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.
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
Comments
Leave a comment