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An agent for a residential real estate company in a suburb located outside a maj

ID: 3178712 • Letter: A

Question

An agent for a residential real estate company in a suburb located outside a major city has the business objective of developing more accurate estimates of the monthly rental cost for apartments. Toward the goal, the agent would like to use the size of an apartment, as defined by square footage to predict the monthly rental cost. The agent selects a sample of 8 one-bedroom apartments and the data are shown. Complete parts below

Monthly Rent ($) Size (Square Feet)

900 750

1,550 1,200

800 1,050

1,600 1,250

2,000 2,000

925 700

1,750 1,350

1,250 950




b. Use the least-squares method to determine the regression coefficients

B0 and B1.

B0=

b1=

Now interpret B0,B1 in this problem

c. Predict the mean monthly rent for an apartment that has 1,000 square feet.

Use the equation

Y^i=b0+b1(Xi)

D. Why would it not be appropriate to use the model to predict the monthly rent for apartments that have 500 square feet?

When using a regression model for prediction purposes, consider only the range that includes all values from the smallest to the largest X-value used in developing the regression model.

f. Two people are considering signing a lease for an apartment in this neighborhood. They are trying to decide between two apartments, one with 1,000 square feet for a monthly rent of $1,275 and the other with 1,200 square feet for a monthly rent of $1,425. Based on (a) through (d), which apartment is a better deal?

Compare the model's predicted rents for the two apartments to their actual rents. The apartment whose rent differs from this prediction by the lesser amount is the better deal.

Calculate the difference between the actual and predicted rent values for a 1,000 square foot apartment.

Actual -predicted=

Explanation / Answer

x = Size

y = Monthly Rent

x

y

750

900

1200

1550

1050

800

1250

1600

2000

2000

700

925

1350

1750

950

1250

We go to excel and the we go to Data option. There we select Data Analysis. Under Data Analysis we select Regression. We select data for x and y, then we click on OK. We get the Regression Output:

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.870088452

R Square

0.757053914

Adjusted R Square

0.716562899

Standard Error

236.4312656

Observations

8

ANOVA

df

SS

MS

F

Significance F

Regression

1

1045148.415

1045148

18.697

0.0050

Residual

6

335398.4601

55899.74

Total

7

1380546.875

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

261.9966

264.4565

0.9907

0.3601

-385.1050

909.0982

X Variable 1

0.9383

0.2170

4.3240

0.0050

0.4073

1.4692

b.

The line of regression is,

y^ = 261.9966 + 0.9383x


Here the value of the intercept is 261.9966 (B0) and the value of the slope (B1) is 0.9383.

The value of the slope tells us that if there is one unit increment is size then the rent amount is going to be increased by 0.9383 units.

The value of the intercept is 261.9966; this is the rent amount when the size of the house is 0.

c)

Here x = 1000

y^ = 261.9966 + (0.9383*1000)

     = 1200.30

The predicted value is $1200.30

d)

The size of 500 is quite far from the smallest number. This is the case of extrapolation where we predict the value of the dependent variable for an independent variable which is outside the range of our data. So only this is not appropriate to use the model to predict the monthly rent for apartments that have 500 square feet.

f)

Here we find the Residuals:

X = 1000

Residual = 1275 – 1200.30 = 74.70

X = 1200

y^ = 261.9966 + (0.9383*1200)

     = 1387.96

Residual = 1425 – 1387.96 = 37.04

Here the value of the residual is smaller in case of the 1200 square feet house. So the best deal is 1200 square feet house for a monthly rent of $1425.

x

y

750

900

1200

1550

1050

800

1250

1600

2000

2000

700

925

1350

1750

950

1250

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