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Losielle, Cappella & Gleitz (2022)

Losielle, Cappella & Gleitz (2022)

Losielle, Cappella & Gleitz (2022)

Losielle, Cappella & Gleitz (2022)

Rosencrans, Unrath, Henriquez & Bhalla (2022)

Rosencrans, Unrath, Henriquez & Bhalla (2022)

Rosencrans, Unrath, Henriquez & Bhalla (2022)

Rosencrans, Unrath, Henriquez & Bhalla (2022)

Rosencrans, Unrath, Henriquez & Bhalla (2022)

Regression Review: Intercept

Regression Review: Intercept + Dummy

Regression Review: Intercept + Slope

Regression Review: Intercept + Slope + Dummy

Regression Overview: Four Components

Corn yields vs rainfall

Corn yields vs rainfall

Source: Statistical Sleuth, Display 9.6

Multiple Regression: Quadratic term

Case: Corn Data

corn <- Sleuth3::ex0915 %>% clean_names()
head(corn, 15)

   year yield rainfall
1  1890  24.5      9.6
2  1891  33.7     12.9
3  1892  27.9      9.9
4  1893  27.5      8.7
5  1894  21.7      6.8
6  1895  31.9     12.5
7  1896  36.8     13.0
8  1897  29.9     10.1
9  1898  30.2     10.1
10 1899  32.0     10.1
11 1900  34.0     10.8
12 1901  19.4      7.8
13 1902  36.0     16.2
14 1903  30.2     14.1
15 1904  32.4     10.6

Visualizing Corn Data

ggplot(data = corn) +
 aes(x = rainfall, y = yield) + 
    geom_point() + geom_smooth(method = "loess")

Regression Table

lm1 <-lm(yield ~ rainfall, data = corn)
lm2 <-lm(yield ~ rainfall + I(rainfall^2), data = corn)

Equation with rainfall squared

$$\begin{eqnarray*} \mu \{ \operatorname{yield} | \operatorname{rainfall} \} = \beta_{0} + \beta_{1}(\operatorname{rainfall}) + \beta_{2}(\operatorname{rainfall^2}) \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 6.004(\operatorname{rainfall}) -0.229(\operatorname{rainfall^2}) \end{eqnarray*}$$

• Example: rainfall = 9.6

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 6.004(\operatorname{9.6}) -0.229(\operatorname{9.6^2}) \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 57.638 - 21.105 \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = 31.523 \end{eqnarray*}$$

Visualizing rainfall = 9.6

plot_model(lm2, type = "pred", terms = "rainfall", show.data = TRUE) + geom_vline(xintercept = 9.6, col = "purple", linetype = "dashed")

Equation with rainfall = 12.9

$$\begin{eqnarray*} \mu \{ \operatorname{yield} | \operatorname{rainfall} \} = \beta_{0} + \beta_{1}(\operatorname{rainfall}) + \beta_{2}(\operatorname{rainfall^2}) \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 6.004(\operatorname{rainfall}) -0.229(\operatorname{rainfall^2}) \end{eqnarray*}$$

• Example: rainfall = 12.9

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 6.004(\operatorname{12.9}) -0.229(\operatorname{12.9^2}) \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 77.452 - 38.108 \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = 34.329 \end{eqnarray*}$$

Visualizing rainfall = 12.9

plot_model(lm2, type = "pred", terms = "rainfall", show.data = TRUE) + geom_vline(xintercept = 12.9, col = "purple", linetype = "dashed")

Equation with rainfall = 16.5

$$\begin{eqnarray*} \mu \{ \operatorname{yield} | \operatorname{rainfall} \} = \beta_{0} + \beta_{1}(\operatorname{rainfall}) + \beta_{2}(\operatorname{rainfall^2}) \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 6.004(\operatorname{rainfall}) -0.229(\operatorname{rainfall^2}) \end{eqnarray*}$$

• Example: rainfall = 16.5

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 6.004(\operatorname{16.5}) -0.229(\operatorname{16.5^2}) \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = -5.015 + 99.066 - 62.345 \end{eqnarray*}$$

$$\begin{eqnarray*} \mu \{\operatorname{yield} | \operatorname{rainfall}\} = 31.706 \end{eqnarray*}$$

Visualizing rainfall = 16.5

plot_model(lm2, type = "pred", terms = "rainfall", show.data = TRUE) + geom_vline(xintercept = 16.5, col = "purple", linetype = "dashed")

What do we mean by “slope shift”?

# set up some plausible rainfall values 
rainfall_values <- 7:16
rainfall_values

 [1]  7  8  9 10 11 12 13 14 15 16

term1 <- 6.004 * rainfall_values
term1

 [1] 42.03 48.03 54.04 60.04 66.04 72.05 78.05 84.06 90.06 96.06

term2 <- 0.229 * rainfall_values^2
term2

 [1] 11.22 14.66 18.55 22.90 27.71 32.98 38.70 44.88 51.52 58.62

-5.015 + term1 - term2

 [1] 25.79 28.36 30.47 32.12 33.32 34.06 34.34 34.16 33.52 32.42

interactions package

lm3 <- lm(yield ~ rainfall + I(rainfall^2) + year, data = corn)
interactions::interact_plot(
  lm3,                 # pick a model to plot
  pred = "rainfall",   # this variable will be your x-axis
  modx = "year",       # a moderator, i.e. a control we think is important
  plot.points = TRUE  # plot the data points
)

When to include quadratic terms?

• As with interaction terms, quadratic terms should not routinely be included.

• Consider in four situations:

  • When the analyst has good reason to suspect that the response is nonlinear in some explanatory variable (through knowledge of the process or by graphical examination)

  • When the question of interest calls for finding the values that maximize or minimize the mean response;

  • When careful modeling of the regression is called for by the questions of interest (and presumably this is only the case if there are just a few explanatory variables);

  • Or when inclusion is used to produce a rich model for assessing the fit of an inferential model.

Statistical Sleuth 3e, 10.4.4, p 295