library(tidyverse)
library(scales)
options(scipen=999)
Make sure to include the unit of the values whenever appropriate.
Hint: The variables are available in the CPS85 data set from the mosaicData package.
data(CPS85, package = "mosaicData")
wages_lm <- lm(wage ~ sex + exper + educ,
data = CPS85)
# View summary of model 1
summary(wages_lm)
##
## Call:
## lm(formula = wage ~ sex + exper + educ, data = CPS85)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.571 -2.746 -0.653 1.893 37.724
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -6.50451 1.20985 -5.376 0.0000001141795 ***
## sexM 2.33763 0.38806 6.024 0.0000000031877 ***
## exper 0.11330 0.01671 6.781 0.0000000000319 ***
## educ 0.94051 0.07886 11.926 < 0.0000000000000002 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.454 on 530 degrees of freedom
## Multiple R-squared: 0.2532, Adjusted R-squared: 0.2489
## F-statistic: 59.88 on 3 and 530 DF, p-value: < 0.00000000000000022
The coefficient of education is statistically significant at 5 because the p value of education is less than 5%
Hint: Discuss both its sign and magnitude.
For every change in an additional unit of ‘education’ which is in years an additional 94 cents per hour will be added to the wage.
Hint: Discuss all three aspects of the relevant predictor: 1) statistical significance, 2) sign, and 3) magnitude.
There is evidence for gender discrimination in wages because the sign of the coefficient is positive, with male gender being statistically significant and the magnitude of the coefficient being over 2.
The predicated wage of a women with 15 years of education and 5 years of experience would be 8.15 dollars per hour. You get this number by multiplying the two coefficients by the years and then subtracting the intercept.
Hint: Provide a technical interpretation.
When all predictors are at 0 the intercept is the value the wage will become, which is -6.50451 dollars per hour.
Hint: Discuss in terms of both residual standard error and reported adjusted R squared.
data(CPS85, package = "mosaicData")
wages_lm <- lm(wage ~ sex + exper + educ +
union,
data = CPS85)
# View summary of model 1
summary(wages_lm)
##
## Call:
## lm(formula = wage ~ sex + exper + educ + union, data = CPS85)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.496 -2.708 -0.712 1.909 37.784
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -6.48023 1.20159 -5.393 0.00000010459 ***
## sexM 2.14765 0.39097 5.493 0.00000006145 ***
## exper 0.10692 0.01674 6.387 0.00000000037 ***
## educ 0.93495 0.07835 11.934 < 0.0000000000000002 ***
## unionUnion 1.47111 0.50932 2.888 0.00403 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.423 on 529 degrees of freedom
## Multiple R-squared: 0.2648, Adjusted R-squared: 0.2592
## F-statistic: 47.62 on 4 and 529 DF, p-value: < 0.00000000000000022
The second model is better because the difference between the actual wage per hour and the predicted wage per hour is smaller shown to us by the Residual Standard Error. This model can also explain more variability in the model with an adjusted r-squared of 25.92%.
Hint: Use message, echo and results in the chunk options. Refer to the RMarkdown Reference Guide.