Question

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However, they calculate that average sale price for their 15 listings is 400.

(a) Given that the standard deviation of sale prices is known to be 80, calculate a 99 percent confidence interval for the average sale price in Guelph.
(b) Given that the standard deviation of sale prices is unknown but estimated to be 80, calculate a 99 percent confidence interval for the average sale price in Guelph.

Answer 1

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Explanation:

1) 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=400` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma=80` represent the population standard deviation

n=15 represent the sample size

2) Part a

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that
`z_(\alpha/2)=2.58`

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

`400-2.58(80)/(√(15))=346.708`

`400+2.58(80)/(√(15))=453.292`

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

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

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

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that
`t_(\alpha/2)=2.98`

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

`400-2.98(80)/(√(15))=338.445`

`400+2.98(80)/(√(15))=461.555`

So on this case the 99% confidence interval would be given by (338.445;461.555)