Question

A survey was conducted of the age​ (in years) of 30 randomly selected customers. The mean was 29.61 years and the standard deviation was 10.19 years. How large would the sample size have to be to cut the margin of error of a confidence interval based on this data in​ half?

Answer 1

`ME= t_(\alpha/2)(s)/(√(n))`

No matter the confidence level used we can express the margin of error like this:

`ME = t (10.19)/(√(30))= 1.8604 t`

And if we want to reduce this margin of error to the half we need a new margin of error of
`ME= 0.9302 t`

And using the formula for the margin of error we have:

`0.9302 t = t (10.19)/(√(n))`

And solving for n we got:

`n = ((10.19)/(0.9302))^2 = 120`

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=29.61` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=10.19 represent the sample standard deviation

n=30 represent the original sample size

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

Solution to the problem

For this case the margin of error is given by:

`ME= t_(\alpha/2)(s)/(√(n))`

No matter the confidence level used we can express the margin of error like this:

`ME = t (10.19)/(√(30))= 1.8604 t`

And if we want to reduce this margin of error to the half we need a new margin of error of
`ME= 0.9302 t`

And using the formula for the margin of error we have:

`0.9302 t = t (10.19)/(√(n))`

And solving for n we got:

`n = ((10.19)/(0.9302))^2 = 120`