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 each crew was rated as working either successfully or unsuccessfully as a team. Then, during a simulated flight, the number of leadership functions exhibited per minute was determined for each individual crew member. One objective was to compare the mean leadership scores for successful and unsuccessful teams. How many crew members would need to be sampled from successful and unsuccessful teams to estimate the difference in means to within .05 with 99% confidence? Assume you will sample twice as many members from successful teams as from unsuccessful teams. Also, assume that the variance of the leadership scores for both groups is approximately .04.

Short answer

The required sample size for unsuccessful team is 160.

The required sample size for successful team is 320.

Step-by-step solution

Given Information

The variance of two groups are given below

${\sigma }_1^2={\sigma }_2^2=0.04$

The sampling error is

$SE=.05$

Z-value

For $\alpha =0.01$

The z-value is given by

$$\begin{gathered} z_{\frac{\alpha }{2}}=z_{0.005} \\ =2.58 \end{gathered}$$

Compute the sample

For, $z=2.58,{\sigma }_1^2={\sigma }_2^2=0.04 andSE=.05$

The sample is calculated as

$$\begin{gathered} 2n_1=n_2 \\ =\frac{{\left(z_{0.005}\right)}^2\times \left[2{\sigma }_1^2+{\sigma }_2^2\right]}{2{\left(SE\right)}^2} \\ =\frac{{2.58}^2\times \left[2\times 0.04+0.04\right]}{2\times {.05}^2} \\ =\frac{6.6564\times 0.12}{.005} \\ =\frac{0.79876}{.005} \\ =159.752 \\ \approx 160 \end{gathered}$$

Therefore, the required sample size for unsuccessful team is 160.

And

$$\begin{gathered} n_2=2n_1 \\ =2\times 160 \\ =320 \end{gathered}$$

Therefore, the required sample size for successful team is 320.