STAT3032_004_HW5_S2023_Solution (Shared)

STAT 3032 Regression and Correlated Data Homework 5 (Solution) Please show your work on each problem for full credit. A correct answer, unsupported by the necessary explanation , R code or output will receive very little if any credit. Your work needs to be organized in a reasonably neat and coherent way, and submitted as a pdf file on Canvas. Please do not share this handout outside the class. Problem 1 [10 pts] The vapor.csv file contains two variables: temp The temperature in degrees Celsius mercury the vapor pressure of mercury measured in mmHg (millimeters of mercury) Interest centers on modeling the vapor pressure (the response variable) as a function of temperature (the predictor variable). (a)_[1 pt] Use R to produce a scatterplot of mercury (y-axis) vs. temp (x-axis). Describe the shape of the points in the scatterplot. Reminder: you should provide the R code of the plot. Solution: > plot(mercury~temp,data = vapor) Examples of the description: -The points form a “J” shape. -As the temperature increases, the pressure is going up at an increasing rate. -The vapor pressure shows exponential growth when the temperature goes up. -The relationship between mercury and temperature is not a line.

STAT 3032 Regression and Correlated Data (b)_[3 pts] Fit the following three models using R. For each model, provide the model equation, the degrees of freedom (of the residuals), and the R 2 . You should provide the relevant R code and output. ● Model (1): mercury ~ 1 + temp ● Model (2): mercury ~ 1 + temp + tem p 2 ● Model (3): mercury ~ 1 + temp + tem p 2 + tem p 3 Solution: ● Model (1): ^ mercury =− 146.653 + 1.539 ×temp , df = 17, R 2 = 0.5642 ● Model (2): mercury = ¿ 104.208 − 2.888 ×temp + 0.0123 ×tem p 2 ¿ , df = 16, R 2 = 0.9073 ● Model (3): mercury = ¿ − 12.69 + 1.629 ×temp − 0.01994 ×tem p 2 + 5.969 × 10 − 5 ×tem p 3 ¿ , df = 15, R 2 = 0.9804 > m1 = lm(mercury ~ 1 + temp, data = vapor) > summary(m1) Call: lm(formula = mercury ~ 1 + temp, data = vapor) Residuals: Min 1Q Median 3Q Max -163.39 -112.01 -26.93 64.44 422.20 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -146.653 69.111 -2.122 0.04883 * temp 1.539 0.328 4.691 0.00021 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 156.6 on 17 degrees of freedom Multiple R-squared: 0.5642, Adjusted R-squared: 0.5386 F-statistic: 22.01 on 1 and 17 DF, p-value: 0.0002101 > m2 = lm(mercury ~ 1 + temp + I(temp^2), data = vapor) > summary(m2)

STAT 3032 Regression and Correlated Data Not a part of the homework: some of you may want to see what the fitted models look like in the scatterplot. Here they are: These are the R codes I used to generate the 3rd plot (for your information only and NOT required by this class): # the third plot x_seq = seq(from = 0, to = 360, by = 0.01) mod3 = lm(mercury ~ 1+ temp + I(temp^2) + I(temp^3), data = vapor) fitted_seq = predict(mod3, newdata = data.frame(temp = x_seq)) plot(mercury ~ 1+ temp, data=vapor) lines(fitted_seq ~ x_seq) (c)_[2 pts] Interpret the estimated intercept of the fitted model of mercury ~ 1 + temp + tem p 2 in the context.

