Assume that the current date is February 1,2003.The linear regression model was applied to a monthly time series based on the last 24 months' sales (from January 2000 through December 2002).The following partial computer output summarizes the results. $$\begin{array} { l c c }
\text { Coefficient } & \text { Estimate } & \mathrm { t } \
\text { Intercept } & 4.3 & 2.07 \
\text { Slope } & 1.6 & 2.98
\end{array}$$
At a significance level of .05,what is the value of the rejection point in testing the slope for significance?
A)1.717
B)1.96
C)2.074
D)1.645
E)2.064

Question

Assume that the current date is February 1,2003.The linear regression model was applied to a monthly time series based on the last 24 months' sales (from January 2000 through December 2002).The following partial computer output summarizes the results. $$\begin{array} { l c c } 
\text { Coefficient } & \text { Estimate } & \mathrm { t } \
\text { Intercept } & 4.3 & 2.07 \
\text { Slope } & 1.6 & 2.98
\end{array}$$
At a significance level of .05,what is the value of the rejection point in testing the slope for significance?
A)1.717
B)1.96
C)2.074
D)1.645
E)2.064

Quizplus · Accepted Answer

The rejection point for the slope is determined by the t-distribution with n-2 degrees of freedom. With 24 months of data, n = 24, so degrees of freedom = 22. At a significance level of 0.05 for a two-tailed test, the critical t-value is approximately 2.074.

Assume That the Current Date Is February 1,2003 At a Significance Level of