Simple Linear Regression
Energy can be produced from wind using windmills. Choosing a site for a wind farm (i.e. the location of the windmills), however, can be a multi-million dollar gamble. If wind is inadequate at the site, then the energy produced over the lifetime of the wind farm can be much less than the cost of building the operation. Hence, accurate prediction of wind speed at a candidate site can be an important component in the decision to build or not to build. Since energy produced varies as the square of the wind speed, even small errors in prediction can have serious consequences.
One possible solution to help predict wind speed at a candidate site is to use wind speed at a nearby reference site. A reference site is a nearby location where the wind speed is already being monitored and should, theoretically, be similar to the candidate site. Using information from the reference site will allow windmill companies to estimate the wind speed at the candidate site without going through a costly data collection period, if the reference site is a good predictor.
The Windmill data set contains measurements of wind speed (in meters per second m/s) at a candidate site (CSpd) (column 1) and at an accompanying reference site (RSpd) (column 2) for 1,116 areas. Download the Windmill.txt file from Canvas (which can be found in Files -> Data Sets), and put it in the same folder as this quarto file.
Simple linear regression could be a helpful tool in figuring out how correlated the candidate site is to the reference site. Knowing how the two different sites are related could help us decide if it would be a good idea to invest and build a wind farm at the candidate site.
wind <- read.table('Windmill.txt', header = TRUE)
summary(wind) CSpd RSpd
Min. : 0.400 Min. : 0.2221
1st Qu.: 6.100 1st Qu.: 4.7769
Median : 8.800 Median : 7.5477
Mean : 9.019 Mean : 7.7773
3rd Qu.:11.500 3rd Qu.:10.2096
Max. :22.400 Max. :21.6015 The response variable will be the CSpd (Candidate Site wind speed), because this is the variable we want to predict, based on what the RSpd (Reference Site wind speed) is.
The explanatory variable must then be RSpd, as it is what we are using to explain how we predict the CSpd. In other words, we mesured the reference site’s wind speed many times in order to try to predict the wind speed at the candidate site.
ggplot(data = wind) +
geom_point(mapping = aes(x = RSpd, y = CSpd)) + labs(title = "Scatterplot of RSpd and CSpd",
x = "Reference site",
Y = "Candidate site")
saved_plot <- ggplot(data = wind) +
geom_point(mapping = aes(x = RSpd, y = CSpd)) + labs(title = "Scatterplot of RSpd and CSpd",
x = "Reference site",
Y = "Candidate site")There is a noticeable upward trend in CSpd, indicating a positive correlation between RSpd and CSpd in this scatterplot.
r <- cor(wind$RSpd, wind$CSpd)There is a strong positive correlation between RSpd and CSpd.
< your response here. Note that you can write math in R markdown by surrounding the math in dollar signs. For example:
\(\beta_0\) (intercept - beta with zero subscript)
\(\times\) (multiplication symbol)
\(\text{Weight}_i\) (i subscript on variable name not italicized)
\(\epsilon_i\) (error term with i subscript) >
\(\text{CSpd} = \beta_0 + beta_1\) \(\times\) \(\text{RSpd} + \epsilon_i\)
ggplot with geom_smooth, You can remove the standard error line with the option se = FALSE).# <your code here>
saved_plot +
geom_smooth(mapping = aes(x = RSpd, y = CSpd),
method = "lm",
se = FALSE)`geom_smooth()` using formula = 'y ~ x'
lm function. (c) Save the residuals and fitted values to the wind tibble. (d) Print the first few rows of the wind tibble.#a
lm_wind <- lm(CSpd ~ RSpd, data = wind)
#b
summary(lm_wind)
Call:
lm(formula = CSpd ~ RSpd, data = wind)
Residuals:
Min 1Q Median 3Q Max
-7.7877 -1.5864 -0.1994 1.4403 9.1738
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.14123 0.16958 18.52 <2e-16 ***
RSpd 0.75573 0.01963 38.50 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 2.466 on 1114 degrees of freedom
Multiple R-squared: 0.5709, Adjusted R-squared: 0.5705
F-statistic: 1482 on 1 and 1114 DF, p-value: < 2.2e-16#c
wind$residuals <- residuals(lm_wind)
wind$fitted_values <- fitted(lm_wind)
#d
print(head(wind)) CSpd RSpd residuals fitted_values
1 6.9 5.9666 -0.7503908 7.650391
2 7.1 7.2176 -1.4958132 8.595813
3 7.8 7.9405 -1.3421328 9.142133
4 6.9 6.0174 -0.7887821 7.688782
5 5.5 6.1646 -2.3000260 7.800026
6 3.1 1.7687 -1.3778979 4.477898Our goal is to find the best-fitting model. In our case, what does best mean? Closes to all the lines, should we connect the dots? No, we want to find the general trends and find how the data are mostly correlated. It leverages our knowledge of calculus to computationally efficiently find the line that best fits the general trend of the data.
< your response here. Note that you can write math in R markdown by surrounding the math in dollar signs. >
The coefficient for the slope in a simple linear regression model represents the change in the dependent variable (or CSpd) for a one-unit change in the independent(RSpd) variable.
The coefficient for the intercept in a simple linear regression model represents the estimated value of the dependent variable (CSpd) when the independent variable (RSpd) is zero.
wind_coefs <- coef(lm_wind)
speed_at_12 <- wind_coefs[1] + wind_coefs[2] * 12
print(speed_at_12)(Intercept)
12.21003 It would be risky because we are extrapolating on the data
wind_sig <- sigma(lm_wind)
print(wind_sig)[1] 2.466234design_mat <- model.matrix(lm_wind)
print(head(design_mat)) (Intercept) RSpd
1 1 5.9666
2 1 7.2176
3 1 7.9405
4 1 6.0174
5 1 6.1646
6 1 1.7687lm) using matrix multiplication. You should use the following in your computations: t() [tranpose], solve() [inverse], %*% [matrix multiplicaiton].# <your code here>< your response here >
< your response here >