bike sharing data analysis

a) Fit a regression model to dependent variable cnt using independent variable month_name

model1 <- lm(cnt ~ month_name, data = bsd)

b) Get the summary of the fitted model

summary(model1)

## 
## Call:
## lm(formula = cnt ~ month_name, data = bsd)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -5177.2 -1095.2  -249.3  1290.0  4669.7 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     4484.9      196.7  22.799  < 2e-16 ***
## month_nameAug   1179.5      275.9   4.275 2.17e-05 ***
## month_nameDec  -1081.1      275.9  -3.918 9.79e-05 ***
## month_nameFeb  -1829.6      281.8  -6.492 1.58e-10 ***
## month_nameJan  -2308.6      275.9  -8.366 3.09e-16 ***
## month_nameJul   1078.8      275.9   3.909 0.000101 ***
## month_nameJun   1287.5      278.2   4.628 4.38e-06 ***
## month_nameMar   -792.6      275.9  -2.873 0.004192 ** 
## month_nameMay    864.9      275.9   3.134 0.001793 ** 
## month_nameNov   -237.7      278.2  -0.854 0.393113    
## month_nameOct    714.3      275.9   2.589 0.009829 ** 
## month_nameSep   1281.6      278.2   4.607 4.83e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1524 on 719 degrees of freedom
## Multiple R-squared:  0.3906, Adjusted R-squared:  0.3813 
## F-statistic:  41.9 on 11 and 719 DF,  p-value: < 2.2e-16

As we can observe the R-squared in our case is 0.3906 which means that ~39% of the variation in cnt is explained by the month name variable. R-squared basically is the square of the correlation coefficient and informs us how much percent of the variation of dependent variable is explained by the indepenedent variable.

c) Reference month

It can be observed that the month names are appearing in lexicographic order. So it is easy to identify that month of April is set as the reference. This can be confirmed also by going through the list and finding the missing month from the 12 months.

In our case, the intercept represents the estimate of cnt for the month of April, and the coefficient estimates for the other variables represent the difference between that month from the month of April.

Hence, 4484.9 is the reference month’s prediction.

d) Predictions for the month of January and June

2176.339 for the month of January and 5772.367 for the month of June

newObservations = data.frame(month_name=c('Jan', 'Jun'))

newObservations$predictedValues = predict(model1, newObservations)

# model1$fitted.values

There can be multiple ways of seeing the predicted values for the month of January and june, all of which arrive at the same answer.

The first one is theoretical where we can see that our model is cnt = 4484.9 + (-2308.6)(1 if month if june, 0 otherwise) + (coefficient of Feb)(1 if month if june, 0 otherwise) + ……

So basically the above linear model uses the coefficients that we find out by fitting our regression model. So the prediction for the month of January would be cnt = 4484 - 2308.6 = 2176.339 Similarly, for the month of June, it would be cnt = 4484 + 1287.5 = 5772.367

This can also be achieved with the predict() function as used in the code chunk above or the fitted.values column of the model1.

a) Fit a multiple linear regression model where the dependent variable is cnt and the independent variables are temp and month_name.

model2 = lm(cnt ~ month_name + temp, data = bsd)

b) The summary of the fitted model is reported below

summary(model2)

## 
## Call:
## lm(formula = cnt ~ month_name + temp, data = bsd)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4896.6 -1080.0  -228.4  1245.2  3372.9 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    1554.39     390.76   3.978 7.66e-05 ***
## month_nameAug  -308.08     315.42  -0.977   0.3290    
## month_nameDec  -170.96     283.80  -0.602   0.5471    
## month_nameFeb  -764.81     296.15  -2.582   0.0100 *  
## month_nameJan  -852.31     313.41  -2.719   0.0067 ** 
## month_nameJul  -701.18     335.50  -2.090   0.0370 *  
## month_nameJun   -47.47     307.78  -0.154   0.8775    
## month_nameMar  -297.20     269.38  -1.103   0.2703    
## month_nameMay    86.73     278.37   0.312   0.7555    
## month_nameNov   390.66     275.22   1.419   0.1562    
## month_nameOct   620.72     263.30   2.357   0.0187 *  
## month_nameSep   368.25     285.93   1.288   0.1982    
## temp           6235.14     729.40   8.548  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1453 on 718 degrees of freedom
## Multiple R-squared:  0.4469, Adjusted R-squared:  0.4377 
## F-statistic: 48.35 on 12 and 718 DF,  p-value: < 2.2e-16

First, we notice that the R-squared for model2 is .4469, which is higher than the R-squared of 0.3906 in model1. This shows that a lot more variance of cnt variable is explained by our current model since it combines two variables month_name and temp to perform the prediction rather than using just a single variable to predict cnt. This means that we can have a little more confidence in the accuracy of these predictions than in the predictions from model1.

tempmodel = lm(cnt ~ temp, data = bsd)
summary(tempmodel)

## 
## Call:
## lm(formula = cnt ~ temp, data = bsd)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4615.3 -1134.9  -104.4  1044.3  3737.8 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   1214.6      161.2   7.537 1.43e-13 ***
## temp          6640.7      305.2  21.759  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1509 on 729 degrees of freedom
## Multiple R-squared:  0.3937, Adjusted R-squared:  0.3929 
## F-statistic: 473.5 on 1 and 729 DF,  p-value: < 2.2e-16

Also, we can observe with the help of the above code chunk that the R-squared is not simply a sum of the R-squared from the simple regression models regressed using month_name or temp. This could be because month_name and temp are correlated to each other.

c) The coefficient estimates for model 1 and model 2 are compared below.

export_summs(model1, model2)

It can be seen that the coefficient estimated for the month_nameJan is -2308.56 for model1 whereas it increase to -852.31 in the case of model2.

This could be because month_name and temp are correlated with each other, and the coefficient estimates represent the unique effect of each predictor variable after considering the effect(s) of the other predictor variables.

d) Predict

newObservations2 = data.frame(month_name=c('Jan'), temp=0.25)

newObservations2$predictedValues = predict(model2, newObservations2)

The predicted value for count is 2260.863 when the month is January and the temp is 0.25