In: Statistics and Probability
A Realtor examines the factors that influence the price of a house in Arlington, Massachusetts. He collects data on recent house sales (Price) and notes each house’s square footage (Sqft) as well as its number of bedrooms (Beds) and number of bathrooms (Baths). A portion of the data is shown in the accompanying table.
Price | Sqft | Beds | Baths |
840,000 | 2,768 | 4 | 3.5 |
822,000 | 2,500 | 4 | 2.5 |
⋮ | ⋮ | ⋮ | ⋮ |
307,500 | 850 | 1 | 1 |
Click here for the Excel Data File
a. Estimate: Price = β0 + β1Sqft + β2Beds + β3Baths + ε. (Round your answers to 2 decimal places.)
PriceˆPrice^ = + Sqft + Beds + Baths
b-1. Choose the appropriate hypotheses to test
whether the explanatory variables are jointly significant in
explaining price.
H0: β1 = β2 = β3 = 0; HA: At least one βj < 0
H0: β1 = β2 = β3 = 0; HA: At least one βj ≠ 0
H0: β1 = β2 = β3 = 0; HA: At least one βj > 0
b-2. Calculate the value of the test statistic.
(Round your answer to 3 decimal places.)
b-3. At the 5% significance level, what is the conclusion to the test? Are the explanatory variables jointly significant in explaining Price?
Reject H0; the explanatory variables are jointly significant in explaining Price.
Reject H0; the explanatory variables are not jointly significant in explaining Price.
Do not reject H0; the explanatory variables are jointly significant in explaining Price.
Do not reject H0; the explanatory variables are not jointly significant in explaining Price.
c-1. Choose the appropriate hypotheses to test whether each of the explanatory variables are individually significant in explaining Price.
H0: βj = 0; HA: βj > 0
H0: βj = 0; HA: βj < 0
H0: βj = 0; HA: βj ≠ 0
c-2. At the 5% significance level, are all explanatory variables individually significant in explaining Price?
EXCEL > DATA > DATA ANALYSIS > REGRESSION
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.850689001 | |||||||
R Square | 0.723671776 | |||||||
Adjusted R Square | 0.697766005 | |||||||
Standard Error | 74984.98417 | |||||||
Observations | 36 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 3 | 4.71211E+11 | 1.5707E+11 | 27.93477086 | 4.59052E-09 | |||
Residual | 32 | 1.79928E+11 | 5622747851 | |||||
Total | 35 | 6.51138E+11 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 153348.2664 | 57141.79374 | 2.683644603 | 0.011433126 | 36954.24141 | 269742.2914 | 36954.24141 | 269742.2914 |
Sqft | 95.85594255 | 35.39974687 | 2.707814349 | 0.010779404 | 23.74901779 | 167.9628673 | 23.74901779 | 167.9628673 |
Beds | 556.8906656 | 20280.31276 | 0.027459669 | 0.978263646 | -40752.75462 | 41866.53595 | -40752.75462 | 41866.53595 |
Baths | 92022.91259 | 25012.29756 | 3.679106742 | 0.000854652 | 41074.52969 | 142971.2955 | 41074.52969 | 142971.2955 |
a)
Price^ = 153348.27 + 95.86*Sqft + 556.89*Beds + 92022.91*Baths
b1)
H0: β1 = β2 = β3 = 0
HA: At least one βj ≠ 0
b2)
F stat = 27.935
b3)
P value = 0 < 0.05
Reject H0; the explanatory variables are jointly significant in explaining Price.
c1)
H0: βj = 0
HA: βj ≠ 0
c2)
Except beds remaining all explanatory variables individually significant in explaining Price
beds p value = 0.9783 > 0.05 (Remaining all explanatory variables P value < 0.05)