In this video, answering these questions and completing an ANOVA table
for this multiple regression model scenario. Here are the formulas we need for the ANOVA
table. First, we have 40 observations, so n = 40
and 5 independent variables means k = 5. So, we have df regression k = 5 here.
And for df total, n – 1 is 40 – 1 which equals 39.
df Residual or Error will be 39 minus 5 which gives 34.
We’re given the unexplained variation SS residual
or SSE to be 93.5 here. Total variation or SSTotal is given as 150.3
Thus, SSRegression or SSR will be SST – SSE which gives 56.8.
Now MS = SS/df So, MSR = 56.8/5 which gives 11.36
and MSE is 93.5/34 which gives 2.75 The F statistic is MSR/MSE which gives 4.13
The ANOVA table is now complete. To compute the coefficient of determination,
we will use SSR/SST. That is 56.8 divided by 150.3.
And that gives 0.3779 That is, about 38% of the variation in the
dependent variable can be explained by the variation in the independent variables.
The remaining 62% is not explained by the model.
Next, we can compute the adjusted R-square by using this formula:
n – 1 is 39, n – k – 1 is 34
and R-square is 0.3779. Substituting those in the formula
gives an adjusted R-square of 0.2864. For the standard error, we can simply take
the square root of the mean square error (MSE) of 2.75 here, and that gives 1.658.
Finally, we conduct an overall significance test for the model at alpha = .05.
The null hypothesis for the test is that the regression coefficients of the 5 independent
variables are all zero. That is, there is no significant relationship
between the dependent variable and any of the independent variables.
The alternative hypothesis will be that at least one of the coefficients in not zero.
That is, at least one independent variable is linearly related to the dependent variable.
The test statistic is the F-stat computed here to be 4.13.
The critical F-value for this test can be obtained from the F-table with numerator degrees
of freedom k = 5 and the denominator n – k – 1 = 34.
At alpha = .05, this critical value turns out to be approximately 2.5.
In other words, we will reject the null hypothesis if the test statistic is greater than 2.5.
Since our test statistic is greater than the critical value, we reject the null hypothesis
ay alpha = .05. Alternatively, by using software, we can also
obtain the p-value to be .0049. Since this p-value is less than our alpha
of .05, we will again reject the null hypothesis using the p-value approach.
In essence both the critical value and p-value methods should lead to the same decision.
We can thus conclude that the overall model is significant or valid.
That is, there is a significant relationship between the dependent variable and at least
of the independent variables. And that’s it for this video.
Thanks for watching.