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Good evening,

I am attempting the following question where I was able to successfully find the R-square (the coefficient of determination) at 0.8945.

While I am able to identify this, my weak foundation in regression leave me puzzled to what this actually means. Appreciate some advice please.

An accountant for a large department store would like to develop a model to predict the amount of time it takes to process invoices. Data are collected from the past 30 working days. The number of invoices processed and the completion time (in hours) are recorded. All the computer outputs are given in the end of this exam paper. You may directly use the computer outputs to answer the following questions if necessary.

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2 Answers 2

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R-squared measures the extent that a model explains variations in the response variable, i.e., Number of Hours.

Consider a baseline model with only an intercept. That model can't really explain variations - it can only predict one value every time. The mean is typically taken to be that value. Adding up the squared differences between the response values and the mean gives you a baseline variability, or level of uncertainty in the mean-only model.

Now consider a better model that depends on Invoices Processed. Adding up the squared differences between the response values and this model's prediction will also give you a variability, but it will be less than before, since this model can adapt based on differences in Invoices Processed.

R-squared is the ratio of these two variabilities (the better model is in the numerator). It can be interpreted of a fraction of the available variation that the model explains.

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$R^2$, coefficient of determination, measures the relative improvement of your regression based on previous model regarding the sum of squares between observed value and estimated value.

Previous model: original data without regression. The estimated value is the mean $\bar{Y}$.

$SSTO=\sum(Y_i-\bar{Y})^2$

Current model: regression model. The estimated value is $\hat{Y}=b_0+b_1X$.

$SSE=\sum(Y-\hat Y)^2$

$R^2=1-\frac{SSE}{SSTO}$, measures the improvement.

That means your regression model reduce the sum of squares from $SSTO$ to $SSE$.

$R^2=0.89$ means your regression reduces $89\%$ of the sum of squares of the original model $SSTO$.


A larger $R^2$ does not imply regarding the following things.

  1. Better estimation doesn't qualify better prediction
  2. It does not qualify any relationship between predictors and responses
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