{data(iris)}
x: the 3 species (categorical), y: sepal length (continuous) Hence, do a 1-way ANOVA
hist(iris$Sepal.Length) # check for normality
model <- lm(Sepal.Length ~ Species, data = iris)
summary(model)
##
## Call:
## lm(formula = Sepal.Length ~ Species, data = iris)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.6880 -0.3285 -0.0060 0.3120 1.3120
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.0060 0.0728 68.762 < 2e-16 ***
## Speciesversicolor 0.9300 0.1030 9.033 8.77e-16 ***
## Speciesvirginica 1.5820 0.1030 15.366 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5148 on 147 degrees of freedom
## Multiple R-squared: 0.6187, Adjusted R-squared: 0.6135
## F-statistic: 119.3 on 2 and 147 DF, p-value: < 2.2e-16
F-statistic: 119.3 on 2 and 147 DF, p-value: < 2.2e-16. There is a statistically significant difference in mean sepal length between at least two of the three iris species.
Prediction –> do linear regression
regression_model <- lm(Petal.Length ~ Petal.Width, data = iris)
summary(regression_model)
##
## Call:
## lm(formula = Petal.Length ~ Petal.Width, data = iris)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.33542 -0.30347 -0.02955 0.25776 1.39453
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.08356 0.07297 14.85 <2e-16 ***
## Petal.Width 2.22994 0.05140 43.39 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4782 on 148 degrees of freedom
## Multiple R-squared: 0.9271, Adjusted R-squared: 0.9266
## F-statistic: 1882 on 1 and 148 DF, p-value: < 2.2e-16
R-squared: 0.927, F statistic: 1882 on 1 and 148 DF, p-value: < 2.2e-16. 92.7% of variation in petal length can be explained by petal width.