Question

Question: Shared leadership in airplane crews. Refer to the Human Factors (March 2014) study of shared leadership by the cockpit and cabin crews of a commercial airplane, Exercise 8.14 (p. 466). Recall that simulated flights were taken by 84 six-person crews, where each crew consisted of a 2-person cockpit (captain and first officer) and a 4-person cabin team (three flight attendants and a purser.) During the simulation, smoke appeared in the cabin and the reactions of the crew were monitored for teamwork. One key variable in the study was the team goal attainment score, measured on a 0 to 60-point scale. Multiple regression analysis was used to model team goal attainment (y) as a function of the independent variables job experience of purser (x₁), job experience of head flight attendant (x₂), gender of purser (x₃), gender of head flight attendant (x₄), leadership score of purser (x₅), and leadership score of head flight attendant (x₆).

a. Write a complete, first-order model for E(y) as a function of the six independent variables.

b. Consider a test of whether the leadership score of either the purser or the head flight attendant (or both) is statistically useful for predicting team goal attainment. Give the null and alternative hypotheses as well as the reduced model for this test.

c. The two models were fit to the data for the n = 60 successful cabin crews with the following results: R² = .02 for reduced model, R² = .25 for complete model. On the basis of this information only, give your opinion regarding the null hypothesis for successful cabin crews.

d. The p-value of the subset F-test for comparing the two models for successful cabin crews was reported in the article as p 6 .05. Formally test the null hypothesis using α = .05. What do you conclude?

e. The two models were also fit to the data for the n = 24 unsuccessful cabin crews with the following results: R² = .14 for reduced model, R² = .15 for complete model. On the basis of this information only, give your opinion regarding the null hypothesis for unsuccessful cabin crews.

f. The p-value of the subset F-test for comparing the two models for unsuccessful cabin crews was reported in the article as p < .10. Formally test the null hypothesis using α = .05. What do you conclude?

Short answer

Answer

a. A first-order model equation in 6 independent variables can be written as$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{4}}{\mathbf{x}}_4\mathbf{+}{\mathit{\beta }}_{\mathbf{5}}{\mathbf{x}}_5\mathbf{+}{\mathit{\beta }}_{\mathbf{6}}{\mathbf{x}}_{\mathbf{6}}\mathbf{+}\mathbf{\epsilon }$

b. To test whether the leadership score of either the purser or the head flight attendant (or both) is statistically useful for predicting team goal attainment can be done using H₀: β₅ = β ₆ = 0 and H_a: Atleast one of β parameters are nonzero.

The reduced model for this test becomes$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{4}}{\mathbf{x}}_4\mathbf{+}\mathbf{\epsilon }$

c. The value of R²is 0.25 for the complete model. Higher value of R²denotes that the model is a good fit for the data and that approximately 25% of the variation in the variables is explained by the model. This number is very less indicating that the model is not a good fit hence the added variables might not be statistically significant for the model.

d. At 95% confidence interval there is enough evidence to reject H₀. Hence at least one of the β parameters are nonzero.

e. The value of R²is 0.15 for the complete model. Higher value of R²denotes that the model is a good fit for the data and that approximately 15% of the variation in the variables is explained by the model. This number is very less indicating that the model is not a good fit hence the added variables might not be statistically significant for the model.

f. At 95% confidence interval there is not enough evidence to reject H₀. Hence, β₅ = β₆ = 0.

Step-by-step solution

First-order model equation

A first-order model equation in 6 independent variables can be written as$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{4}}{\mathbf{x}}_4\mathbf{+}{\mathit{\beta }}_{\mathbf{5}}{\mathbf{x}}_5\mathbf{+}{\mathit{\beta }}_{\mathbf{6}}{\mathbf{x}}_{\mathbf{6}}\mathbf{+}\mathbf{\epsilon }$

Hypotheses

To test whether the leadership score of either the purser or the head flight attendant (or both) is statistically useful for predicting team goal attainment can be done using

H₀: β₅ = β₆ = 0and H_a: At least one of β parameters are nonzero.

The reduced model for this test becomes.$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{4}}{\mathbf{x}}_4\mathbf{+}\mathbf{\epsilon }$

Interpretation of R2

The value of R²is 0.25 for the complete model. Higher value of R²denotes that the model is a good fit for the data and that approximately 25% of the variation in the variables is explained by the model. This number is very less indicating that the model is not a good fit hence the added variables might not be statistically significant for the model.

Hypothesis testing

H₀: β₅ = β₆ = 0 and H_a: At least one of β parameters are nonzero

For α = .05, F-test statistic p-value< 0.05. H₀ is rejected if p-value< 0.05

Therefore, at 95% confidence interval there is enough evidence to reject H₀. Hence at least one of the β parameters are nonzero.

Clarification of R2 

The value of R²is 0.15 for the complete model. Higher value of R²denotes that the model is a good fit for the data and that approximately 15% of the variation in the variables is explained by the model. This number is very less indicating that the model is not a good fit hence the added variables might not be statistically significant for the model.

Hypothesis testing

H₀: β₅ = β₆ = 0and H_a: At least one of β parameters are nonzero

For α = .05, F-test statistic p-value < 0.10. H₀ is rejected if p-value < 0.05

Therefore, at 95% confidence interval there is not enough evidence to reject H₀.

Hence, β₅ = β₆ = 0.