Question

The weekly earnings of students in one age group are normally distributed with a standard deviation of 10 dollars. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 2 dollars.

Answer 1

We need a sample size of at least 97.

Explanation:

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

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

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.

Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 2 dollars.

We need a sample size of at least n.

n is found when
`M = 2, \sigma = 10`.

So

`M = z*(\sigma)/(√(n))`

`2 = 1.96*(10)/(√(n))`

`2√(n) = 1.96*10`

`√(n) = (1.96*10)/(2)`

`(√(n))^(2) = ((1.96*10)/(2))^(2)`

`n = 96.04`

Rouding up

We need a sample size of at least 97.