Interpret R^2 (adjusted) for Equation 1, and indicate how this is possible given your above interpretations.

use the below R console for the quesiton.

> dim(ab.survey.csv)

[1] 1144 15

>

> sex2 <- rep(9,1144)

> sex2[sex=="Male"]<- 0

> sex2[sex=="Female"]<- 1

>

> table(sex)

sex

0 1

5 5

> table(sex2)

sex2

9

1144

>

> print(mean(sex2))

[1] 9

> print(mean(yrs.educ))

[1] 15.2736

> print(mean(age))

[1] 49.98951

>

> senior <- rep(9, 1144)

> senior [age < 65] <- 0

> senior [age >= 65] <- 1

> table (senior)

senior

0 1

927 217

>

> equation1 <- lm(yrs.educ ~ age + sex2)

> summary(equation1)

Call:

lm(formula = yrs.educ ~ age + sex2)

Residuals:

Min 1Q Median 3Q Max

-13.7684 -2.4257 -0.2163 2.0122 26.7266

Coefficients: (1 not defined because of singularities)

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

(Intercept) 16.22539 0.33592 48.30 < 2e-16 ***

age -0.01904 0.00641 -2.970.00304 **

sex2 NA NA NA NA

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3.408 on 1142 degrees of freedom

Multiple R-squared:0.007666, Adjusted R-squared: 0.006797

F-statistic: 8.822 on 1 and 1142 DF, p-value: 0.003039

>

> equation2 <- lm(yrs.educ ~ age + senior + sex2)

> summary(equation2)

Call:

lm(formula = yrs.educ ~ age + senior + sex2)

Residuals:

Min 1Q Median 3Q Max

-13.5664 -2.4600 -0.4083 2.4227 26.5753

Coefficients: (1 not defined because of singularities)

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

(Intercept) 15.697180 0.408420 38.434 <2e-16 ***

age -0.005450 0.008771 -0.621 0.5345

senior -0.796690 0.351697 -2.265 0.0237 *

sex2 NA NA NA NA

---

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 3.402 on 1141 degrees of freedom

Multiple R-squared:0.01211, Adjusted R-squared: 0.01038

F-statistic: 6.993 on 2 and 1141 DF, p-value: 0.0009585

>

> print(num <- rnorm(1))

[1] 2.077669

> #End of R syntax.

>