In: Statistics and Probability
2. The estimated regression equation for a model involving two independent variables and 65 observations is: yhat = 55.17+1.2X1 -0.163X2 Other statistics produced for analysis include: SSR = 12370.8, SST = 35963.0, Sb1 = 0.33, Sb2 = 0.20. (16 points)
a. Interpret b1 and b2 in this estimated regression equation
b. Predict y when X1 = 65 and X2 = 70.
c. Compute R-square and Adjusted R-Square.
d. Comment on the goodness of fit of the model.
e. Compute MSR and MSE.
f. Compute F and use it to test whether the overall model is significant using a p-value (α = 0.05).
g. Perform a t test using the critical value approach for the significance of β1. Use a level of significance of 0.05. h. Perform a t test using the critical value approach for the significance of β2. Use a level of significance of 0.05.
a.
Keeping X2 constant, with increase in one unit in X1, the dependent variable y increases by 1.2 unit.
Keeping X1 constant, with increase in one unit in X2, the dependent variable y decreases by 0.163 unit.
b.
yhat = 55.17+1.2X1 -0.163X2
For X1 = 65 and X2 = 70,
yhat = 55.17+1.2 * 65 -0.163 * 70 = 121.76
c,
R-square = SSR / SST = 12370.8 / 35963.0 = 0.344
Adjusted R-Square = 1 - (1 -R-square) * (n-1) / (n-k-1)
= 1 - (1 - 0.344) * (65 - 1) / (65 - 2 - 1)
= 0.3228
d.
As, the R-square is less than 0.50, the model is not a good fit of the data.
e.
MSR = SSR / k = 12370.8 / 2 = 6185.4
MSE = SSE / (n-k-1) = 35963.0 / (65 - 2 - 1) = 580.05
f.
H1: At least one of the coefficients is not zero.
F = [R-squared / k] / [(1-R-squred)/(n-k-1)]
= [0.344 / 2] / [(1-0.344)/(65-2-1)]
= 16.256
Degree of freedom for F statistic, = k, n-k-1 = 2, 62
P-value = P(F > 16.256) = 0.0000
Since, p-value is less than 0.05 significance level, we reject
null hypothesis H0 and conclude that there is strong evidence
that
overall model is significant
g.
Degree of freedom = n-k-1 = 65-2-1 = 62
Critical value of t at α = 0.05 and df = 62 is 2.00
Test statistic, t = b1 / Sb1 = 1.2 / 0.33 = 3.64
Since the observed t (3.64) is greater than the critical value, we reject the null hypothesis H0 and conclude that there is strong evidence that β1 is significant.
Degree of freedom = n-k-1 = 65-2-1 = 62
Critical value of t at α = 0.05 and df = 62 is 2.00
Test statistic, t = b2 / Sb2 = -0.163 / 0.20 = -0.815
Since the absolute value of t (0.815) is less than the critical value, we fail to reject the null hypothesis H0 and conclude that there is no strong evidence that β2 is significant.