library(tidverse) library
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 ~ educ + exper + sex, data = CPS85)
#view summary of model 1
summary(Wages_lm)
##
## Call:
## lm(formula = wage ~ educ + exper + sex, 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 1.14e-07 ***
## educ 0.94051 0.07886 11.926 < 2e-16 ***
## exper 0.11330 0.01671 6.781 3.19e-11 ***
## sexM 2.33763 0.38806 6.024 3.19e-09 ***
## ---
## 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: < 2.2e-16
It is statistically significant because the P value of education is less than 5%
Hint: Discuss both its sign and magnitude. For every unit of education which is defined in years there is an additional 94 cents added to the wage ## Q4 Is there evidence for gender discrimination in wages? Make your argument using the relevant test results. Hint: Discuss all three aspects of the relevant predictor: 1) statistical significance, 2) sign, and 3) magnitude.
There is infact evidence for gender discrimiation in wages because the coifficent sign is postive. Males are statistically significant the magnitude of the coeficcent is over 2
The predicted wage of a women with 15 years of education and 5 years experience would be 8.15 dollars per hour. To recieve this number one must multiply the two coefiicents by the years then subtract the intercept.
Hint: Provide a technical interpretation. If all the predictors are at 0 then the intercept is the value the wage will become, in which this number 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 1.05e-07 ***
## sexM 2.14765 0.39097 5.493 6.14e-08 ***
## exper 0.10692 0.01674 6.387 3.70e-10 ***
## educ 0.93495 0.07835 11.934 < 2e-16 ***
## 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: < 2.2e-16
The second model is better because the residual standard error is lower then the first becasue it is 4.42 compared to 4.45 This means that the real wage and the wage predicted by the model are closer then the actual wage. ```
Hint: Use message, echo and results in the chunk options. Refer to the RMarkdown Reference Guide.