Need help with the codings in R Markdown (in R Studio)
Sol: Since sample mean and sample variance is unbiased estimator of population mean and population variance respectively.
and
Where,
is a sample mean and is a population mean.
is sample variance.
is population variance.
Therefore, R-code result gives approximately same value of sample variance (dividing by n-1) and population variance (dividing by n)
> red=sample(c(1:6),repl=T,size=10000)
> green=sample(c(1:6),repl=T,size=10000)
> sum=red+green
> max=ifelse(red>green, red, green)
> diff=abs(red-green)
> table(sum)
sum
2 3
4 5 6
7 8 9
10 11 12
270 555 850 1132 1380 1649 1372 1084 848 591 269
> mean(sum)
[1] 7.0029
> mean(sum^2)
[1] 54.9115
> var(sum)
[1] 5.871479
> sd(sum)
[1] 2.423113
> sum((sum-mean(sum))^2)/9999 # This case provides sample
variance.
[1] 5.871479
> mean(sum^2)-mean(sum)^2 # This case provides
population variance.
[1] 5.870892
> hist(sum)
Need help with the codings in R Markdown (in R Studio) Preview of Homework 1 R.1...
PLEASE HELP WITH THE FOLLOWING R CODE! I NEED HELP WITH PART C AND D, provided is part a and b!!!! a) chiNum <- c() for (i in 1:1000) { g1 <- rnorm(20,10,4) g2 <- rnorm(20,10,4) g3 <- rnorm(20,10,4) g4 <- rnorm(20,10,4) g5 <- rnorm(20,10,4) g6 <- rnorm(20,10,4) mse <- (var(g1)+var(g2)+var(g3)+var(g4)+var(g5)+var(g6))/6 M <- (mean(g1)+mean(g2)+mean(g3)+mean(g4)+mean(g5)+mean(g6))/6 msb <- ((((mean(g1)-M)^2)+((mean(g2)-M)^2)+((mean(g3)-M)^2)+((mean(g4)-M)^2)+((mean(g5)-M)^2)+((mean(g6)-M)^2))/5)*20 chiNum[i] <- msb/mse } # plot a histogram of F statistics h <- hist(chiNum,plot=FALSE) ylim <- (range(0, 0.8)) x <- seq(0,6,0.01) hist(chiNum,freq=FALSE, ylim=ylim)...
Need help for the coding using the R Markdown in R Studio for question2 and question2. Please provide a detailed solution with an original R code, outputs and a clear statement of the final answer. First, verify by typing out the terms in the appropriate formulas for the t-statistic and the confidence limits that you and R agree about how these should be calculated. For instance, your R code should be structured the way the Excel commands for confidence limits...
R studio #Exercise : Calculate the following probabilities : #1. Probability that a normal random variable with mean 22 and variance 25 #(i)lies between 16.2 and 27.5 #(ii) is greater than 29 #(iii) is less than 17 #(iv)is less than 15 or greater than 25 #2.Probability that in 60 tosses of a fair coin the head comes up #(i) 20,25 or 30 times #(ii) less than 20 times #(iii) between 20 and 30 times #3.A random variable X has Poisson...
Consider the number of days absent from a random sample of six students during a semester: A= {2, 3, 2, 4, 2, 5} Compute the arithmetic mean, geometric mean, median, and mode by hand and verify the results using R Arithmetic Mean: X=i=1nXin=2+3+2+4+2+56=3 mean(data2$absent) [1] 3 Geometic Mean: GMx=Πi=1nX11n=2∙3∙2∙4∙2∙516=2.79816 >gmean <- prod(data2$absent)^(1/length(data2$absent)) > gmean [1] 2.798166 Median: X=12n+1th, Xi2,2,2,3,4,5, n=6=126+1th ranked value=3.5, value=2.5 days absent >median(data2$absent) [1] 2.5 Mode: Most frequent value=2 > mode <- names(table(data2$absent)) [table(data2$absent)==max(table(data2$absent))] > mode [1]...
For expert using R , I solve it but i need to figure out what I got is correct or wrong. Thank you # Simple Linear Regression and Polynomial Regression # HW 2 # # Read data from csv file data <- read.csv("C:\data\SweetPotatoFirmness.csv",header=TRUE, sep=",") head(data) str(data) # scatterplot of independent and dependent variables plot(data$pectin,data$firmness,xlab="Pectin, %",ylab="Firmness") par(mfrow = c(2, 2)) # Split the plotting panel into a 2 x 2 grid model <- lm(firmness ~ pectin , data=data) summary(model) anova(model) plot(model)...