Question

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,119. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.

Answer 1

`n=((1.640(2119))/(500))^2 =48.31 \approx 49`

So the answer for this case would be n=49 rounded up to the nearest integer

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=37500` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma=2119` represent the sample standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =500 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got
`z_(\alpha/2)=1.640`, replacing into formula (b) we got:

`n=((1.640(2119))/(500))^2 =48.31 \approx 49`

So the answer for this case would be n=49 rounded up to the nearest integer

Answer 2

The smallest sample size needed is 49.

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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
`1-0.05 = 0.95`, so
`z = 1.645`

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 smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.

This is n for which
`M = 500, \sigma = 2119`

So

`M = z*(\sigma)/(√(n))`

`500 = 1.645*(2119)/(√(n))`

`500√(n) = 1.645*2119`

`√(n) = (1.645*2119)/(500)`

`(√(n))^(2) = ((1.645*2119)/(500))^(2)`

`n = 48.6`

Rounding up

The smallest sample size needed is 49.