Question

A realtor would like to create a 95% confidence interval for the average price of a house in Franklin County, Ohio. To accomplish this, she collected a random sample of 529 houses and calculated a sample mean price of $320,128. She knows that the population standard deviation for house prices in Franklin County is $50,324. What is her 95% confidence interval?

Answer 1

Answer:

$320128-1.96\frac{50324}{\sqrt{529}}=315839.52$    

$320128+1.96\frac{50324}{\sqrt{529}}=324416.48$    

So on this case the 95% confidence interval would be given by (315839.52;324416.48)    

Step-by-step 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".

$\bar X=320128$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=50324$ represent the population standard deviation

n=529 represent the sample size  

Solution to the problem

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$320128-1.96\frac{50324}{\sqrt{529}}=315839.52$    

$320128+1.96\frac{50324}{\sqrt{529}}=324416.48$    

So on this case the 95% confidence interval would be given by (315839.52;324416.48)