HWK10_Soln

Stat 371 Homework #10 SOLUTIONS *Submit your homework to Canvas by the due date and time. Email your instructor if you have extenuating circumstances and need to request an extension. *If an exercise asks you to use R, include a copy of the code and output. Please edit your code and output to be only the relevant portions. *If a problem does not specify how to compute the answer, you many use any appropriate method. I may ask you to use R or use manually calculations on your exams, so practice accordingly. *You must include an explanation and/or intermediate calculations for an exercise to be complete. *Be sure to submit the HWK 10 Auto grade Quiz which will give you ~20 of your 40 accuracy points. *50 points total: 40 points accuracy, and 10 points completion Least Squares Linear Regression Exercise 1. Suppose we are interested in exploring the relationship between city air particulate and rates of childhood asthma. We sample 15 cities for particulate (X) measured in parts-per-million (ppm) of large particulate matter and for the rate of childhood asthma (Y) measured in percents. The data is given in the summary table and R vectors below. variable size mean variance X 15 11.42 13.05 Y 15 14.513 2.636 a. Plot the data as you see fit and summarize the pattern’s shape, direction, and strength in the context of the problem. There appears to be a strong, positive, linear pattern between level of particulate and percent asthma for the range of values observed in this data set. particulate <- c( 11.6 , 15.9 , 15.7 , 7.9 , 6.3 , 13.7 , 13.1 , 10.8 , 6.0 , 7.6 , 14.8 , 7.4 , 16.2 , 13.1 , 11.2 ) asthma <- c( 14.5 , 16.6 , 16.5 , 12.6 , 12.0 , 15.8 , 15.1 , 14.2 , 12.2 , 13.1 , 16.0 , 12.9 , 16.4 , 15.4 , 14.4 ) plot(particulate, asthma) 1

6 8 10 12 14 16 12 13 14 15 16 particulate asthma b. Calculate the correlation coefficient (you can use an R function) and explain how the value corresponds to what you observed in the graph in part (a). r = 0.993 calculated in R. This value is very close to 1, which is not surprising based on how tight and linear the x, y data points were in the scatterplot. cor(particulate, asthma) ## [1] 0.9931873 c. Build a linear regression model with least squares estimators for slope and y intercept for the data (i) First, build a regression model by hand using the correlation computed in (b) and summary statistics given above. Our estimated slope is given by ˆ β 1 = r s y s x = 0 . 993 √ 2 . 636 √ 13 . 05 = 0 . 446 We can find the intercept as ¯ y − ˆ β 1 ¯ x which in this case is 14 . 513 − 0 . 446(11 . 42) = 9 . 42 . Our linear model is y = 0 . 446 x + 9 . 42 (ii) Check your computations using lm in R. asthma_mod <- lm(asthma ~ particulate) asthma_mod ## ## Call: ## lm(formula = asthma ~ particulate) ## ## Coefficients: ## (Intercept) particulate 2