Question

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample mean is $800 with a known standard deviation of $160. The Error for this calculation, using a 95% CI is $78.40. You decide you cannot have this large an error and want to reduce the error to $60. What size sample should you take? Group of answer choices

Answer 1

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Explanation:

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".

Assuming the X follows a normal distribution

`X \sim N(\mu, \sigma=160)`

And the distribution for
`\bar X` is:

`\bar X \sim N(\mu, (160)/(√(n)))`

We know that the margin of error for a confidence interval is given by:

`Me=z_(\alpha/2)(\sigma)/(√(n))` (1)

The next step would be find the value of
`\z_(\alpha/2)`,
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`

Using the normal standard table, excel or a calculator we see that:

`z_(\alpha/2)=\pm 1.96`

If we solve for n from formula (1) we got:

`√(n)=(z_(\alpha/2) \sigma)/(Me)`

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

And we have everything to replace into the formula:

`n=((1.96(160))/(78.4))^2 =16`

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

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

And now we can replace the new value of Me and see what we got, like this:

`n=((1.96*160)/(60))^2 =27.32`

And if we round up the answer we see that the value of n to ensure the margin of error required
`Me=\pm 60` $ is n=28.