Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Homework 2 Solutions - Applied Linear Regression | 22S 152, Assignments of Statistics

Material Type: Assignment; Professor: DeCook; Class: 22S - Applied Linear Regression; Subject: Statistics and Actuarial Science; University: University of Iowa; Term: Fall 2008;

Typology: Assignments

Pre 2010

Uploaded on 03/19/2009

koofers-user-fcq
koofers-user-fcq 🇺🇸

10 documents

1 / 5

Toggle sidebar

Related documents


Partial preview of the text

Download Homework 2 Solutions - Applied Linear Regression | 22S 152 and more Assignments Statistics in PDF only on Docsity! 22s:152 Homework 2: Solutions Assigned Wednesday, September 10 Due Wednesday, September 17 at classtime 1a) > library(car) > data(Angell) > names(Angell) [1] "moral" "hetero" "mobility" "region" > attach(Angell) ## hetero > y = hetero > par(mfrow = c(1, 2)) > hist(y) > boxplot(y) Histogram of y y F re q u e n c y 20 40 60 80 0 5 1 0 1 5 ! ! ! 2 0 4 0 6 0 8 0 1 > stem(y) The decimal point is 1 digit(s) to the right of the | 1 | 1122334666678899 2 | 01122444789 3 | 12378 4 | 069 5 | 078 6 | 5 7 | 14 8 | 35 The boxplot for hetero does show us that there are some outliers (unusual points) at high values, and that a large portion of the data falls between 0 and 24 (scrunched up toward 0). Also the mean is greater than the median, so this distribution is right-skewed, and is non-normal. It appears to have one mode. ## mobility > y = mobility > par(mfrow = c(1, 2)) > hist(y) > boxplot(y) Histogram of y y F re q u e n c y 10 20 30 40 50 0 2 4 6 8 2 0 3 0 4 0 5 0 2 > stem(y) The decimal point is 1 digit(s) to the right of the | 1 | 24 1 | 555688999 2 | 0022344 2 | 55567778 3 | 12234 3 | 5566789 4 | 23 4 | 57 5 | 0 > detach(Angell) The histogram and stem-n-leaf plot suggest the distribution has something of a bell-shape, but it has a slight right-skew (mean is greater than median). It might fall under a near- normality description depending on how closely the spread gets to a normal distribution. No unusual values. 1b) > data(Leinhardt) > names(Leinhardt) [1] "income" "infant" "region" "oil" > attach(Leinhardt) ## income > y = income > par(mfrow = c(1, 2)) > hist(y) > boxplot(y) 3 Histogram of y y F re q u e n c y 0 2000 4000 6000 0 1 0 2 0 3 0 4 0 5 0 6 0 !! ! ! ! ! ! ! ! ! ! 0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 5 0 0 0 > stem(y) The decimal point is 3 digit(s) to the right of the | 0 | 11111111111111111111111111111111222222222223333333333333344444444 0 | 55666777788 1 | 0022333 1 | 58 2 | 03 2 | 55 3 | 00333444 3 | 7 4 | 11 4 | 8 5 | 00 5 | 56 The income variable is very right-skewed and is non-normal. The far right tail values may be considered unusual or outliers. 4