STAT 3032 Regression and Correlated Data Solution: When temperature is 0 celsius degrees, the vapor pressure of mercury is 104.208 mmHg on average. (d)_[2 pts] In the fitted model of mercury ~ 1 + temp + tem p 2 + tem p 3 , if we conduct the hypothesis test of H 0 : β 3 = 0 vs. H A : β 3 ≠ 0 , where β 3 is the slope of tem p 3 . What is the test statistic value based on this sample? What distribution does the test statistic follow under the null hypothesis? What is the p value? Use 0.05 as the significance level, do we have evidence to include tem p 3 in the model? Solution: Based on the coefficient table in the summary of Model (3) in Part (b), the test statistic value is 7.487. Under the null hypothesis it follows a t distribution with 15 degrees of freedom. The p- value is 1.93 × 10 − 6 , which is less than 0.05, we reject the null hypothesis, thus we have evidence to include tem p 3 in our model. (e)_[2 pts] Do you think a fourth order polynomial term ( tem p 4 ) is necessary? Why or why not? This is an open-ended question. Potential answers: -No, the 4th order polynomial term is not necessary. When we add it to the model, the R- squared value becomes 0.9924, which shows only a small improvement from 0.9801. -No, the 4th order polynomial term is not necessary. Based on the scatterplot with the fitted model of mercury ~ 1 + temp + tem p 2 + tem p 3 , the model with up to the 3rd order polynomial terms was fitting the dataset well. There is no need to make the model more complicated by adding a 4th order polynomial term. -Yes, the 4th order polynomial term is necessary based on the t test of its slope. When testing whether the slope is zero, the p value is 0.000334 (see the model summary). There is strong evidence that the slope of temp^4 is nonzero, which means that we need this term. > m4 = lm(mercury ~ 1 + temp + I(temp^2) + I(temp^3) + I(temp^4), data = vapor) > summary(m4) Call: lm(formula = mercury ~ 1 + temp + I(temp^2) + I(temp^3) + I(temp^4), data = vapor) Residuals: Min 1Q Median 3Q Max -35.726 -16.762 1.896 9.332 48.695

STAT 3032 Regression and Correlated Data Min 1Q Median 3Q Max -8.031 -3.568 -1.253 2.010 36.456 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 8.9757 0.2951 30.414 < 2e-16 *** unionYes 2.0555 0.5729 3.588 0.000364 *** statusSingle -0.9319 0.4629 -2.013 0.044613 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 5.062 on 531 degrees of freedom Multiple R-squared: 0.03354, Adjusted R-squared: 0.0299 F-statistic: 9.215 on 2 and 531 DF, p-value: 0.0001164 (b)_[1 pt] Based on the model in Part (a), sort the fitted wages of the following four groups from the lowest to the highest. There is no need to explain. ● Married worker with union membership ● Married worker without union membership ● Single worker with union membership ● Single worker without union membership Solution: From the lowest wage to the highest wage: single workers without union membership < married workers without union membership < single workers with union membership < married workers with union membership Explanation: (Students don’t need to provide explanation) For married workers with union membership , we set statusSingle = 0 and unionYes = 1, the fitted wage is 8.98 + 2.06 x 1 - 0.93 x 0 = 11.04. For married workers without union membership , we set statusSingle = 0 and unionYes = 0, the fitted wage is 8.98 + 2.06 x 0 - 0.93 x 0 = 8.98. For single workers with union membership , we set statusSingle = 1 and unionYes = 1, the fitted wage is 8.98 + 2.06 x 1 - 0.93 x 1 = 10.11. For single workers without union membership , we set statusSingle = 1 and unionYes = 0, the fitted wage is 8.98 + 2.06 x 0 - 0.93 x 1 = 8.05 (c)_[2 pts] Fit the model wage ~ 1 + exper + educ . Provide the summary of the fitted model.

STAT 3032 Regression and Correlated Data Solution: > mod2 = lm(wage ~ 1 + exper + educ, data = cps) > summary(mod2) Call: lm(formula = wage ~ 1 + exper + educ, data = cps) Residuals: Min 1Q Median 3Q Max -8.351 -2.857 -0.599 1.994 36.336 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -4.9045 1.2189 -4.024 6.56e-05 *** exper 0.1051 0.0172 6.113 1.89e-09 *** educ 0.9260 0.0814 11.375 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 4.599 on 531 degrees of freedom Multiple R-squared: 0.202, Adjusted R-squared: 0.199 F-statistic: 67.22 on 2 and 531 DF, p-value: < 2.2e-16 (d)_[2 pts] Based on the model in Part (c), interpret the slope of exper in the context. Solution: When the education level of the workers is fixed (When controlling for the education level), increasing the work experience by 1 year is associated with an average increase of $0.1051 in the wage per hour. (e)_[2 pts] Can you interpret the intercept in Part (c) in the context? If yes, please provide the interpretation. If not, please explain. Solution: The intercept means that the workers who have no work experience and no education receive an average wage of -4.90 dollar per hour. However, a negative wage per hour does not make sense.