Assignment_Unit4
SilaAda
29 November 2020
Task 1: Correlation
- For the following sets of variables produce a scatterplot,
Comment on the relationship between the variables (strength and direction of the relationship).
- Number of ads and number of impressions.
## Plot
plot(x = dat$number_ads, y = dat$impressions)
## Covariance
# Covariance using built-in function:
SxSy <- cov(dat$number_ads, dat$impressions)
# Covariance, hipster-style:
RESx <- dat$number_ads - mean(dat$number_ads)
RESy <- dat$impressions - mean (dat$impressions)
(SxSy <- sum(RESx * RESy)/(N-1))## [1] 5974098## Correlation
# Built-in function
r <- cor(dat$number_ads, dat$impressions)
# Hipster version
(r <- SxSy/(sd(dat$number_ads)*sd(dat$impressions)))## [1] 0.6975701There’s positive (70%) correlation between variables, so as number of ads increase on a webpage, number of impressions increase as well. But direction is unclear.
- Number of impressions and number of clicks.
## Plot
plot(x = dat$impressions, y = dat$clicks)
## Covariance
# Covariance using built-in function:
SxSy <- cov(dat$clicks, dat$impressions)
# Covariance, hipster-style:
RESx <- dat$clicks - mean(dat$clicks)
RESy <- dat$impressions - mean (dat$impressions)
(SxSy <- sum(RESx * RESy)/(N-1))## [1] 122209677## Correlation
# Built-in function
r <- cor(dat$clicks, dat$impressions)
# Hipster version
(r <- SxSy/(sd(dat$clicks)*sd(dat$impressions)))## [1] 0.6717794There’s positive (67%) correlation between variables, so as number of impressions increase on a webpage, number of clicks increase as well. But direction is unclear.
Task 2: Simple linear regression
You are curious about the relationships discussed earlier in Task 1, so you decide to dive deeper.
- Run a linear regression on the two sets of variables and interpret the coefficients.
Number of ads - impressions
Assumptions check
Looking at the scatter plot as well as model diagnostics below, you’d prefer to use a more proper and advanced regression model, however for simplicity we go with the linear regression model first.
x <- dat$number_ads
y <- dat$impressions
## hipster style:
X <- cbind(1, x)
coeffs <- solve(t(X) %*% X) %*% (t(X) %*% y)
coeffs <- solve(crossprod(X)) %*% crossprod(X, y)
lr <- lm(impressions ~ number_ads + number_ads, data = dat)
summaryLR <- summary(lr); summaryLR##
## Call:
## lm(formula = impressions ~ number_ads + number_ads, data = dat)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2589730 -1499885 -218481 805872 9582015
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2153615 100183 -21.50 <2e-16 ***
## number_ads 474641 13998 33.91 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1730000 on 1213 degrees of freedom
## Multiple R-squared: 0.4866, Adjusted R-squared: 0.4862
## F-statistic: 1150 on 1 and 1213 DF, p-value: < 2.2e-16plot(y ~ x, pch = 16)
abline(lr, col = "darkred", lwd = 2)
legend("topleft", paste0("R-square = ", round(summaryLR$r.squared, 2)))
par(mfrow = c(2, 2))
plot(lr)
Number of impressions - clicks
Assumptions check
Looking at the scatter plot as well as model diagnostics below, you’d prefer to use a more proper and advanced regression model, however for simplicity we go with the linear regression model first.
y <- dat$clicks
x <- dat$impressions
reg <- lm(clicks ~ impressions, data = dat)
summaryLR <- summary(reg); summaryLR##
## Call:
## lm(formula = clicks ~ impressions, data = dat)
##
## Residuals:
## Min 1Q Median 3Q Max
## -172.80 -1.75 -1.69 -0.76 828.21
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.684e+00 1.687e+00 0.998 0.318
## impressions 2.097e-05 6.640e-07 31.585 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 55.85 on 1213 degrees of freedom
## Multiple R-squared: 0.4513, Adjusted R-squared: 0.4508
## F-statistic: 997.6 on 1 and 1213 DF, p-value: < 2.2e-16plot(y ~ x, pch = 16)
abline(reg, col = "darkred", lwd = 2)
legend("topleft", paste0("R-square = ", round(summaryLR$r.squared, 2)))
par(mfrow = c(2, 2))
plot(reg)
- Prediction:
- How many impressions would a website owner sell on average if she published 5 ad slots on her website?
num_ads <- 5
lr$coefficients[1] + lr$coefficients[2]*num_ads## (Intercept)
## 219592- How many clicks would a website owner generate on average if she sold 1M impressions on her website?
num_imps <- 1e+6
reg$coefficients[1] + reg$coefficients[2]*num_ads## (Intercept)
## 1.684169Task 3: Multiple linear regression
- Run an appropriate test.
mr <- lm(clicks ~ impressions + number_ads, data = dat)
summary(mr)##
## Call:
## lm(formula = clicks ~ impressions + number_ads, data = dat)
##
## Residuals:
## Min 1Q Median 3Q Max
## -170.32 -3.36 -1.44 -0.01 830.50
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.687e+00 3.800e+00 -0.444 0.657
## impressions 2.033e-05 9.267e-07 21.941 <2e-16 ***
## number_ads 6.243e-01 6.305e-01 0.990 0.322
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 55.85 on 1212 degrees of freedom
## Multiple R-squared: 0.4517, Adjusted R-squared: 0.4508
## F-statistic: 499.3 on 2 and 1212 DF, p-value: < 2.2e-16- Compare the obtained results with above regression 2 results and comment on whether the results are improved.
- Interpret the effect of number of impressions on number of clicks.
Effect of number of impressions doesn’t change (significant / positive) when we include number of ads in the regression. However effects gets slightly lower. Number of ads has no significant effect on number of clicks, despite its significant effect on number of impressions.
- How many clicks would a website owner generate on average if she published 5 ad slots and sold 1M impressions on her website?
mr$coefficients[1] + mr$coefficients[2]*num_imps + mr$coefficients[3]*num_ads## (Intercept)
## 21.76686