Question

In the following table is a partial computer output based on a sample of 21 observations, relating an independent variable (X) and a dependent variable:

Predictor	Coefficient	Standard Error
Constant	30.139	1.181
X	‐0.2520	0.022

SOURCE	SS
Regression	1,759.481
Error	259.186

a. Develop the estimated regression line.

b. At α = 0.05, test for the significance of the slope.

c. At α = 0.05, perform an F‐test.  

d. Determine the coefficient of determination.

e. Determine the coefficient of correlation.

Answer 1

a) Estimated regression line is given by

$\hat{Y} = b_0 + b_1\hat{X}$

where $b_0,b_1$ are the estimated coefficients of original coefficients $\beta_0,\beta_1$ and $\hat{Y}$ is the fitted value value.

Thus, $\hat{Y} = 30.139 - 0.2520\hat{X}$

b)

• Hypothesis: H_o: The slope of the regression line is equal to zero.

H_a: The slope of the regression line is not equal to zero.

If the relationship between Y and X is significant, the slope will not equal zero.

• Given the significance level is 0.05, we will conduct a t-test using sample data.
• To apply the t-test to sample data, we require the standard error of the slope, the slope of the regression line, the degrees of freedom, the t statistic test statistic, and the P-value of the test statistic.

  From the table, we can see that-

  b₁ = -0.2520 SE = 0.022

  We compute the degrees of freedom and the t statistic test statistic, using the following equations.

  DF = n - 2 = 21 - 2 = 19

  t = b₁/SE = -0.2520/0.022 = -11.45

• Based on the t statistic test statistic and the degrees of freedom, we determine the P-value. The P-value is the probability that a t statistic having 19 degrees of freedom is more extreme than -11.45. Since this is a two-tailed test, using the table, we find that at 5% significance level, P-value < 0.0001

• Since the P-value is less than the significance level (0.05), we cannot accept the null hypothesis and the result is significant.

c)

To test the hypothesis
Ho 61 0
, the statistic used is based on the distribution. It can be shown that if the null hypothesis is true, then the statistic:

MSR SSR/1 MS SSE/ (n 2) Fo

follows the distribution with degree of freedom in the numerator and
(n-2)
degrees of freedom in the denominator. is rejected if the calculated statistic,
Fo
, is such that:

where is the percentile of the distribution corresponding to a cumulative probability of () and is the significance level.

From the table, we can see that regression sum of squares

$SS_R = \sum_{i=1}^{n} (\hat{y_i} - \bar{y})^2 = 1759.481$

The error sum of squares can be calculated as:

$SS_E = \sum_{i=1}^{n} ( y_i- \hat{y_i})^2 = 259.186$

Knowing the sum of squares, the statistic to test can be calculated as follows:

Ho 61 0

$f_0 = \frac{MS_R}{MS_E} = \frac{SS_R/ 1}{SS_E/(n-2)}$

$= \frac{1759.481/ 1}{259.186/(19)} = \frac{1759.481}{13.641} = 128.985$

The critical value at a significance level of 0.05 is $f_{0.05, 1, 19}$ = 4.381. Since ,
Ho 61 0
is rejected and it is concluded that is not zero, implying that a relation does exist between Y and X for the data in the preceding table.

d)

The coefficient of determination is a measure of the amount of variability in the data accounted for by the regression model.

The coefficient of determination is the ratio of the regression sum of squares to the total sum of squares.

5 SSR SST

can take on values between 0 and 1 .

$SS_T = \sum_{i=1}^{n} (y_i - \bar{y})^2 = SS_R &plus; SS_E = 1759.481 &plus; 259.186 = 2018.667$

  $= \frac{1759.481}{2018.667} = 0.872$

5 SSR SST