R Hints for Chapter : Enter the numbers of men as a data vector and give it a name, e.g.,

Transcription

1 R Hints for Chapter 2 2.1: Enter the numbers of men as a data vector and give it a name, e.g., > nomen=c(13,150,442,...) (You complete the entries.) Give the components of the vector names. > names(nomen)=c( , , ,...) (Don t omit the quotes) Check to see how it looks. > nomen Since this data is already grouped and not in raw form, you have to use the barplot function to produce the histogram. > barplot(nomen,space=0) (space=0 is so that there won t be gaps between the bars.) To plot the cumulative relative frequency graph, first you have to find the cumulative frequencies. > relfreqs=cumsum(nomen)/1067 > relfreqs By inspection, determine which is the first interval whose cumulative frequency is at least 0.5. The median is in that interval. It will be equal to the left endpoint of the interval plus a certain fraction of the width of the interval. Use the interpolation procedure I showed you to get an estimate of the median. You want to plot the end points of the intervals on the horizontal x-axis and the cumulative relative frequencies on the vertical y-axis with the points connected by line segments. > xs=c(80,120,160,...,400) > ys=c(0,relfreqs) (Append 0 at the beginning because the relative frequency is 0 at the first value of xs.) > plot(xs,ys,type= l ) (type=l ( ell ) means to plot a line graph and not just points.) 2.3: You will have to enter the data in R. You can do it with the c function like this: > ages=c(15.0, 17.1, 14.6,...) An easier way is to use the scan function: > ages=scan( )

2 After you type this, enter the numbers one at a time, separated by blank spaces, not commas. You can hit enter only once at any time to start a new line. Hit enter twice to signal the end of the input. The result will look like this (abbreviated): > ages=scan( ) 1: : : Read 16 items > ages [1] To get the frequency distribution use the histogram function. > ages.hist=hist(ages) > ages.hist$breaks will give you the endpoints of the intervals, > relfreqs=ages.hist$counts/96 > relfreqs will give you the relative frequencies (there are 96 data values). The cumulative relative frequencies are > cumrelfreqs=cumsum(relfreqs) Now to plot the cumulative relative frequency diagram, > xs=ages.hist$breaks > ys=c(0,cumrelfreqs) > plot(xs,ys,type= l ) To find the median and 95 th percentiles, use the median and/or the quantile function. > median(ages) > quantile(ages,.95) 2.5: Use the same functions. The data file is available on the web site, so you don t have to enter it. 2.6: Use the mean function to find the mean of the raw data. For the approximation if your data frame is called, for example, ex2.5 with the single variable fat.

3 > attach(ex2.5) > fat.hist=hist(fat) > relfreqs=fat.hist$counts/150 (150 data values) > sum(fat.hist$mids*relfreqs) 2.9: Again, you have to enter the data. > bile=scan() or > bile=c(...) For the frequency distribution you can use the histogram function, but you have to specify the breaks. > bile.hist=hist(bile,breaks=seq(40,100,10)) (Means the sequence of numbers from 40 to 100 in steps of 10) Omit part (b). For part (d) create a function for calculating the coefficient of variation (the base package doesn t have one.) > cv=function(x) sd(x)/mean(x) Once you have created it you can use it forever if you save it when you quit. > cv(bile) 2.24: If days is the name of the variable, the geometric mean can be found like this: > exp(mean(log(days))) R doesn t have a geometric mean function, but you can create one: > geomean=function(x) exp(mean(log(x))) 2.29: Create 4 data vectors: > mht=c(162, 168, 174, 176, 180, 180, 182, 184, 186, 186) > mwt=c(65, 65,...)

4 and do the same for women > wht=... > wwt=... Then do the scatter diagram for men like this: > plot(mwt,mht) and the correlations by > cor(mwt,mht,- > cor(mwt,mht,method= s ) (For Spearman) > cor(mwt,mht,method= k ) (For Kendall) Do the same for the women. Do one of the Pearson correlations by hand calculator, just so you know how. 2.30: The data is available at the web site. 2.32: After attaching the data frame the mean of Age for 0 nodal involvement is > mean(age[nodes==0]) and for nodal involvement > mean(age[nodes==1]) Do the same for the variable Acid. To get the variances and standard deviations use the var function and the sd function. 2.33: For all of the data, > cor(age,acid) For the group with nodal involvement, > cor(age[nodes==1],acid[nodes==1])

5 2.34: The data is in the file ED.visits.txt. Import it as a data frame named ED.visits. > attach(ed.visits) To get the mean no of complaints per visit for female physicians, > mean(complaint[gender== F ]/NVisits[Gender== F ]) Do something similar for male physicians and for each of the two Residency levels. For parts (b) and (c), > cor(complaint/nvisits,revenue) > plot(hours,complaint/nvisits) 2.35: Enter the variables one at a time: > ndeaths=c(`112,140,143,...) > smoke=c(0.30,0.49,0.61,...) >SO2=c(0.09,0.16,0.22,...) and then find the correlations, as in > cor(ndeaths,smoke) However, you can collect all three of these variables in a data frame. > tablee2.35=data.frame(ndeaths,smoke,so2) Then get all the correlations at once, > cor(tablee2.35)