a sample variance of 100. Find a 95% confidence interval for ļ1 ļ ļ2 , the difference of the population means of the maximum sustainable clock speeds for brand 1 and 2
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Step 1
sp^2 = [ (n1ā1)s1^2 + (n2ā1)s2^2 ] / (n1 + n2 ā 2) n1 = 41, s1^2 = 200; n2 = 21, s2^2 = 100 sp^2 = [40*200 + 20*100] / 60 = (8000 + 2000) / 60 = 10000/60 ā 166.6667 sp = sqrt(sp^2) ā 12.9099 Show moreā¦
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