7.1.1 Examples

We return to the birth weight data in example 6.2.1. The full data set is given in Table 7.1. We will fit the simple linear regression for birth weight $Y_i$ with gestational age $x_i$ as explanatory variable,

The response vector and design matrix are

and

Obtain the least squares estimate $\widehat{\beta }$.

To find $\widehat{\beta }$ we use the formula

From above,

1. 1

2. 2

3. 3

   $X^{\prime }y=\left[\begin{array}{c}\hfill 71224\hfill \\ \hfill 2753867\hfill \end{array}\right].$

Therefore,

The fitted model for birth weight, given gestational age at birth is,

We can interpret this as follows,

• 1

  For every additional week of gestation, expected birth weight increases by $116$ grams.

• 2

  If a child was born at zero weeks of gestation, their birth weight would be $-1485$ grams.

Why does the second result not make sense?

Load the data set bwt into R. To obtain the size of the data set,

This tells us that there are 24 subjects and 3 variables. The variables are,

To fit the simple linear regression of the previous example ‘by hand’,

1. 1

   Set up the design matrix,

   ⬇

   > X <- matrix(cbind(rep(1,24),bwt$Age),ncol=2)
2. 2

   Calculate $\widehat{\beta }$ using equation (7.3),

   ⬇

   > beta <- solve(t(X)%*%X)%*%t(X)%*%bwt$Weight
3. 3

   View results

   ⬇

   > beta

   [,1]

   [1,] -1484.9846

   [2,]   115.5283

To fit the same model using lm,

1. 1

   Specify the required model. Note R assumes that you want to include an intercept term, so this need not be explicitly included,

   ⬇

   > bwtlm <- lm(bwt$Weight~bwt$Age)
2. 2

   To view the estimates $\widehat{\beta }$,

   ⬇

   > bwtlm$coefficient

   (Intercept)     bwt$Age

   -1484.9846    115.5283

Recall example 6.4.2 in which we investigated the relationship between gas consumption and external temperature. To measure the effect of changes in the external temperature on gas consumption, we fit the multiple linear regression model 6.6. We will allow a different relationship between gas consumption and outside temperature before and after the installation of cavity wall insulation. The model has four regression coefficients

Here $x_{i,1}$ is outside temperature and $x_{i,2}$ is an indicator variable taking the value 1 after installation.

The data are shown in Table 7.2.

To estimate the parameters by hand, we first set up the response vector and design matrix,

and

Since $X^{\prime }X$ will be a $4\times 4$ matrix, it is easier to do our calculations in R. First load the data set gas.

• 1

  Insulate contains Before or After to indicate whether or not cavity wall insulation has taken place;

• 2

  Temp contains outside temperature;

• 3

  Gas contains gas consumption;

• 4

  Insulate2 contains a 0 or 1 to indicate before (0) or after (1) cavity wall insulation.

To set up the design matrix

Then to obtain $\widehat{\beta }$,

Thus the fitted model is

• 1

  Before cavity wall insulation, when the outside temperature is 0${}^{\circ }$C, the expected gas consumption is 6.85 1000’s cubic feet.

• 2

  Before cavity wall insulation, for every increase in temperature of 1${}^{\circ }$C, the expected gas consumption decreases by 0.393 1000’s cubic feet.

• 3

  After cavity wall insulation, for every increase in temperature of 1${}^{\circ }$C, the expected gas consumption decreases by 0.249 1000’s cubic feet.

Where does the figure 0.249 come from?

Substitute $x_{i,2}=1$ into the fitted model; -0.393+0.144 is the overall rate of change of gas consumption with temperature.

What is the expected gas consumption after cavity wall insulation, when the outside temperature is $0^{\circ }$C?

We can alternatively fit this model in R using lm,