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Georgia Institute Of Technology MGT 6203X_Data Analytics in Business Graded Hw2 Part2.-GRADED A+

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MGT6203 Data Analytics in Business Grade Homework#2 Part 2 Instructions for Q1 For parts A and B: PlantGrowth is a dataset in R that contains crop weights of a control group and two treatment grou... ps. Clear the environment and get data > rm(list = ls()) > set.seed(1) > library(datasets)> > data(PlantGrowth) > force(PlantGrowth) Perform the following operations: (i) Create two separate datasets, one with datapoints of treatment 1 group along with control group and other with datapoints of treatment 2 group with the control group. A) Now compute the difference estimator for treatment 1 and treatment 2 datasets that were created, in comparison with the control group? Clear the environment and get data > rm(list = ls()) > set.seed(1) > library(datasets) > data(PlantGrowth) > force(PlantGrowth) weight group 1 4.17 ctrl 2 5.58 ctrl 3 5.18 ctrl 4 6.11 ctrl 5 4.50 ctrl 6 4.61 ctrl 7 5.17 ctrl 8 4.53 ctrl 9 5.33 ctrl 10 5.14 ctrl 11 4.81 trt1 12 4.17 trt1 13 4.41 trt1 14 3.59 trt1 15 5.87 trt1 16 3.83 trt1 17 6.03 trt1 18 4.89 trt1 19 4.32 trt120 4.69 trt1 21 6.31 trt2 22 5.12 trt2 23 5.54 trt2 24 5.50 trt2 25 5.37 trt2 26 5.29 trt2 27 4.92 trt2 28 6.15 trt2 29 5.80 trt2 30 5.26 trt2 Now create dummy variable for control group > Plantdata <- PlantGrowth %>% mutate(Treat_ind = ifelse(group == "ctrl", 0,1)) Create two data sets one for treatment group 1 and one for treatment group 2 > Plantdata1 <- Plantdata[1:20,] > Plantdata2 <- Plantdata[c(1:10,21:30),] Use linear regression to calculate the difference estimator lmt1 <- lm(weight ~ Treat_ind, data = Plantdata1) > summary(lmt1) Call: lm.default(formula = weight ~ Treat_ind, data = Plantdata1) Residuals: Min 1Q Median 3Q Max -1.0710 -0.4938 0.0685 0.2462 1.3690 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 5.0320 0.2202 22.850 9.55e-15 *** Treat_ind -0.3710 0.3114 -1.191 0.249 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.6964 on 18 degrees of freedom Multiple R-squared: 0.07308, Adjusted R-squared: 0.02158 F-statistic: 1.419 on 1 and 18 DF, p-value: 0.249 Here the difference estimator is -0.3710 for treatment group 1. (meaning weight is 0.3710 less for treatment group1 vs the control group) Use linear regression to calculate the difference estimator for group 2 > lmt2 <- lm(weight ~ Treat_ind, data = Plantdata2) > summary(lmt2) Call: lm.default(formula = weight ~ Treat_ind, data = Plantdata2) Residuals: Min 1Q Median 3Q Max -0.862 -0.410 -0.006 0.280 1.078 Coefficients:Estimate Std. Error t value Pr(>|t|) (Intercept) 5.0320 0.1637 30.742 <2e-16 *** Treat_ind 0.4940 0.2315 2.134 0.0469 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.5176 on 18 degrees of freedom Multiple R-squared: 0.2019, Adjusted R-squared: 0.1576 F-statistic: 4.554 on 1 and 18 DF, p-value: 0.04685 Here the difference estimator is 0.4940 for treatment group 2. (meaning weight is 0.4940 higher for treatment group 2 vs the control group). B) [Show More]

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