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3. university dean of students wishes to estimate the average number of hours students spend doing homework per week. The standard deviation from a previous study is 4 hours. How large a sample must be selected if he wants to be 96% confident of finding whether the true mean differs from the sample mean by 2 hours

Sagot :

Answer:

A sample of 17 must be selected.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.96}{2} = 0.02[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.02 = 0.98[/tex], so Z = 2.054.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

The standard deviation from a previous study is 4 hours.

This means that [tex]\sigma = 4[/tex]

How large a sample must be selected if he wants to be 96% confident of finding whether the true mean differs from the sample mean by 2 hours?

A sample of n is required.

n is found for M = 2. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]2 = 2.054\frac{4}{\sqrt{n}}[/tex]

[tex]2\sqrt{n} = 2.054*4[/tex]

Simplifying both sides by 2:

[tex]\sqrt{n} = 2.054*2[/tex]

[tex](\sqrt{n})^2 = (2.054*2)^2[/tex]

[tex]n = 16.88[/tex]

Rounding up:

A sample of 17 must be selected.

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