Question

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

Answer 1

Answer:

A sample of 17 must be selected.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.96}{2} = 0.02$

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

That is z with a pvalue of $1 - 0.02 = 0.98$, so Z = 2.054.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ 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 $\sigma = 4$

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

$M = z\frac{\sigma}{\sqrt{n}}$

$2 = 2.054\frac{4}{\sqrt{n}}$

$2\sqrt{n} = 2.054*4$

Simplifying both sides by 2:

$\sqrt{n} = 2.054*2$

$(\sqrt{n})^2 = (2.054*2)^2$

$n = 16.88$

Rounding up:

A sample of 17 must be selected.