##sum of squares #########Variance: A Worked Example########## ozone<- span=""> read.table('gardens.txt',header=T)->attach(ozone) ozone## gardenA gardenB gardenC ## 1 3 5 3 ## 2 4 5 3 ## 3 4 6 2 ## 4 3 7 1 ## 5 2 4 10 ## 6 3 4 4 ## 7 1 3 3 ## 8 3 5 11 ## 9 5 6 3 ## 10 2 5 10mean(gardenA)## [1] 3mean(ozone$gardenA)## [1] 3mean(ozone[,1])## [1] 3gardenA-mean(gardenA)## [1] 0 1 1 0 -1 0 -2 0 2 -1(gardenA-mean(gardenA))^2## [1] 0 1 1 0 1 0 4 0 4 1sum((gardenA-mean(gardenA))^2) #sum of squares## [1] 12sum((gardenA-mean(gardenA))^2)/(length(gardenA-1)) #the sum of squares #degree of 9## [1] 1.2mean(gardenB)## [1] 5gardenB-mean(gardenB)## [1] 0 0 1 2 -1 -1 -2 0 1 0(gardenB-mean(gardenB))^2## [1] 0 0 1 4 1 1 4 0 1 0sum((gardenB-mean(gardenB))^2)## [1] 12sum((gardenB-mean(gardenB))^2)/(length(gardenB)-1)## [1] 1.333333mean(gardenC)## [1] 5gardenC-mean(gardenC)## [1] -2 -2 -3 -4 5 -1 -2 6 -2 5(gardenC-mean(gardenC))^2## [1] 4 4 9 16 25 1 4 36 4 25sum((gardenC-mean(gardenC))^2)## [1] 128sum((gardenC-mean(gardenC))^2)/(length(gardenC)-1)## [1] 14.22222#F test var(gardenC)/var(gardenB)## [1] 10.666672*(1-pf(10.667,9,9)) #The F Distribution## [1] 0.001624002help(pf)## starting httpd help server ... donevar.test(gardenB,gardenB) #built-in F test:## ## F test to compare two variances ## ## data: gardenB and gardenB ## F = 1, num df = 9, denom df = 9, p-value = 1 ## alternative hypothesis: true ratio of variances is not equal to 1 ## 95 percent confidence interval: ## 0.2483859 4.0259942 ## sample estimates: ## ratio of variances ## 1gardenC## [1] 3 3 2 1 10 4 3 11 3 10#########Variance and Sample Size plot(c(0,32),c(0,15),type='n',xlab='Simple size',ylab='Variance') for(n in seq(3,31,2)){ for(i in 1:30){ x<- span=""> rnorm(n,mean=10,sd=2) points(n,var(x)) } }->#########A Measure of Unreliability sqrt(var(gardenA)/10)## [1] 0.3651484sqrt(var(gardenB)/10)## [1] 0.3651484sqrt(var(gardenC)/10)## [1] 1.19257########Confidence Intervals #The first argument in qt() is the probability and the second is the #degrees of freedom #left 95% confidence interval value of Student’s t associated with α qt(.025,9)#Sample size 10,d.f 10-1## [1] -2.262157qt(.025,df=9)## [1] -2.262157##right 95% confidence interval value of Student’s t associated with α qt(.975,9)## [1] 2.262157#values of t for 99% are bigger than these (0.005 in each tail): qt(.995,9)## [1] 3.249836##99.5% confidence are bigger still (0.0025 in each tail): qt(.9975,9)## [1] 3.689662#Bootstrap data<- span=""> read.csv('skewdata.txt') attach(data) names(data)->## [1] "values"#This draws the plain graph plot(c(0,30),c(0,60),type='n',xlab='Sample size',ylab='Confidence interval') for(k in seq(5,30,3)){ a<- span=""> numeric(10000) for(i in 1:10000){ a[i]<- span=""> mean(sample(values,k,replace=T)) } points(c(k,k),quantile(a,c(.025,.975)),type='b',pch=21,bg='red') }->->#At n= 30, the bootstrapped CI based on 10 00 simulations quantile(a,c(.025,.975))## 2.5% 97.5% ## 24.90652 37.90855#bootstrapped intervals compared with the intervals calculated from the normal (blue solid line xv <- span=""> seq(5,30,0.1) yv <- span=""> mean(values)+1.96*sqrt(var(values)/xv) plot(c(0,30),c(0,60),type='n',xlab='Sample size',ylab='Confidence interval') for(k in seq(5,30,3)){ a<- span=""> numeric(10000) for(i in 1:10000){ a[i]<- span=""> mean(sample(values,k,replace=T)) } points(c(k,k),quantile(a,c(.025,.975)),type='b',pch=21,bg='red') } lines(xv,yv,col="blue") yv <- span=""> mean(values)-1.96*sqrt(var(values)/xv) lines(xv,yv,col="blue") #Student’s t distribution (green dotted line): yv <- span=""> mean(values)-qt(.975,xv-1)*sqrt(var(values)/xv) lines(xv,yv,lty=2,col="green") yv <- span=""> mean(values)+qt(.975,xv-1)*sqrt(var(values)/xv) lines(xv,yv,lty=2,col="green")->->->->->->->######################### points<- span=""> c(28,26,10,27,20,38,23,28,25,2) x.m<- span=""> function(x) { #sprintf('x-mean=%f ',x-mean(x)); x-mean(x) } x.m.t<- span=""> function(x) { #x.m() #sprintf('x-mean(x)/length(x)=%f ',x-mean(x)/length(x)) ((x-mean(x))^2) } x_mean<- span=""> x.m(points) x_mean_sqrt<- span=""> x.m.t(points) x_mean->->->->->## [1] 5.3 3.3 -12.7 4.3 -2.7 15.3 0.3 5.3 2.3 -20.7x_mean_sqrt## [1] 28.09 10.89 161.29 18.49 7.29 234.09 0.09 28.09 5.29 428.49sum(x_mean)/length(points)## [1] 7.105427e-16sum(x_mean_sqrt)/length(points)## [1] 92.21sqrt(sum(x_mean_sqrt)/length(points))## [1] 9.602604mean(points)## [1] 22.7(mean(points))^2## [1] 515.29var(points)## [1] 102.4556
Variance
Nikkei225
28000-28550 up in the early session, down lately.
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ECBによる資金供給や公的資金の投入が奏功し、欧州金融機関の融資余力が増しているようだ。 source- nikkei net
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Japanese yen retreated yesterday. US durable goods number unexpectedly up,bond auction was so good and stocks were up. So why not to ...