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

Answer:

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

n=28

Step-by-step 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:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$  

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

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (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:  

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$  

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$  

And we have everything to replace into the formula:  

$n=(\frac{1.96(160)}{78.4})^2 =16$  

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:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$  

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

$n=(\frac{1.96*160}{60})^2 =27.32$

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.