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

A sample of 17 must be selected.

Explanation:

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

`\alpha = (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(\sigma)/(√(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(\sigma)/(√(n))`

`2 = 2.054(4)/(√(n))`

`2√(n) = 2.054*4`

Simplifying both sides by 2:

`√(n) = 2.054*2`

`(√(n))^2 = (2.054*2)^2`

`n = 16.88`

Rounding up:

A sample of 17 must be selected.