This is my data
x = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22)
y = c(1, 6, 2, 5, 4, 7, 9, 6, 8, 4, 5, 6, 5, 5, 6, 7, 5, 8, 9,
5, 4, 7)
plot(x, y)
fit <- lm(y ~ x)
fit
abline(fit, col = "black", lwd = "1")
I would like to the plot to split the data into two groups, observations above the regression line and and those under the regression line. How can I do this?
You can use predict to get the fitted value at each x, and then a logical comparison between the observed and fitted to test if they're above or below the line. Then set the colors you plot based on this logical comparison.
prediction <- predict(fit)
colors<-ifelse(y>prediction,1,2)
plot(x,y,col=colors)
abline(fit, col= "black",lwd="1")
Related
I would like to perform a Quade test with more than one covariate in R. I know the command quade.test and I have seen the example below:
## Conover (1999, p. 375f):
## Numbers of five brands of a new hand lotion sold in seven stores
## during one week.
y <- matrix(c( 5, 4, 7, 10, 12,
1, 3, 1, 0, 2,
16, 12, 22, 22, 35,
5, 4, 3, 5, 4,
10, 9, 7, 13, 10,
19, 18, 28, 37, 58,
10, 7, 6, 8, 7),
nrow = 7, byrow = TRUE,
dimnames =
list(Store = as.character(1:7),
Brand = LETTERS[1:5]))
y
quade.test(y)
My question is as follows: how could I introduce more than one covariate? In this example the covariate is the Store variable.
I have a vector of calls made on each days of a certain month.
callsperDayforMonth <- c(3, 1, 2, 1, 1, 3, 9, 1, 4, 2, 6, 4, 9, 13, 15, 2, 5, 5, 2, 7, 3, 0, 1, 2, 7, 1, 8, 6, 9, 4)
I also have a vector of factors which spans the range of the "callsperDayforMonth" vector.
"0-2" "3-5" "6-8" "9-11" "12-14" "16+"
I need to create a histogram, with the factors on the horizontal axis.
How can this be done.
The hist command has an argument breaks that can be a vector of the breakpoints to be used. That should do what you want.
Or you could use table and cut to do the counts yourself and create a barplot from the result.
For example:
library(ggplot2)
cuts <- cut(callsperDayforMonth,
breaks = c(-Inf,2, 5, 8, 11, 14, 16, Inf),
labels = c("0-2", "3-5", "6-8", "9-11", "12-14", "15-16", "16+"))
df <- data.frame(cuts, callsperDayforMonth)
ggplot(df, aes(x=cuts)) + geom_bar(stat = "count")
I have this data:
x1 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22)
y1 = c(1, 6, 2, 5, 4, 7, 9, 6, 8, 4, 5, 6, 5, 5, 6, 7, 5, 8, 9,
5, 4, 7)
plot(x1, y1)
fit <- lm(y1 ~ x1)
fit
abline(fit, col = "black")
prediction <- predict(fit)`
I want to separate the data in two groups using the next condition:
if (y1 < prediction) {print("Negative number")}
else if (y1 > prediction) {print("Positive number")}
But it appears: Warning messages:
1: In if (y1 < prediction) { :
the condition has length > 1 and only the first element will be used
2: In if (y1 > prediction) { :
the condition has length > 1 and only the first element will be used
Can someone tell me how to fix it?
R code:
x <- c(9, 5, 9 ,10, 13, 8, 8, 13, 18, 30)
y <- c(10, 6, 9, 8, 11, 4, 1, 3, 3, 10)
library(exactRankTests)
wilcox.exact(y,x, paired = TRUE, alternative = "two.sided")
The results: V = 3, p-value = 0.01562
SAS code:
data aaa;
set aaa;
diff=x-y;
run;
proc univariate;
var diff;
run;
The results: S=19.5 Pr >= |S| 0.0156
How to get statistics S in R?
If n<=20 the exact P was same in SAS and R,but if n>20 the results were different.
x <- c(9, 5, 9 ,10, 13, 8, 8, 13, 18, 30,9, 5, 9 ,10, 13, 8, 8, 13, 18, 30,9,11,12,10)
y <- c(10, 6, 9, 8, 11, 4, 1, 3, 3, 10,10, 6, 9, 8, 11, 4, 1, 3, 3, 10,10,12,11,12)
wilcox.exact(y,x,paired=TRUE, alternative = "two.sided",exact = FALSE)
The results: V = 34, p-value = 0.002534
The SAS results:S=92.5 Pr >= |S| 0.0009
How to get the same statistics S and P value in SAS and R? Thank you!
I am trying to fit a curved line segment to a dataset. While I can create the line it is always connected back to the starting point. I can't figure out how to get rid of this. I would really appreciate any help. Here is the code
mscF25=c(-12.94382785, -11.0281518, -9.186403952, -7.691576905, -6.470229134, -5.43000796, -4.559074508, -12.87271022, -10.0646268, -6.796208225, -4.433351598, -2.928135666, -1.979265556, -1.38936463, -11.05819006, -7.785838826, -5.297330858, -3.674159165, -2.64702678, -1.980973252, -1.533714976, -11.83971039, -9.168353808, -6.89192172, -5.23424594, -4.033326594, -3.148798626, -2.480469911)
bscF25=c(4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10)
df25 <- data.frame(bscF25,mscF25)
plot(mscF25 ~ bscF25, data = df25)
ls25 <- loess(mscF25 ~ bscF25, data = df25, span = 3)
lines(df25$bscF25, ls25$fitted)
You might try the scatter.smooth function: "Plot and add a smooth curve computed by loess to a scatter plot"
scatter.smooth(x = df25$bscF25, y = df25$mscF25, span = 3)