the average grade of a group of 24 alegbra students is 85%, and the standard deviation...
Homework Answers
n = 24
x-bar = 85
s = 4
% = 95
Standard Error, SE = s/√n = 4/√24 = 0.816496581
Degrees of freedom = n - 1 = 24 -1 = 23
t- score = 2.068657599
Width of the confidence interval = t * SE = 2.06865759861054 * 0.816496580927726 = 1.689051856
Lower Limit of the confidence interval = x-bar - width = 85 - 1.68905185637567 = 83.31094814
Upper Limit of the confidence interval = x-bar + width = 85 + 1.68905185637567 = 86.68905186
The confidence interval is [83.31%, 86.69%]
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