Question: Job performance under time pressure. Refer to the Academy of Management Journal (October 2015) study of how time pressure affects team job performance, Exercise 12.89 (p. 765). Recall that the researchers hypothesized a complete second-order model relating team performance (y) to perceived time pressure (x₁), and whether or not the team had an effective leader (x₂ = 1 if yes, 0 if no):

$E\left(Y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x_2}^2+{\beta }_3{x}_2+{\beta }_4{x}_1{x}_2+{\beta }_5{x_1}^2{x}_2$

a) How would you determine whether the rate of increase of team performance with time pressure depends on effectiveness of the team leader?

b) For fixed time pressure, how would you determine whether the mean team performance differs for teams with effective and non-effective team leaders?

a. The rate of increase of team performance with time pressure depends on effectiveness of the team leader will be determined by the interaction term. The interaction term explains the relation between the independent variables (here time pressure and effectiveness of the team leader) and how that relationship affects the dependent variable (team performance).

b. For given fixed time pressure, the changes in mean team performance for teams with effective and non-effective team leaders can be explained for effective team leader by

.$$\begin{gathered} byE\left(y\right)=({\beta }_0+{\beta }_3)+({\beta }_1+{\beta }_4)x_1+({\beta }_2+〖{\beta }_5)x_1{〗}^2 \\ andfornon-effectiveteamleaderby \\ E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2〖x_1{〗}^2. \end{gathered}$$

For given fixed time pressure, the changes in mean team performance for teams with effective and non-effective team leaders can be explained by putting x₂ = 1 and 0 for effective and non-effective leader respectively.

Mathematically, for effective team leader

$$\begin{gathered} E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2〖x_1{〗}^2+{\beta }_3{x}_2+{\beta }_4{x}_1{x}_2+{\beta }_5〖x_1{〗}^2{x}_{2}for{x}_2=1 \\ E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2〖x_1{〗}^2+{\beta }_3\left(1\right)+{\beta }_4{x}_1\left(1\right)+{\beta }_5〖x_1{〗}^2\left(1\right) \\ E\left(y\right)=({\beta }_0+{\beta }_3)+({\beta }_1+{\beta }_4)x_1+({\beta }_2+〖{\beta }_5)x_1{〗}^2 \\ Andfornon-effectiveteamleader \\ E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2〖x_1{〗}^2+{\beta }_3{x}_2+{\beta }_4{x}_1{x}_2+{\beta }_5〖x_1{〗}^2{x}_{2}for{x}_2=0 \\ E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2〖x_1{〗}^2+{\beta }_3\left(0\right)+{\beta }_4{x}_1\left(0\right)+{\beta }_5〖x_1{〗}^2\left(0\right) \\ E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2〖x_1{〗}^2 \end{gathered}$$