Assignment # 10
Fareeha Imam
Last compiled on March 02, 2023 at 7:18 PM - CST
dat <- read.csv('C:\\Users\\18067\\Documents\\Fareeha Imam\\TTU R11767331\\Spring 2023\\SDA\\Assignment 10\\data-SoftDrinkDeliveryTime.csv')
dat## Observation DeliveryTime.min. NumCases Distance.ft.
## 1 1 16.68 7 560
## 2 2 11.50 3 220
## 3 3 12.03 3 340
## 4 4 14.88 4 80
## 5 5 13.75 6 150
## 6 6 18.11 7 330
## 7 7 8.00 2 110
## 8 8 17.83 7 210
## 9 9 79.24 30 1460
## 10 10 21.50 5 605
## 11 11 40.33 16 688
## 12 12 21.00 10 215
## 13 13 13.50 4 255
## 14 14 19.75 6 462
## 15 15 24.00 9 448
## 16 16 29.00 10 776
## 17 17 15.35 6 200
## 18 18 19.00 7 132
## 19 19 9.50 3 36
## 20 20 35.10 17 770
## 21 21 17.90 10 140
## 22 22 52.32 26 810
## 23 23 18.75 9 450
## 24 24 19.83 8 635
## 25 25 10.75 4 1501 Part 1:
What are your estimates of the regression parameters and what is the associated value of R2?
head(dat)## Observation DeliveryTime.min. NumCases Distance.ft.
## 1 1 16.68 7 560
## 2 2 11.50 3 220
## 3 3 12.03 3 340
## 4 4 14.88 4 80
## 5 5 13.75 6 150
## 6 6 18.11 7 330dat<- dat[,-1]
head(dat)## DeliveryTime.min. NumCases Distance.ft.
## 1 16.68 7 560
## 2 11.50 3 220
## 3 12.03 3 340
## 4 14.88 4 80
## 5 13.75 6 150
## 6 18.11 7 330NumCases as x1
Distance.ft. as x2
dat$y<-dat$DeliveryTime.min.
dat$x1<-dat$NumCases
dat$x2<-dat$Distance.ft.head(dat)## DeliveryTime.min. NumCases Distance.ft. y x1 x2
## 1 16.68 7 560 16.68 7 560
## 2 11.50 3 220 11.50 3 220
## 3 12.03 3 340 12.03 3 340
## 4 14.88 4 80 14.88 4 80
## 5 13.75 6 150 13.75 6 150
## 6 18.11 7 330 18.11 7 330model1<-lm(y~x1+x2+x1:x2,data= dat)
summary(model1)##
## Call:
## lm(formula = y ~ x1 + x2 + x1:x2, data = dat)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.7316 -1.5387 0.0606 1.4375 4.7841
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.1390846 1.3997413 5.100 4.73e-05 ***
## x1 1.0144063 0.1912517 5.304 2.93e-05 ***
## x2 0.0058273 0.0033825 1.723 0.099622 .
## x1:x2 0.0007419 0.0001750 4.240 0.000366 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.449 on 21 degrees of freedom
## Multiple R-squared: 0.9782, Adjusted R-squared: 0.9751
## F-statistic: 314.6 on 3 and 21 DF, p-value: < 2.2e-16model1$coefficients## (Intercept) x1 x2 x1:x2
## 7.1390845734 1.0144062540 0.0058273479 0.0007419211summary(model1)$r.squared## [1] 0.9782308b1 = 1.0144063
b2 = 0.0058273
b3 = 0.0007419
As you can see the value of R squared is 0.9782308
2 Part 2:
What do you notice in the diagnostic plots (all of them)
plot(model1)
we observed that Point 9 in Residual vs Leverage plot is significant and 2 points have leverage
Also, we noticed that the scale location plot is normal and the values are less than 1.7
3 Part 3:
Remove the observation(s) that appears to be the most influential
dat<-dat[-9,]4 Part 4:
What are your estimates of the regression parameters what is the associated value of R2?
model2<-lm(y~x1+x2+x1:x2,data = dat)
summary(model2)##
## Call:
## lm(formula = y ~ x1 + x2 + x1:x2, data = dat)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.8495 -1.3509 -0.0835 1.6174 4.9098
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.7984402 1.9709874 2.942 0.008062 **
## x1 1.2660217 0.3229617 3.920 0.000848 ***
## x2 0.0080441 0.0040895 1.967 0.063212 .
## x1:x2 0.0003480 0.0004432 0.785 0.441497
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.452 on 20 degrees of freedom
## Multiple R-squared: 0.9502, Adjusted R-squared: 0.9428
## F-statistic: 127.3 on 3 and 20 DF, p-value: 3.368e-13summary(model2)$r.squared## [1] 0.9502353model1$coefficients## (Intercept) x1 x2 x1:x2
## 7.1390845734 1.0144062540 0.0058273479 0.0007419211model2$coefficients## (Intercept) x1 x2 x1:x2
## 5.7984402036 1.2660216932 0.0080440801 0.0003480246b1 = 1.0144062540
b2 = 0.0058273479
b3 = 0.0007419211
The values of each parameters for model2 are given below; b0 = 5.7984402036
b1 = 1.2660216932
b2 = 0.0080440801
b3 = 0.0003480246
5 Part 5:
What do you notice in the diagnostic plots (all of them) after removing this point?
plot(model2)
we observed that Point 9 in Residual vs Leverage plot is significant and 1 points have leverage
Also, we noticed that the scale location plot is normal and the values are less than 1.7, also there is no need to eliminate any point just like above we did in previous model, because there is no point we found significant. These plots looks good.
6 Part 6:
Is there now another point that might be influential or that has leverage?
--> Yes there is a point that has leverage and it is not influential.7 Part 7:
Do you believe that this point should have be removed? (explain)
--> I believe there is no need to eliminate any point, since we didn't found any influential point