2025-10-25
To determine the linear fit manually, two data points can be taken which can estimate the slope and intercept.
For example, taking (1,3) and (5,17) gets slope, \({b_1} = (17-3)/(5-1) = 3.5\) and using either point, (5,17) for example, \(b_0 = 17-3.5*5 = -0.5\) with the linear model being,
\(\hat{y} = -0.5 + 3.5*x\)
The graph below graphs the fit with the data, showing that the line is very close to all the points meaning it is accurate in determining unknown dependent variables.
It can also be inaccurate as shown below. This depends on the data set where the independent and dependent variables having a linear relationship creates an accurate model but if not then it would create an inaccurate model.
Equation: \(\hat{y} = 3 + 9.2*x\)
These models can be created without any manual calculations useful for more complex data sets.
In R, a simple linear regression can be created using the lm and fitted functions as shown below using the mtcars data set.
data(mtcars)
mod <- lm(hp ~ disp, data = mtcars)
x = mtcars$disp; y = mtcars$hp
xax <- list(
title = "Displacement",
titlefont = list(family = "Modern Computer Roman")
)
yax <- list(
title = "Horsepower",
titlefont = list(family = "Modern Computer Roman"),
range= c(0, 300)
)
fig <- plot_ly(x = x, y = y, type = "scatter", mode = "markers", name = "data",
width = 800, height = 430) %>%
add_lines(x=x, y = fitted(mod), name = "fitted") %>%
layout(xaxis = xax, yaxis = yax) %>%
layout(margin = list(
l = 150,
r = 50,
b = 20,
t = 40
))
config(fig, displaylogo = T)
As mentioned earlier, the accuracy of these models can be determined by finding the {\(R^2\)} where the closer it is to 1 the more accurate the model is.
This can be found using many programs like R.
Using Example 3, the summary function can be used to determine the {\(R^2\)}. It is shown that {\(R^2=0.6131\)}, revealing that there is a slight correlation between the two variables but that the linear regression model would not be that accurate.
summary(mod)
## ## Call: ## lm(formula = hp ~ disp, data = mtcars) ## ## Residuals: ## Min 1Q Median 3Q Max ## -48.623 -28.378 -6.558 13.588 157.562 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 45.7345 16.1289 2.836 0.00811 ** ## disp 0.4375 0.0618 7.080 7.14e-08 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 42.65 on 30 degrees of freedom ## Multiple R-squared: 0.6256, Adjusted R-squared: 0.6131 ## F-statistic: 50.13 on 1 and 30 DF, p-value: 7.143e-08
\(R^2\) for simple regression can be calculated using the formula:
\(R^2 = \frac{(n\sum xy - (\sum x)(\sum y))^2}{(n\sum x^2 - (\sum x)^2)(n\sum y^2 - (\sum y)^2)}\)
Where n is the number of observations in the data set.
This is a tedious computation which becomes more time consuming as more observations are added
Therefore, it is recommended to use online tools to compute \(R^2\) for large data sets
x <- c(1, 2, 3, 4, 5) y <- c(3, 7, 10, 14, 17) n <- length(x) numerator <- (n * sum(x * y) - sum(x) * sum(y))^2 denominator <- (n * sum(x^2) - (sum(x))^2) * (n * sum(y^2) - (sum(y))^2) R2 <- numerator / denominator R2
## [1] 0.997557
The two methods can be compared.
Going back to Example 1, the previous slide calculated \(R^2 = 0.9976\).
The summary function can also be used with the lm function.
For this example, both values are the same but may differ depending on rounding and missing data.
summary(lm(y ~ x))$r.squared
## [1] 0.997557