Homework 4

hw1

desriptive statistics

probability

HW4

Author

HW4

Published

November 11, 2022

Question 1

a

For recent data in Jacksonville, Florida, on y = selling price of home (in dollars), x1 = size of home (in square feet), and x2 = lot size (in square feet), the prediction equation is y^ = ???10,536 + 53.8x1 + 2.84x2.

A particular home of 1240 square feet on a lot of 18,000 square feet sold for $145,000. Find the predicted selling price and the residual, and interpret.

Code

-10536+53.8*1240+2.84*18000

[1] 107296

$107,297 is the predictive price.

Code

145000-107296

[1] 37704

The House was overpaid by 37,704.

B.For fixed lot size, how much is the house selling price predicted to increase for each square-foot increase in home size? Why?

Code

x1240 = 53.8*(1240) + 2.84*(18000) - 10536
x1241 = 53.8*(1241) + 2.84*(18000) - 10536
x1242 = 53.8*(1242) + 2.84*(18000) - 10536
x1241-x1240

[1] 53.8

$53.8

Code

53.8/2.84

[1] 18.94366

The Lot Size is 19 Square Feet

According to this prediction equation, for fixed home size, how much would lot size need to increase to have the same impact as a one-square-foot increase in home size?

Code

library(dplyr)


Attaching package: 'dplyr'

The following objects are masked from 'package:stats':

    filter, lag

The following objects are masked from 'package:base':

    intersect, setdiff, setequal, union

Code

library(alr4)

Loading required package: car

Loading required package: carData


Attaching package: 'car'

The following object is masked from 'package:dplyr':

    recode

Loading required package: effects

lattice theme set by effectsTheme()
See ?effectsTheme for details.

Code

data("salary")
"Female" <- salary %>%
  filter(sex == "Female")
"Male" <- salary %>%
  filter(sex == "Male")
t.test(Female$salary, Male$salary)


    Welch Two Sample t-test

data:  Female$salary and Male$salary
t = -1.7744, df = 21.591, p-value = 0.09009
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -7247.1471   567.8539
sample estimates:
mean of x mean of y 
 21357.14  24696.79

Based on the T test we can conclude that it’s a Null Hypothesis.

B.Run a multiple linear regression with salary as the outcome variable and everything else as predictors, including sex. Assuming no interactions between sex and the other predictors, obtain a 95% confidence interval for the difference in salary between males and females.

Code

"Fit" <- lm(salary ~ rank + sex + degree + ysdeg + year, data = salary)
summary(Fit)


Call:
lm(formula = salary ~ rank + sex + degree + ysdeg + year, data = salary)

Residuals:
    Min      1Q  Median      3Q     Max 
-4045.2 -1094.7  -361.5   813.2  9193.1 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 15746.05     800.18  19.678  < 2e-16 ***
rankAssoc    5292.36    1145.40   4.621 3.22e-05 ***
rankProf    11118.76    1351.77   8.225 1.62e-10 ***
sexFemale    1166.37     925.57   1.260    0.214    
degreePhD    1388.61    1018.75   1.363    0.180    
ysdeg        -124.57      77.49  -1.608    0.115    
year          476.31      94.91   5.018 8.65e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2398 on 45 degrees of freedom
Multiple R-squared:  0.855, Adjusted R-squared:  0.8357 
F-statistic: 44.24 on 6 and 45 DF,  p-value: < 2.2e-16

Code

confint(Fit, level = .95)

                 2.5 %      97.5 %
(Intercept) 14134.4059 17357.68946
rankAssoc    2985.4107  7599.31080
rankProf     8396.1546 13841.37340
sexFemale    -697.8183  3030.56452
degreePhD    -663.2482  3440.47485
ysdeg        -280.6397    31.49105
year          285.1433   667.47476

C.Interpret your finding for each predictor variable; discuss (a) statistical significance, (b) interpretation of the coefficient / slope in relation to the outcome variable and other variables

The only predictive variables are Rank and Year according to the model.

Change the baseline category for the rank variable. Interpret the coefficients related to rank again.

Code

"relevel_sex" <- relevel(salary$sex, ref = "Female")

"New_Fit" <- lm(salary ~ rank + relevel_sex + degree + ysdeg + year, data = salary)
summary(New_Fit)


Call:
lm(formula = salary ~ rank + relevel_sex + degree + ysdeg + year, 
    data = salary)

Residuals:
    Min      1Q  Median      3Q     Max 
-4045.2 -1094.7  -361.5   813.2  9193.1 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)     16912.42     816.44  20.715  < 2e-16 ***
rankAssoc        5292.36    1145.40   4.621 3.22e-05 ***
rankProf        11118.76    1351.77   8.225 1.62e-10 ***
relevel_sexMale -1166.37     925.57  -1.260    0.214    
degreePhD        1388.61    1018.75   1.363    0.180    
ysdeg            -124.57      77.49  -1.608    0.115    
year              476.31      94.91   5.018 8.65e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2398 on 45 degrees of freedom
Multiple R-squared:  0.855, Adjusted R-squared:  0.8357 
F-statistic: 44.24 on 6 and 45 DF,  p-value: < 2.2e-16

When using Relevel, we see a significant change in the coefficient, the coefficient of Gender is negative showing a negative relationship.

Finkelstein (1980), in a discussion of the use of regression in discrimination cases, wrote, “a variable may reflect a position or status bestowed by the employer, in which case if there is discrimination in the award of the position or status, the variable may be ‘tainted.’” Thus, for example, if discrimination is at work in promotion of faculty to higher ranks, using rank to adjust salaries before comparing the sexes may not be acceptable to the courts.

Exclude the variable rank, refit, and summarize how your findings changed, if they did.

