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Fifty students are enrolled in a Business Statistics class. After the first examination, a random sample of 5 papers was selected. The grades were 60, 75, 80, 70, and 90. a) Determine the standard error of the mean

Answer :

Answer:

The standard error  S.E of the mean is 5

Step-by-step explanation:

From the given data;

Fifty students are enrolled in a Business Statistics class.

After he first examination, a random sample of 5 papers was selected.

Now; let consider a random sample of 5 papers was selected. with the following grades : 60, 75, 80, 70, and 90

The objective of this question is to determine the standard error of the mean

In order to achieve this ; we need to find the mean and the standard deviation from the given data.

TO start with the mean;

Mean [tex]\overline X[/tex] = [tex]\dfrac{1}{n} \sum x_i[/tex]

Mean [tex]\overline X[/tex] = [tex]\dfrac{1}{5} (60+75+80+70+90)[/tex]

Mean [tex]\overline X[/tex] = 0.2(375)

Mean [tex]\overline X[/tex] = 75

On the other hand; the standard deviation is :

[tex]s = \sqrt{\dfrac{1}{n-1}\sum(x_i - \overline X)^2}[/tex]

[tex]s = \sqrt{\dfrac{1}{5-1}((60-75)^2+(75-75)^2+(80-75)^2+(70-75)^2+(90-75)^2 )}[/tex]

[tex]s = \sqrt{\dfrac{1}{4}(225+0+25+25+225 )}[/tex]

[tex]s = \sqrt{\dfrac{1}{4}(500 )}[/tex]

[tex]s = \sqrt{125}[/tex]

s = 11.18

Finally; the standard error  S.E of the mean is:

[tex]S.E = \dfrac{s}{\sqrt{n}}[/tex]

[tex]S.E = \dfrac{11.18}{\sqrt{5}}[/tex]

[tex]S.E = \dfrac{11.18}{2.236}[/tex]

[tex]S.E = 5[/tex]

The standard error  S.E of the mean is 5

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