Lab 08: Regression and Correlation
Markdown Author: Jessie Bell, 2023
Libraries Used: ggplot2
Part 1: Regression
1
Include the excel graph in your write-up.
2
predictor <- c(4.2,5.6,5.1,4.6,9.2,6.9,10.0,3.4,7.0,11.2,1.2,8.7)
response <- c(6.9,8.3,7.3,6.3,11.2,8.1,13.6,5.7,8.0,13.8,3.3,12.8)
data <- data.frame(predictor, response)
fit <- lm(response ~ predictor, data = data)3
\(H_0\): The slope of the regression line is not significantly different from zero.
\(H_a\): The slope of the regression line is significantly different from zero.
4
fit <- lm(response ~ predictor, data = data)
anova(fit)## Analysis of Variance Table
##
## Response: response
## Df Sum Sq Mean Sq F value Pr(>F)
## predictor 1 115.298 115.298 146.98 2.653e-07 ***
## Residuals 10 7.844 0.784
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1summary(fit)##
## Call:
## lm(formula = response ~ predictor, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.4045 -0.5110 0.1064 0.4615 1.5342
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.74045 0.63407 2.745 0.0207 *
## predictor 1.09487 0.09031 12.124 2.65e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8857 on 10 degrees of freedom
## Multiple R-squared: 0.9363, Adjusted R-squared: 0.9299
## F-statistic: 147 on 1 and 10 DF, p-value: 2.653e-075
ggplot(data, aes(predictor, response))+
geom_point(color="#ed68ed")+
geom_smooth(formula = y ~ x, method = "lm", color="#00c19a")+
labs(title="Response vs. Predictor")
6
plot(fit$fitted.values, fit$residuals, xlab = "Predicted scores", ylab = "Residuals")|>
abline(0, 0)
Part 2: Correlations
7
correlation <- cor(response, predictor, method = "pearson")8
\(H_0\): There is no correlation between the variables.
\(H_a\): There is a significant correlation between the variables.
9
cor.test(response, predictor)##
## Pearson's product-moment correlation
##
## data: response and predictor
## t = 12.124, df = 10, p-value = 2.653e-07
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.8854130 0.9911306
## sample estimates:
## cor
## 0.967625510
cor.test(response, predictor, alternative = "less")##
## Pearson's product-moment correlation
##
## data: response and predictor
## t = 12.124, df = 10, p-value = 1
## alternative hypothesis: true correlation is less than 0
## 95 percent confidence interval:
## -1.0000000 0.9890686
## sample estimates:
## cor
## 0.967625511
Changing the method from “pearson” to “spearman” may alter the test results. Discuss how this affects the interpretation.
Part 3
12
Include the plot with additional variables in your write-up.
#Create some more fake data
data2 <- data.frame(data, Predictor2 = rpois(12, 3), Response2 = sort(rnorm(12, 23)), CatPredictor = gl(3, 4, labels = c("Low", "Med", "High")))
#Plot the all the fake data
plot(data2)
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data.frame(data, Predictor2 = rpois(12, 3), Response2 = sort(rnorm(12, 23)), CatPredictor = gl(3, 4, labels = c("Low", "Med", "High")))## predictor response Predictor2 Response2 CatPredictor
## 1 4.2 6.9 3 21.49213 Low
## 2 5.6 8.3 5 21.94792 Low
## 3 5.1 7.3 1 23.14765 Low
## 4 4.6 6.3 1 23.25382 Low
## 5 9.2 11.2 3 23.34147 Med
## 6 6.9 8.1 3 23.45784 Med
## 7 10.0 13.6 4 23.57566 Med
## 8 3.4 5.7 1 23.72436 Med
## 9 7.0 8.0 0 23.84807 High
## 10 11.2 13.8 4 23.98014 High
## 11 1.2 3.3 2 23.99908 High
## 12 8.7 12.8 4 24.91393 High#Plot the all the fake data
plot(data2)
14
corInfo <- cor.test(data$predictor, data$response)
plot(response ~ predictor, data = data, pch = 16)
text(2, 13, paste("r =", signif(corInfo[["estimate"]], 3), "\nP =", signif(corInfo[["p.value"]], 3)), pos = 4)
abline(fit, col = "grey")
#I pull information out of fit to display on the graph
i=round(fit$coef[[1]], 2)
s=round(fit$coef[[2]], 2)
text(2, 13, paste("y=", i, "+", s, "*x"), pos = 4)