Code

"Fit_No_Rank" <- lm(salary ~ sex + degree + ysdeg + year, data = salary)
summary(Fit_No_Rank)


Call:
lm(formula = salary ~ sex + degree + ysdeg + year, data = salary)

Residuals:
    Min      1Q  Median      3Q     Max 
-8146.9 -2186.9  -491.5  2279.1 11186.6 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 17183.57    1147.94  14.969  < 2e-16 ***
sexFemale   -1286.54    1313.09  -0.980 0.332209    
degreePhD   -3299.35    1302.52  -2.533 0.014704 *  
ysdeg         339.40      80.62   4.210 0.000114 ***
year          351.97     142.48   2.470 0.017185 *  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3744 on 47 degrees of freedom
Multiple R-squared:  0.6312,    Adjusted R-squared:  0.5998 
F-statistic: 20.11 on 4 and 47 DF,  p-value: 1.048e-09

When Rank is excluded from Predictor variables,variables degree, ysdeg, and year all find an increase in significance.

Everyone in this dataset was hired the year they earned their highest degree. It is also known that a new Dean was appointed 15 years ago, and everyone in the dataset who earned their highest degree 15 years ago or less than that has been hired by the new Dean. Some people have argued that the new Dean has been making offers that are a lot more generous to newly hired faculty than the previous one and that this might explain some of the variation in Salary.

Code

salary$New_Hire <- ifelse(salary$ysdeg <= 15, 1, 0)

New_Dean <- lm(salary ~ New_Hire + sex + degree , data = salary)
summary(New_Dean)


Call:
lm(formula = salary ~ New_Hire + sex + degree, data = salary)

Residuals:
    Min      1Q  Median      3Q     Max 
-8260.4 -3557.7  -462.6  3563.2 12098.5 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)    28663       1155  24.821  < 2e-16 ***
New_Hire       -7418       1306  -5.679 7.74e-07 ***
sexFemale      -2716       1433  -1.896    0.064 .  
degreePhD      -1227       1372  -0.895    0.375    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 4558 on 48 degrees of freedom
Multiple R-squared:  0.4416,    Adjusted R-squared:  0.4067 
F-statistic: 12.65 on 3 and 48 DF,  p-value: 3.231e-06

The current Model actually shows that New Hires are actually being paid. So being a new hire does not explain the difference.

##QUESTION 3

A.Using the house.selling.price data, run and report regression results modeling y = selling price (in dollars) in terms of size of home (in square feet) and whether the home is new (1 = yes; 0 = no). In particular, for each variable; discuss statistical significance and interpret the meaning of the coefficient.

Code

library("smss")

Warning: package 'smss' was built under R version 4.2.2

Code

data("house.selling.price")
Price_Model <- lm(Price ~ Size + New, data = house.selling.price)
summary(Price_Model)


Call:
lm(formula = Price ~ Size + New, data = house.selling.price)

Residuals:
    Min      1Q  Median      3Q     Max 
-205102  -34374   -5778   18929  163866 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) -40230.867  14696.140  -2.738  0.00737 ** 
Size           116.132      8.795  13.204  < 2e-16 ***
New          57736.283  18653.041   3.095  0.00257 ** 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 53880 on 97 degrees of freedom
Multiple R-squared:  0.7226,    Adjusted R-squared:  0.7169 
F-statistic: 126.3 on 2 and 97 DF,  p-value: < 2.2e-16

B.Report and interpret the prediction equation, and form separate equations relating selling price to size for new and for not new homes.

Price Increases are done by 116.32 for every additional square foot.

Find the predicted selling price for a home of 3000 square feet that is (i) new, (ii) not new.

New Home

Code

116.132*(3000) + 17505.416

[1] 365901.4

Code

116.132*(3000) - 40230.867

[1] 308165.1

D.Fit another model, this time with an interaction term allowing interaction between size and new, and report the regression results

Code

New_Price_Model <- lm(Price ~ Size + New + Size*New, data = house.selling.price)
summary(New_Price_Model)


Call:
lm(formula = Price ~ Size + New + Size * New, data = house.selling.price)

Residuals:
    Min      1Q  Median      3Q     Max 
-175748  -28979   -6260   14693  192519 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) -22227.808  15521.110  -1.432  0.15536    
Size           104.438      9.424  11.082  < 2e-16 ***
New         -78527.502  51007.642  -1.540  0.12697    
Size:New        61.916     21.686   2.855  0.00527 ** 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 52000 on 96 degrees of freedom
Multiple R-squared:  0.7443,    Adjusted R-squared:  0.7363 
F-statistic: 93.15 on 3 and 96 DF,  p-value: < 2.2e-16

E.Report the lines relating the predicted selling price to the size for homes that are (i) new, (ii) not new.

The New Size variable has a high significance and Positive Coefficient.This shows both variables together are is a much better combination.

F. Find the predicted selling price for a home of 3000 square feet that is (i) new, (ii) not new.

Code

104.438*(3000) + 61.916*(3000) - 100755.31

[1] 398306.7

Code

104.438*(3000) - 22227.808

[1] 291086.2

G.Find the predicted selling price for a home of 1500 square feet that is (i) new, (ii) not new. Comparing to (F), explain how the difference in predicted selling prices changes as the size of home increases.

Code

# New Home
104.438*(1500) + 61.916*(1500) - 100755.31

[1] 148775.7

Code

# Old Home 
104.438*(1500) - 22227.808

[1] 134429.2

Newer and Larger homes are the most expensive whereas old smaller houses are least Expensive.

3H.Do you think the model with interaction or the one without it represents the relationship of size and new to the outcome price? What makes you prefer one model over another?

The second model I think is better as it has more variability and accounts for the fact that combining Size and New is much better, the signifiance is greatly improved.