Answer to Question #216410 in Microeconomics for israel Bereket

Question

Suppose that a researcher estimates a consumption function and obtains the

following results:

C=15+0.81Yd n=19 R2=0.99

(3.1) (18.7)

where C=Consumption, Yd=disposable income, and numbers in the parenthesis are the

‘t-ratios’

a. Test the significant of Yd statistically using t-ratios

b. Determine the estimated standard deviations of the parameter estimates

Answer 1

a

For a fitted regression model $Y=\hat{\beta_1}+\hat{\beta_2}x$ with Y as response and X as prediction variable, the test statistic for testing the significance of X is given by,

$T=\frac{\hat{\beta_2}-{\beta^0_2}}{\sqrt{\frac{\hat{\sigma^2}}{s_{xx}}}}$ $\utilde{H_0}\space t_n-2$

where, $\hat{\sigma^2}=\frac{Rss}{n-2}, s_{xx}=\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2, \bar{x}=\frac{1}{n}\displaystyle\sum_{i=1}^nxi,$

$\beta^0_2$ is the hypothesized value of $\beta_2,$ here it is 0.

a.

In this case, $Y=C,x=Y_d, n=19, R^2=0.99$

T-ratio for $\beta_1=3.1,$ T-ratio for $\beta_2=18.7$

Therefore we are to test the null hypothesis, $H_1:\beta_2=0.$

against the alternative hypothesis, $H_1:\beta_2\not=0.$

The test statistic for this test is given by, $t_{\beta_2}=$ T-ratio for $\beta_2=18.7$ [ ∵ under $H_0, \beta^0_2=0]$

The p-value for this test can be computed for t-distribution with $f=n-2=19-2=17$ using R code:

$2^Hpt(18.7,17,lower.tail=F)$

which gives p-value as 0 for which we reject the null hypothesis.

∵ we can conclude that $Y_d$ is statistically significant.

b.

The t-ratio is basically the estimate divided by the standard error. Again the standard error is the standard deviation of the estimates.

∵ $t-ratio(\beta_1)=3.1=\frac{\hat{\beta^1}}{se\hat({\beta_1})}=\frac{15}{se\hat({\beta_1})}$

$\implies se({\beta_1})=\frac{15}{3.1}=4.839$

$t-ratio(\beta_2)=18.7=\frac{\hat{\beta^2}}{se\hat({\beta_2})}=\frac{0.81}{se\hat({\beta_2})}$

$\implies se({\beta_2})=\frac{0.81}{18.7}=0.043$

Therefore the estimated standard deviations of the parameter estimates are 4.389 and 0.043 respectively.