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The value of [tex]\mu[/tex] = 29.6, M = 31.9, [tex]\sigma[/tex] = 16 , s = 22.4, [tex]\sigma_m[/tex] = 3.121, and [tex]\rm S_m[/tex] = 2.473.
It is given that the mean break time per day from the most recent census is 29.6 minutes with a standard deviation of 16 minutes.
It is required to organize the information in a table if the sample size is 82.
What is the margin of error(MOE)?
It is defined as an error that gives an idea about the percentage of errors that exist in the real statistical data.
The formula for finding the MOE:
[tex]\rm MOE = Z_{score}\frac{s}{\sqrt[]{n} }[/tex]
Where is the z score at the confidence interval
s is the standard deviation
n is the number of samples.
We know:
[tex]\rm \sigma_m= \frac{\sigma^2}{n}[/tex]
We have,
[tex]\rm \sigma = 16 \ and \ n = 82[/tex]
[tex]\rm \sigma_m= \frac{16^2}{82}[/tex]
[tex]\rm \sigma_m= 3.121[/tex]
For [tex]\rm S_m[/tex]
[tex]\rm S_m = \frac{s}{\sqrt{n} }[/tex]
We have,
s = 22.4 and n = 82
[tex]\rm S_m = \frac{22.4}{\sqrt{82} }[/tex]
[tex]\rm S_m = 2.473[/tex]
Thus, the value of [tex]\mu[/tex] = 29.6, M = 31.9, [tex]\sigma[/tex] = 16 , s = 22.4, [tex]\sigma_m[/tex] = 3.121, and [tex]\rm S_m[/tex] = 2.473.
Learn more about the Margin of error here:
brainly.com/question/13990500
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