HW_5

HW_5 2023-09-24 8.1 A linear regression model would be great for trying to predict what a fair price should be for a home vs what an inflated price is. It would also be great to try and predict my grades based off of factors like study time sleep or attendance. Another that most people should be familar with is Climate Change and the relation between global temperature and greenhouse gas emissions as well as looking at historical data. Another would be exercising and understanding what help contribute to weight loss and looking at dieting types of weighlifting or types of cardio. Finally businesses would love to know what directly correlates to an increase in sales whether that be advertising, seasonality, or something like inventory management. 1

library (ggplot2) library (tidyverse) ## -- Attaching core tidyverse packages ------------------------ tidyverse 2.0.0 -- ## v dplyr 1.1.3 v readr 2.1.4 ## v forcats 1.0.0 v stringr 1.5.0 ## v lubridate 1.9.2 v tibble 3.2.1 ## v purrr 1.0.2 v tidyr 1.3.0 ## -- Conflicts ------------------------------------------ tidyverse_conflicts() -- ## x dplyr::filter() masks stats::filter() ## x dplyr::lag() masks stats::lag() ## i Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors library (caret) ## Loading required package: lattice ## ## Attaching package: ’caret’ ## ## The following object is masked from ’package:purrr’: ## ## lift crime <- read.table ( crime.txt , stringsAsFactors = FALSE , header = TRUE ) vice <- data.frame ( M = 14.0 , So = 0 , Ed = 10.0 , Po1 = 12.0 , Po2 = 15.5 , LF = 0.640 , M.F = 94.0 , Pop = 15 set.seed ( 42 ) # as simple linear model model_1 <- lm (Crime ~ ., data= crime) summary (model_1) ## ## Call: ## lm(formula = Crime ~ ., data = crime) ## ## Residuals: ## Min 1Q Median 3Q Max ## -395.74 -98.09 -6.69 112.99 512.67 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -5.984e+03 1.628e+03 -3.675 0.000893 *** ## M 8.783e+01 4.171e+01 2.106 0.043443 * ## So -3.803e+00 1.488e+02 -0.026 0.979765 ## Ed 1.883e+02 6.209e+01 3.033 0.004861 ** ## Po1 1.928e+02 1.061e+02 1.817 0.078892 . ## Po2 -1.094e+02 1.175e+02 -0.931 0.358830 ## LF -6.638e+02 1.470e+03 -0.452 0.654654 ## M.F 1.741e+01 2.035e+01 0.855 0.398995 ## Pop -7.330e-01 1.290e+00 -0.568 0.573845 2

## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 ## ## Residual standard error: 347.5 on 42 degrees of freedom ## Multiple R-squared: 0.2629, Adjusted R-squared: 0.1927 ## F-statistic: 3.745 on 4 and 42 DF, p-value: 0.01077 predict (model_2,vice) ## 1 ## 897.2307 model_3 <- lm (Crime ~ M + Ed + U2 + Ineq + Prob, data= crime, x= TRUE , y= TRUE ) summary (model_3) This prediction works better with our data because we are using values that had a low p-value and in doing that we also saw our Adjusted R-squared value drop significantly. While this is good I am curious to see if the variable U2, which barely missed our p-value cut off of 0.05 makes any difference. ## ## Call: ## lm(formula = Crime ~ M + Ed + U2 + Ineq + Prob, data = crime, ## x = TRUE, y = TRUE) ## ## Residuals: ## Min 1Q Median 3Q Max ## -478.8 -233.6 -46.5 143.2 797.1 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -3336.52 1435.26 -2.325 0.02512 * ## M 85.33 54.39 1.569 0.12437 ## Ed 214.69 73.20 2.933 0.00547 ** ## U2 160.01 65.54 2.441 0.01903 * ## Ineq 29.50 21.56 1.368 0.17880 ## Prob -6897.24 2427.81 -2.841 0.00697 ** ## --- ## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 ## ## Residual standard error: 328.6 on 41 degrees of freedom ## Multiple R-squared: 0.3565, Adjusted R-squared: 0.278 ## F-statistic: 4.542 on 5 and 41 DF, p-value: 0.002186 predict (model_3,vice) ## 1 ## 898.1004 4

model_4 <- lm (Crime ~ Ed + Prob, data= crime, x= TRUE , y= TRUE ) summary (model_4) After looking at the model including those variables with a p-value less then 0.05 and including U2 which bareley missed the cut off we see very little change so our previous model is probably ok. ## ## Call: ## lm(formula = Crime ~ Ed + Prob, data = crime, x = TRUE, y = TRUE) ## ## Residuals: ## Min 1Q Median 3Q Max ## -650.98 -279.57 -14.06 198.00 957.48 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 517.30 588.48 0.879 0.3842 ## Ed 63.67 50.26 1.267 0.2119 ## Prob -6049.00 2472.93 -2.446 0.0185 * ## --- ## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1 ## ## Residual standard error: 351.2 on 44 degrees of freedom ## Multiple R-squared: 0.2115, Adjusted R-squared: 0.1756 ## F-statistic: 5.899 on 2 and 44 DF, p-value: 0.005373 predict (model_4,vice) ## 1 ## 912.0796 Finally I want to highlight one additional model. From our first model prediciton we also saw that the p-values of the predictors used changed again. I wanted to investigate if we only used the lowest p-values from that model to fine tune our model. Note that is bad and not how regression like this should be used because then you do not allow other variables to play a role in the model and ultimatly the decision making process that would stem from this bad practice. By only looking at two variables a decision might be made to increase policing where it is unneccasry simply because not enough varibales were taken into account and poor statistical analysis was done. Also to answer the question that was posed in the homework. We predict that the crim rate in a city with the given data is 897.2307. vice_pred <- data.frame ( M = 14.0 , So = 0 , Ed = 10.0 , Po1 = 12.0 , Po2 = 15.5 , LF = 0.640 , M.F = 94.0 , Pop predicted_value <- predict (model_2, newdata = vice_pred[ 1 ,]) 5