Question

The following data refers to the demand for money (M) and the rate of interest (R)

in for eight different economics:

M (In billions) 56 50 46 30 20 35 37 61

R% 6.3 4.6 5.1 7.3 8.9 5.3 6.7 3.5

a. Assuming a relationship i M   R U , obtain the OLS estimators of

 and 

b. Calculate the coefficient of determination for the data and interpret its value

c. Test the hypothesis that interest rate influences demand for money

d. Compute the standard error of the regression coefficients and conduct test of

significance at the 5% level of significance.

e. If in a 9th economy the rate of interest is R=8.1, predict the demand for money(M)

in this economy.

Answer 1

a) in this case ,Y ,the response variable is demand for moneyn(M) and X, the prediction is interest rate (R)

n =number of observations=8

the regression model is $y_1=\beta_1+\beta_2+x_i,i=1(i)8$

the OLS estimates of $\beta_1$ and $\beta_2$ are given by

$\bar{\beta_1}=\bar{y}-\bar{\beta_2}\bar{X},\bar{\beta_2}=\frac{\displaystyle\sum_{i=1}^8(X_1-\bar{X})(y_1-\bar{y})}{{\displaystyle\sum_{i=1}^8}(X_1-\bar{X})^2}$

$\bar{x}=\frac{1}{8}\displaystyle\sum_{i=1}^8\ x_1=.9625\\\bar{y}=\frac{1}{8}\displaystyle\sum_{i=1}^8y_1=41.875\\\bar{\beta_2}=-6.747,\bar{\beta_1}=-82101$

b)The correlation coefficient between R and M is computed as -0.828 which implies M and R are negatively correlated. That is R increaes , M decreases and vice versa. So there exist a fairly strong negative linear relationship betweeen M and R.

c)

To test whether interest rate influences demand for money

we are to test $H_o:\beta_2=0 \space against \space H_1:\beta_2\not =0$

test statistic for this test is given by

$T=\frac{}{}$

d)The standard errors are given by

e)for x(ie, R)=8.1, the predicted value from te fitted regression line,

$Y=82.101-6.747\times 8.1=27.45$

R code:

> y<-c(56,50,46,30,20,35,37,61)

> x<-c(6.3,4.6,5.1,7.3,8.9,5.3,6.7,3.5)

> xbar<-mean(x)

> xbar

[1] 5.9625

> ybar<-mean(y)

> ybar

[1] 41.875

> m1<-lm(y~x)

> e<-resid(m1)

> sigma2<-sum(e*e)/6

> sigma2

[1] 70.0688

> summary(m1) # all values can be computedby using this particular code

Call:

lm(formula = y ~ x)

Residuals:

   Min   1Q Median   3Q   Max 

-11.3446 -2.2556 -1.3806 0.7033 16.4020 

Coefficients:

      Estimate Std. Error t value Pr(>|t|)  

(Intercept) 82.101  11.498  7.14 0.00038 ***

x      -6.747   1.863 -3.62 0.01109 * 

---

Signif. codes: 

0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 8.371 on 6 degrees of freedom

Multiple R-squared: 0.686, Adjusted R-squared: 0.6337 

F-statistic: 13.11 on 1 and 6 DF, p-value: 0.01109

> ss_x<-sum((x-xbar)^2)

> ss_x

[1] 20.17875

> p_value<-2*pt(3.62,6,lower.tail = F)

> p_value

[1] 0.01109717

> ss_y<-sum((y-ybar)^2)

> r<-sum((x-xbar)*(y-ybar))/sqrt(ss_x*ss_y)

> r

[1] -0.8282484