# Load the iris dataset
data(iris)

# Build the multiple regression model
model <- lm(Sepal.Length ~ Sepal.Width + Petal.Length + Petal.Width + I(Petal.Length^2) + I(Species=="versicolor") + Petal.Width:I(Petal.Length^2), data = iris)

# Interpret the coefficients
summary(model)
## 
## Call:
## lm(formula = Sepal.Length ~ Sepal.Width + Petal.Length + Petal.Width + 
##     I(Petal.Length^2) + I(Species == "versicolor") + Petal.Width:I(Petal.Length^2), 
##     data = iris)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.76917 -0.20083  0.01004  0.17659  0.73369 
## 
## Coefficients:
##                                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)                     3.12056    0.45335   6.883 1.70e-10 ***
## Sepal.Width                     0.55311    0.07470   7.404 1.04e-11 ***
## Petal.Length                   -0.15601    0.34349  -0.454   0.6504    
## Petal.Width                    -0.10300    0.32271  -0.319   0.7501    
## I(Petal.Length^2)               0.11213    0.05318   2.108   0.0368 *  
## I(Species == "versicolor")TRUE  0.25591    0.09483   2.699   0.0078 ** 
## Petal.Width:I(Petal.Length^2)  -0.00924    0.01187  -0.779   0.4375    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3029 on 143 degrees of freedom
## Multiple R-squared:  0.8716, Adjusted R-squared:  0.8662 
## F-statistic: 161.8 on 6 and 143 DF,  p-value: < 2.2e-16
# Create diagnostic plots of the model
par(mfrow=c(2,2))
plot(model)

Indication:

After examining these plots, we can assess whether the linear model is appropriate. One important assumption of linear regression is that the residuals should be normally distributed with mean 0 and constant variance across the range of fitted values.

From the residual plots, it appears that there may be some issues with the normality and constant variance assumptions. The residuals vs. fitted values plot shows a clear pattern, indicating that the variance of the residuals increases with the fitted values. Additionally, the normal probability plot of the residuals deviates from a straight line, suggesting that the distribution of residuals may not be exactly normal.

Overall, while the model may still be useful for making predictions or exploratory data analysis, it may not be fully appropriate for inferential purposes due to the issues with the residuals. We may need to consider using a more complex model or transforming the response variable to improve the fit of the model.