Linear Regression of Ice Cream Revenue (dollars) v. Temperature (degrees Celsius)

Description of Data

The data was sourced from https://www.mathsisfun.com/data/correlation.html, which is an educational website. It is composed of 12 data points, one for each month of the year, starting in January and ending in December.

Analysis of Data

   Revenue<-c(215, 325, 185, 332, 406, 522, 412, 614, 544, 421, 445, 408)
   Temperature<-c(14.2,16.4,11.9,15.2,18.5,22.1,19.4,25.1,23.4,18.1,22.6,17.2)

    IC<-lm(Temperature~Revenue)
    summary(IC)
## 
## Call:
## lm(formula = Temperature ~ Revenue)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.64513 -0.49846 -0.08613  0.41632  2.62743 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 6.412805   1.219376   5.259 0.000369 ***
## Revenue     0.030471   0.002902  10.499 1.02e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.213 on 10 degrees of freedom
## Multiple R-squared:  0.9168, Adjusted R-squared:  0.9085 
## F-statistic: 110.2 on 1 and 10 DF,  p-value: 1.016e-06
    plot(Revenue,Temperature,main="Ice Cream Revenue v. Temperature",xlab="Revenue (dollars)",ylab="Temperature (Celsius)")

    abline(IC,col="red")

    fitted<-coef(IC)
    int<-fitted[1]
    slope<-fitted[2]
    slope
##    Revenue 
## 0.03047139
    residuals(IC)
##           1           2           3           4           5           6 
##  1.23584611  0.08399334 -0.15001223 -1.32930638 -0.28418914 -0.21887024 
##           7           8           9          10          11          12 
##  0.43298252 -0.02223800  0.41075921 -1.14125997  2.62742670 -1.64513192
    res<-residuals(IC)
    res
##           1           2           3           4           5           6 
##  1.23584611  0.08399334 -0.15001223 -1.32930638 -0.28418914 -0.21887024 
##           7           8           9          10          11          12 
##  0.43298252 -0.02223800  0.41075921 -1.14125997  2.62742670 -1.64513192
    plot(Revenue,res)
    abline(a=0,b=0,col="blue")

Discussion (slope and intercept, r^2 value, residuals, question?)

This data was modeled with the linear equation \(Temperature=\epsilon\ revenue + \alpha\) where the slope is equal to 0.030471. The \(R^2\) value is 0.9168, which indicates a strongly linear relationship between ice cream revenue and temperature. As the temperature increased, the ice cream revenue (sales) also increased.