if AJ jones decides to use only \'burger\' as a predictor variable, so that he c

Question

if AJ jones decides to use only 'burger' as a predictor variable, so that he can do the regression on his cheap claculator, will the value of R^2 significantly decrease?

For ten years from 1992 to 2001 A.J.Jones has kept a record of several variables. Here are his data on the average gasoline price in cents per litre over the year (gas), the average burger price in dollars over the year (burger), the average unemployment rate over the year in percentage (unemp), and the consumer price index for the year (cpi) ANOVA MS value Regression Residual Total 2 857 3797 428689847.24201 0.0000862 7 63.52035 9.074336 9 920.9 year gas burger unemp cpi 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 136 141 140 148 147 155 160 157 162 165 0.8 0.8 0.9 7.3 37.8 38.2 45.1 47.4 56.9 60.1 66.3 67.2 65.0 68.5 7.6 8.5 8.3 6.5 5.9 5.5 4.2 Standard 1.2 1.5 1.4 Coefficients Ero Stot p-value Intercept 148.9299 7.89691 18.85926 0.00000293 burger unemp3 1.4 0.000982 N.B. Use a = 0.05 for all hypothesis tests. when answering a question look for a short, possibly indirect, way to get the result before embarking on a long calculation. 3.78271 0.695021 -5.44259 0.000964 1. For the regression of 'cpi, on·burger' and 'unemp? x10.70 53.9308 -3441676 340676 2186 RESIDUAL OUTPUT Observation PredictedY Residuals 1139.7697 3.76969 2 140.5262 0.47377 3 140.9416 0.94158 4 144.4573 3.542747 753930-340676 340.676 2186 53.504 ' 2530.604 14.988 6 152.0227 2.977318 7 161.2124 1.21243 8 160.4188 3.41881 9 1584162 3.58378 753.9308-340,676 340676 2186 53,504 8 1604188 22.97 14.662 26.6436 14.662 46.3496 v 26.6436 34.5924 41.7636 10 165.7146 0.71461

Explanation / Answer

Run Regression in R

with 3 independent variables

and 1 dependent variable

Code:

mod1 <- lm(cpi~ .,data=cpidata)

summary(modl1

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 148.2869 5.9766 24.811 4.41e-08

burger 25.1223 3.2101 7.826 0.000105

unemp -3.9554 0.5807 -6.811 0.000251

(Intercept) ***

burger ***

unemp ***

---

Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.56 on 7 degrees of freedom

Multiple R-squared: 0.9502, Adjusted R-squared: 0.936

R sq value with 3 independent variables=0.9502

Now Build a model with burger

Code

mod2 <- lm(cpi~burger,data=cpidata)
summary(mod2)

Output:

Call:

lm(formula = cpi ~ burger, data = cpidata)

Residuals:

Min 1Q Median 3Q Max

-8.544 -3.465 -1.806 2.305 12.805

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 114.274 9.993 11.435 3.09e-06

burger 31.747 8.434 3.764 0.00551

(Intercept) ***

burger **

---

Signif. codes:  

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 6.445 on 8 degrees of freedom

Multiple R-squared: 0.6391, Adjusted R-squared: 0.594

F-statistic: 14.17 on

With only 1 indpeendent variable

R sq=0.6391

R sq decreased from 0.95 to0.64 from 3 Indepencent variables sto 1 independent variable

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.