2.3 Evaluating and interpreting the model

We are now ready to carry out the simple linear regression analysis. The results of the analysis are as follows:


Call:
lm(formula = happiness_2019 ~ income_2019, data = df)

Residuals:
     Min       1Q   Median       3Q      Max 
-19.4572  -3.5785  -0.1413   3.8410  17.5070 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 4.478e+01  1.559e+00   28.72  < 2e-16 ***
income_2019 5.642e-04  5.489e-05   10.28 4.94e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 5.768 on 76 degrees of freedom
  (2 observations deleted due to missingness)
Multiple R-squared:  0.5816,    Adjusted R-squared:  0.5761 
F-statistic: 105.6 on 1 and 76 DF,  p-value: 4.945e-16

From the above output, we can note the following:

  • The results related to \(\widehat{\beta}_0\) and \(\widehat{\beta}_1\) are under the heading Coefficients:
  • The first row (Intercept) corresponds to the intercept coefficient \(\widehat{\beta}_0\), while the second row income_2019 corresponds to the slope coefficient \(\widehat{\beta}_1\)
  • The estimate for \(\beta_0\) is 4.478e+01. The e+01 tells us to move the decimal point one place to the right, so we have that \(\widehat{\beta}_0 = 44.78\)
  • The estimate for \(\beta_1\) is 5.642e-04. The e-04 tells us to move the decimal point four places to the left, so we have that, rounded to four decimal places, \(\widehat{\beta}_1 = 0.0006\)
  • Knowing the values for \(\widehat{\beta}_0\) and \(\widehat{\beta}_1\), we can write down the estimated model as:
    • \(\widehat{\text{Happiness}} = 44.78 + 0.0006\times\text{Income}\)
  • We can interpret the value of \(\widehat{\beta}_1 = 0.0006\) as follows: "We estimate that, on average, for every $1 increase in GDP per capita, the average happiness score will be 0.0006 higher".
  • Reading from the column labeled Pr(>|t|), the \(p\)-value for the intercept coefficient is < 2e-16, which is very close to zero. This is a test of the form \(H_0 : \beta_0 = 0\) versus \(H_1 : \beta_0 \neq 0\).
  • The \(p\)-value for the slope coefficient is 4.94e-16 which is also very close to zero. This is a test of the form \(H_0 : \beta_1 = 0\) versus \(H_1 : \beta_1 \neq 0\). Since we have \(p < 0.05\), we reject \(H_0\) and conclude that \(\beta_1\) is not zero. This means there is evidence of a significant linear association between income and happiness. (More information on this below)
  • The Multiple R-squared value, which can be found in the second last row, is \(R^2 = 0.5816\). This indicates that 58.16% of the variation in the response can be explained by the model, which is a good fit. (More information on this below)

2.3.1 Testing for \(H_0 : \beta_1 = 0\) versus \(H_1 : \beta_1 \neq 0\)

Recall the simple linear regression model

\[y = \beta_0 + \beta_1x + \epsilon.\]

If the true value of \(\beta_1\) were 0, then the regression model would become

\[y = \beta_0 + \epsilon,\]

meaning \(y\) does not depend on \(x\) in any way. In other words, there would be no association between \(x\) and \(y\). For this reason, the hypothesis test for \(\beta_1\) is very important.

2.3.2 \(R^2\), the Coefficient of Determination

\(R^2\) values are always between 0 and 1. In fact, the \(R^2\) value is simply the correlation squared. To see this, recall that in Section 1.1, we found that the correlation coefficient was \(r = 0.76263\). If we square this number, we get \(R^2 = 0.76263^2 = 0.5816\). Conversely, if we take the square root of \(R^2\), we can find the correlation.

The \(R^2\) value can be used to evaluate the fit of the model. \(R^2\) values close to 0 indicate a poor fit, whereas \(R^2\) values close to 1 indicate an excellent fit. Although the interpretation of the \(R^2\) value can sometimes differ by subject matter, for the purposes of this subject, the below table can be used as a guide when interpreting \(R^2\) values:

\(R^2\) value Quality of the SLR model
\(0.8 \leq R^2 \leq 1\) Excellent
\(0.5 \leq R^2 < 0.8\) Good
\(0.25 \leq R^2 < 0.5\) Moderate
\(0 \leq R^2 < 0.25\) Weak