Chapter 13, Problem 15ERQ

To determine

Identify which predictor differ significantly from zero at $\alpha =0.01$.

Answer to Problem 15ERQ

The predictor X3 differ significantly from zero at $\alpha =0.01$.

Explanation of Solution

Calculations:

The given information is that,

	Coefficient	Std. Error
Intercept	3.210610	0.918974
X1	–0.034719	0.023283
X2	0.026794	0.039741
X3	–0.048533	0.009024
X4	0.000272	0.001145
X5	0.228930	0.103327

The formula for the test statistic for coefficient of predictor $X_j$ is,

$t_{calc}=\frac{b_j-0}{s_j}$.

Where, $b_j$ is the coefficient, and $s_j$ is the standard error.

The formula for degrees of freedom is,

$df=n-k-1$.

Where, n is the total number of observations, k is the number of predictors.

Substitute 40 for n, and 5 for k in the degrees of freedom formula.

$\begin{array}{c}df=40-5-1\\ =40-6\\ =34\end{array}$

Thus, the degrees of freedom is 34.

From the Appendix D: Critical values of $t_{.01}$:

• • Locate the value 0.01 in significance level for Two-Tailed row.
• • Locate the value 34 in degrees of freedom $\left(d.f.\right)$ column.
• • The intersecting value that corresponds to 45 degrees of freedom with level of significance 0.01 is 2.728.

Hence, the critical value for 34 degrees of freedom with 0.01, level of significance is $\pm 2.728$.

Decision rules:

• • If test statistic value lies between positive and negative critical values, then fail to reject null hypothesis.
• • Otherwise, reject the null hypothesis.

Coefficient of X1:

$H_0:{\beta }_1=0$

That is, variable X1 is not related to Y.

Alternative hypothesis:

$H_1:{\beta }_1\ne 0$

That is, variable X1 is related to Y.

Substitute, –0.034719 for $b_1$ and 0.023283 for $s_1$ in the test statistic formula

$\begin{array}{c}{t}_{calc}=\frac{b_1-0}{s_1}\\ =\frac{-0.034719}{0.023283}\\ =-1.4912\end{array}$

The t statistic for X1 is –1.491.

Conclusion:

The value of test statistic is –1.491.

The critical values are –2.728 and 2.728.

The test statistic value lies between positive and negative critical value.

That is, $-2.728\left(=-t_{\frac{\alpha }{2},n-1}\right)<-1.491\left(=t_{\text{cal}}\right)<2.728\left(=t_{\frac{\alpha }{2},n-1}\right)$.

Hence the null hypothesis is not rejected.

Thus, the variable X1 is not significant.

Coefficient of X2:

$H_0:{\beta }_2=0$

That is, variable X2 is not related to Y.

Alternative hypothesis:

$H_1:{\beta }_2\ne 0$

That is, variable X2 is related to Y.

Substitute, 0.026794 for $b_2$ and 0.039741 for $s_2$ in the test statistic formula

$\begin{array}{c}{t}_{calc}=\frac{b_2-0}{s_2}\\ =\frac{0.026794}{0.039741}\\ =0.6742\end{array}$

Thus, the t statistic for X2 is 0.674.

Conclusion:

The value of test statistic is 0.674.

The critical values are –2.728 and 2.728.

The test statistic value lies between positive and negative critical value.

That is, $-2.728\left(=-t_{\frac{\alpha }{2},n-1}\right)<0.647\left(=t_{\text{cal}}\right)<2.728\left(=t_{\frac{\alpha }{2},n-1}\right)$.

Hence the null hypothesis is not rejected.

Thus, the variable X2 is not significant.

Coefficient of X3:

$H_0:{\beta }_3=0$

That is, variable X3 is not related to Y.

Alternative hypothesis:

$H_1:{\beta }_3\ne 0$

That is, variable X3 is related to Y.

Substitute, –0.048533 for $b_3$ and 0.009024 for $s_3$ in the test statistic formula

$\begin{array}{c}{t}_{calc}=\frac{b_3-0}{s_3}\\ =\frac{-0.048533}{0.009024}\\ =-5.3782\end{array}$

The t statistic for X3 is –5.378.

Conclusion:

The value of test statistic is –5.378.

The critical values are –2.728 and 2.728.

The test statistic value is less than the negative critical value.

The test statistic value does not lie between positive and negative critical value.

That is, $-5.378\left(=t_{\text{cal}}\right)<-2.728\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence, the null hypothesis is rejected.

Thus, the variable X3 is significant.

Coefficient of X4:

$H_0:{\beta }_4=0$

That is, variable X4 is not related to Y.

Alternative hypothesis:

$H_1:{\beta }_4\ne 0$

That is, variable X4 is related to Y.

Substitute, 0.000272 for $b_4$ and 0.001145 for $s_4$ in the test statistic formula

$\begin{array}{c}{t}_{calc}=\frac{b_4-0}{s_4}\\ =\frac{0.000272}{0.001145}\\ =0.2376\end{array}$

Thus, the t statistic for X4 is 0.238.

Conclusion:

The value of test statistic is 0.238.

The critical values are –2.728 and 2.728.

The test statistic value lies between positive and negative critical value.

That is, $-2.728\left(=-t_{\frac{\alpha }{2},n-1}\right)<0.238\left(=t_{\text{cal}}\right)<2.728\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence the null hypothesis is not rejected.

Thus, the variable X4 is not significant.

Coefficient of X5:

$H_0:{\beta }_5=0$

That is, variable X5 is not related to Y.

Alternative hypothesis:

$H_1:{\beta }_5\ne 0$

That is, variable X5 is related to Y.

Substitute, 0.228930 for $b_5$ and 0.103327 for $s_5$ in the test statistic formula

$\begin{array}{c}{t}_{calc}=\frac{b_5-0}{s_5}\\ =\frac{0.228930}{0.103327}\\ =2.2156\end{array}$

Thus, the t statistic for X5 is 2.2156.

Conclusion:

The value of test statistic is 2.2156.

The critical values are –2.728 and 2.728.

The test statistic value lies between positive and negative critical value.

That is, $-2.728\left(=-t_{\frac{\alpha }{2},n-1}\right)<2.2156\left(=t_{\text{cal}}\right)<2.728\left(=t_{\frac{\alpha }{2},n-1}\right)$.

Hence the null hypothesis is not rejected.

Thus, the variable X5 is not significant.

Hence, the predictor X3 differ significantly from zero at $\alpha =0.01$.