The objective of this assignment is to introduce you to R and R markdown and to complete some basic data simulation exercises.
Please include all code needed to perform the tasks. This is an R Markdown document. Markdown is a simple formatting syntax for authoring HTML, PDF, and MS Word documents. For more details on using R Markdown see http://rmarkdown.rstudio.com.
To submit this homework you will create the document in Rstudio, using the knitr package (button included in Rstudio) and then submit the document to your Rpubs account. Once uploaded you will submit the link to that document on Moodle. Please make sure that this link is hyperlinked and that I can see the visualization and the code required to create it.
rnorm(n=30, mean=c(1,2,3), sd= c(4,5,6))
## [1] -0.3067032 2.8180971 6.9897292 2.3074020 -1.4061108
## [6] 7.9029913 -6.3842908 -2.1422810 7.3223359 -3.9906920
## [11] 7.9602306 11.5993905 7.9464580 -3.1864483 -0.5861491
## [16] 1.2742506 14.9678922 -11.6288837 4.7243579 0.1304831
## [21] -2.1442739 -1.3278426 -2.1214800 5.0850962 -0.7540866
## [26] 4.1949175 4.8625771 0.7484552 5.5428752 -1.0099523
x=rnorm(n=20, mean = 1, sd=2)
y=rnorm(n=20, mean = 1, sd=2)
plot(x,y, main = "Relationship between x and y", xlab="x", ylab="y")
x1= runif(n=1000, min=0, max =2)
x2= runif(n=1000, min=2, max =4)
y= rnorm(n=1000, mean=3, sd=1)
reg_mod<-lm(y~x1+x2)
reg_mod
##
## Call:
## lm(formula = y ~ x1 + x2)
##
## Coefficients:
## (Intercept) x1 x2
## 2.81859 0.05442 0.04671
summary(reg_mod)
##
## Call:
## lm(formula = y ~ x1 + x2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.10126 -0.63950 0.00546 0.64658 2.85365
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.81859 0.17138 16.446 <2e-16 ***
## x1 0.05442 0.05332 1.021 0.308
## x2 0.04671 0.05307 0.880 0.379
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9859 on 997 degrees of freedom
## Multiple R-squared: 0.00179, Adjusted R-squared: -0.0002122
## F-statistic: 0.894 on 2 and 997 DF, p-value: 0.4093
rep(letters[1:3], each=2, times=2)
## [1] "a" "a" "b" "b" "c" "c" "a" "a" "b" "b" "c" "c"
group= rep(c("yes", "no", "neutral"), length.out=25)
factor= as.factor(rep(c("male", "female"), length.out=25))
response1= rnorm(n=25, mean=10, sd=5)
response2= runif(n=25, min=3, max=300)
data.frame(group,factor, response1, response2 )
## group factor response1 response2
## 1 yes male 8.603269227 267.43980
## 2 no female 12.651119509 149.48640
## 3 neutral male 6.022458818 187.43164
## 4 yes female 5.768209431 285.07984
## 5 no male 6.776919052 27.18542
## 6 neutral female 17.143810734 15.16214
## 7 yes male -0.003513084 16.75086
## 8 no female 3.946159404 254.36478
## 9 neutral male 13.282292344 174.58154
## 10 yes female 12.565481394 112.89924
## 11 no male 7.981242373 136.73615
## 12 neutral female 14.866150678 68.27719
## 13 yes male 14.703748983 145.86205
## 14 no female 13.444493147 120.69169
## 15 neutral male 10.351937444 271.52680
## 16 yes female 19.311115948 99.15760
## 17 no male 4.524498946 220.78543
## 18 neutral female 3.570664210 75.16221
## 19 yes male 15.338909863 204.24108
## 20 no female 8.708006591 123.05323
## 21 neutral male 16.579883972 119.86432
## 22 yes female 13.957163314 241.02280
## 23 no male 9.848620630 172.20234
## 24 neutral female 17.888703111 254.52428
## 25 yes male 20.144823589 25.95670