# I chose to look at the data Iris which is already in R
data("iris")
head(iris)
## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa
attach(iris)
pedal.mod<-lm(Sepal.Length~Sepal.Width,data = iris)
# I want to see how the length and width of the flower species are related
confint(pedal.mod,level = .95)
## 2.5 % 97.5 %
## (Intercept) 5.579865 7.47258038
## Sepal.Width -0.529820 0.08309785
#finding a confidence interval with alpha of .05
plot(Sepal.Length,Sepal.Width)
# this gives a scatter plot of the two variables with length being the predictor
newd<-data.frame(Sepal.Width=5)
# define new variable to predict the length given the width
summary(pedal.mod)
##
## Call:
## lm(formula = Sepal.Length ~ Sepal.Width, data = iris)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.5561 -0.6333 -0.1120 0.5579 2.2226
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.5262 0.4789 13.63 <2e-16 ***
## Sepal.Width -0.2234 0.1551 -1.44 0.152
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8251 on 148 degrees of freedom
## Multiple R-squared: 0.01382, Adjusted R-squared: 0.007159
## F-statistic: 2.074 on 1 and 148 DF, p-value: 0.1519
#statistical summary of the linear model for the sepal length
(pred <- predict(pedal.mod, newd, interval="predict") )
## fit lwr upr
## 1 5.409417 3.668536 7.150298
# create prediction intervals for several wait times with a single call to predict
(conf <- predict(pedal.mod, newd, interval="confidence") )
## fit lwr upr
## 1 5.409417 4.799366 6.019468
conf %*% c(0, -1, 1) #conf interval width
## [,1]
## 1 1.220102
pred %*% c(0, -1, 1) #pred interval width
## [,1]
## 1 3.481762
conf[1] == pred[1]
## [1] TRUE