Homework set 4 - Solutions

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1 Homework set 4 - Solutions Math 3200 Renato Feres 1. (Eercise 4.12, page 153) This requires importing the data set for Eercise You may, if you wish, type the data points into a vector. (a) Calculate the five number summary for the daily rainfall data ( min,q 1,Q 2,Q 3, ma ), as well as the IQR, the sample mean, and the sample variance. (b) Obtain a bo-and-whiskers plot for this data. (c) Obtain a stem-and-leaf plot for this data. (d) Obtain a normal plot for this data. Solution. After importing the data file, for which I used the name rain, I assigned it the class matri using rain=data.matri(rain). (a) The summary numbers are: summary(rain) V1 Min. : 452 1st Qu.: 865 Median :1331 Mean :1369 3rd Qu.:1727 Ma. :3830 (b) The bo plot is obtained with the command> boplot(rain,horizontal=true). Notice that I m drawing it horizontally by addinghorizontal=true inside theboplot function (c) For a stem-and-left plot we use stem(rain):

2 > stem(rain) The decimal point is 3 digit(s) to the right of the The stem function takes an additional argument called scale. The default value is scale=1. Higher values can be used to epand the length of the plot. For eample,scale=2 roughly doubles the length of the plot: > stem(rain,scale=2) The decimal point is 2 digit(s) to the right of the (d) For the normal plot: > qqnorm(rain) > qqline(rain) 2

3 This produces the graph: Normal Q Q Plot Sample Quantiles Theoretical Quantiles 2. (Eercise 4.36, page160) This requires importing the data set for Eercise (a) Make a scatter plot of unemployment rate vs. snowfall, with the regression line superimposed. What relationship is indicated by the scatter plot? (b) Find the equation of the regression line. Also find the correlation coefficient. Is there a strong linear relationship between snowfall and unemployment? Solution. After importing the data into R under the name SnowUnemp I converted the data frame into a data matri usingsnowunemp=data.matri(snowunemp). Here is how the matri looks like: > SnowUnemp Year Snow Unemp [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] [10,] For convenience I define the vectorssnow=snowunemp[,2] andunemp=snowunemp[,3]. (a) For the scatter plot: > plot(snow,unemp) > lm.unemp=lm(unemp~snow) > lines(snow,fitted(lm.unemp)) 3

4 unemp snow There is a clear correlation between the two variables. As snowfall increases, unemployment increases. (b) To obtain the regression line we invoke thelm function: > lm(formula=unemp~snow) Call: lm(formula = unemp ~ snow) Coefficients: (Intercept) snow The intercept is α= and the regression coefficient is β= So the regression line is Unemployment = Snow The correlation coefficient is > cor(snow,unemp) [1] Recall that this coefficient is at most 1, so the two variables are strongly positively correlated. 3. For each of the following set of computer generated data (, y) obtain a scatter plot showing the regression line and find the correlation coefficient of and y. (For the scatterplot, I suggest using small dots; for eample, plot(,y,ce=0.2,asp=1) should produce a nice enough graph.) (a) = rnorm(200,0,1) y=-+rnorm(200,0,0.8) (b) =rnorm(200,0,1) y=rnorm(200,0,0.8) 4

5 (c) a=2*pi*runif(200,0,1) =cos(a)+rnorm(200,0,0.1) y=sin(a)+rnorm(200,0,0.1) (d) a=2*pi*runif(200,0,1) =2*cos(a)-sin(a)+rnorm(200,0,0.5) y=sin(a)+2*cos(a)+rnorm(200,0,0.5) Solution. The R commands and plot are shown below for each case: (a) > = rnorm(200,0,1) > y=-+rnorm(200,0,0.8) y The correlation coefficient in this case (obtained usingcor(,y)) is: (b) > =rnorm(200,0,1) > y=rnorm(200,0,0.8) The correlation coefficient in this case is: (Close to 0.) (c) > a=2*pi*runif(200,0,1) 5

6 y > =cos(a)+rnorm(200,0,0.1) > y=sin(a)+rnorm(200,0,0.1) y The correlation coefficient in this case is: (Close to 0.) (d) > a=2*pi*runif(200,0,1) > =2*cos(a)-sin(a)+rnorm(200,0,0.5) > y=sin(a)+2*cos(a)+rnorm(200,0,0.5) 6

7 y The correlation coefficient in this case is: (Significant positive correlation.) 4. For each of the data sets (, y) of the previous problem, what linear or quadratic formula would and y satisfy in the absence of random errors? (This is to be done by hand.) Solution. We have, in each case, ignoring the normal random errors in front of and y: (a) = y (b) y = 0 (c) 2 + y 2 = 1 (d) This case requires a little more work. First note that = 2cos a sin a and y = sin a+ 2cos a. Therefore, +y = 4cos a and y = 2sin a and ( +y ) 2+ ( y ) 2= cos 2 a+ sin 2 a= This can be simplified to y+ 5y 2 = 16. 7

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