Question

Housing prices in Athens have been researched extensively by faculty at UGA. The current thinking is that housing prices follow an approximately normal model with mean $238,000 and standard deviation $5,041.

(a) What proportion of housing prices in Athens are less than $234,000? (3 decimal places)
(b) A realtor takes a random sample of 134 houses in Athens. Determine the probability the average selling price is greater than $239,000? (3 decimal places)
(c) A realtor in Asheville, NC wants to estimate the mean housing price of houses in Asheville. The realtor believes the distribution of housing prices in Asheville is similar to those in Athens.

If this realtor takes a random sample of 134 homes in Asheville, what is the standard error of the estimate? (3 decimal places)
How many homes in Asheville should the realtor sample to be 98% confident the estimate is within $500 of the true mean price? Use the critical value to exactly 3 decimal places.

Answer 1

Answer:

a) 0.214 or 21.4%

b) P=0.011

c) The realtor should sample at least 551 homes.

Step-by-step explanation:

The current thinking is that housing prices follow an approximately normal model with mean $238,000 and standard deviation $5,041.

a) We need to know the proportion of housing prices in Athens that are less than $234,000. We can calculate this from the z-score for the population distribution.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{234,000-238,000}{5,041}=\dfrac{-4,000}{5.041}=-0.793\\\\\\ P(x<234,000)=P(z<-0.793)=0.214$

The proportion of housing prices in Athens that are less than $234,000 is 0.214.

b) Now, a sample is taken. The size of the sample is n=134.

We have to calculate the probability that the average selling price is greater than $239,000.

In this case, we have to use the standard error of the sampling distribution to calculate the z-score:

$z=\dfrac{\bar x-\mu}{\sigma/\sqrt{n}}=\dfrac{239,000-238,000}{5,041/\sqrt{134}}=\dfrac{1,000}{435.476}= 2.296 \\\\\\P(\bar x>239,000)=P(z>2.296)=0.011$

The probability that the average selling price is greater than $239,000 is 0.011.

c) We have another sample taken from a distribution with the same parameters.

We have to calculate the sample size so that the margin of error for a 98% confidence interval is $500.

The expression for the margin of error of the confidence interval is:

$E=z\cdot \sigma/\sqrt{n}$

We can isolate n from the margin of error equation as:

$E=z\cdot \sigma/\sqrt{n}\\\\\sqrt{n}=\dfrac{z\cdot \sigma}{E}\\\\n=(\dfrac{z\cdot \sigma}{E})^2$

We have to look for the critical value of z for a 98% CI. This value is z=2.327.

Now we can calculate the minimum value for n to achieve the desired precision for the interval:

$n=(\dfrac{z\cdot \sigma}{E})^2\\\\\\n=(\dfrac{2.327*5,041}{500})^2= 23.461 ^2=550.410\approx551$

The realtor should sample at least 551 homes.

Answer 2

Answer:

a) 0.214 or 21.4%

b) P=0.011

c) The realtor should sample at least 551 homes

Step-by-step explanation: