Starting with 3.5, we looked into confidence intervales. To aid with looking into confidence intervales I looked into Sepal Length and Sepal width.
data(iris)
attach(iris)
model <- lm(Sepal.Length ~ Sepal.Width, data=iris)
summary(iris)
## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100
## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300
## Median :5.800 Median :3.000 Median :4.350 Median :1.300
## Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199
## 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800
## Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500
## Species
## setosa :50
## versicolor:50
## virginica :50
##
##
##
To establish a confidence interval we use confident() to find a confidence intervale with 95% confidence. (or 5% alpha)
confint(model) #3.5
## 2.5 % 97.5 %
## (Intercept) 5.579865 7.47258038
## Sepal.Width -0.529820 0.08309785
This shows that there is possibility for there to be a positive corrilation between Sepal width and Sepal length, but it seems like a larger portin of the confidence interval sits below 0.
cor(Sepal.Width,Sepal.Length)
## [1] -0.1175698
This reinforces the though that there is a negative corrilation between Sepal lenght and width.
Next we will look into some confidence intervales. They should both be centered on the same mean, but the main difference is that the confidence intervale that is designed for predicting the next occurance will have a wider interval than the conffidence intervale predicting the mean.
Sepal.Frame <- data.frame( Sepal.Width = 3.15)
Length.mod <- lm(Sepal.Length ~ Sepal.Width)
(p1 <- predict(Length.mod, Sepal.Frame, interval="predict"))
## fit lwr upr
## 1 5.822635 4.186471 7.4588
(c1 <- predict(Length.mod, Sepal.Frame, interval="confidence"))
## fit lwr upr
## 1 5.822635 5.686511 5.95876
The major difference again is that the interval for the prediction of the next point is wider than the interval for predicting the mean. The last thing I will look at is that the middle’s of each interval are equal to eachother.
p1[1]==c1[1]
## [1] TRUE