Answer to Question #289502 in Microeconomics for ermi

Question

A sample of 20 observations corresponding to the regression model

Y i =α + βX i + U i gave the following data.

∑y = 21.9

∑x = 186.2

∑ (x− ¯ x)(y− ¯ y) = 106.2

∑ (y− ¯ y) 2= 86.9

∑ (x − ¯ x) 2= 215.4

(a) Estimate α and β

(b) Calculate the variance of our estimates

Answer 1

We have,

$n=20\\\bar x={\sum x\over n}={186.2\over20}=9.31\\\bar y={\sum y\over n}={21.9\over20}=1.095$

Let,

$S_{xy}=\sum (x− \bar x)(y− \bar y)=106.2\\ S_{xx}=\sum (x − \bar x) ^2=215.4\\ S_{yy}=\sum (y− \bar y)^ 2=86.9$

The values of the Ordinary Least Squares Estimates are given by,

$\hat\beta={S_{xy} \over S_{xx}}={106.2\over215.4}=0.4930$

and,

$\hat\alpha= \bar y-(\bar x\times \hat \beta)=1.095-(9.31\times0.4930)=1.095-4.5902=-3.4952$

Therefore, the values of $\hat\alpha=-3.4952$ and $\hat\beta=0.4930$ both rounded off to 4 decimal places.

The variance of these estimates are derived using the formulas below.

$var(\hat \beta)={s^2\over S_{xx}}$ where $s^2$ is the sample variance. We use the sample variance $s^2$since the population variance is unknown. $s^2$ given by,

$s^2={S_{yy}-\hat \beta \times S_{xy}\over n-2}={86.9-(0.4930\times 106.2)\over 20-2}={34.5395\over18}=1.9189$(4dp)

Therefore,

$var(\hat\beta)={1.9189\over 215.4}=0.00891$(5dp)

and

$var(\hat\alpha)=({1\over n}+{\bar x^2\over S_{xx}})\times s^2=({1\over 20}+{9.31^2\over 215.4})\times 1.9189=0.8681$ (4dp)