Facial structure of CEOs. In Psychological Science (Vol. 22, 2011), researchers reported that a chief executive officer’s facial structure can be used to predict a firm’s financial performance. The study involved measuring the facial width-to-height ratio (WHR) for each in a sample of 55 CEOs at publicly traded Fortune 500 firms. These WHR values (determined by a computer analyzing a photo of the CEO’s face) had a mean of$\overline{x}=1.96$ and a standard deviation of $s=0.15$.

a. Find and interpret a 95% confidence interval for $\mu$, the mean facial WHR for all CEOs at publicly traded Fortune 500 firms.

b. The researchers found that CEOs with wider faces (relative to height) tended to be associated with firms that had greater financial performance. They based their inference on an equation that uses facial WHR to predict financial performance. Suppose an analyst wants to predict the financial performance of a Fortune 500 firm based on the value of the true mean facial WHR of CEOs. The analyst wants to use the value of $\mu =2.2$. Do you recommend he use this value?

Step by step solution

01

Given information

$\overline{x}=1.96,n=55$and$s=0.15$

02

Calculating 95% confidence interval for μ

For the confidence level 95%, the level of significance is 0.95.

$\begin{array}{rcl}\left(1-\alpha \right)& =& 0.95\\ \alpha & =& 0.05\\ \frac{\alpha }{2}& =& 0.025\\ & & \end{array}$

From z score table, the value of $z_{\frac{\alpha }{2}}$is given below:

$\begin{array}{rcl}{z}_{\frac{\alpha }{2}}& =& z_{0.025}\\ & =& 1.96\\ & & \end{array}$

Thus, the value of$z_{\frac{\alpha }{2}}$ is 1.96.

The confidence interval is obtained below:

$\overline{x}\pm \left(z_{\frac{\alpha }{2}}\right){\sigma }_{\overline{x}}=\overline{x}\pm z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$

$$\begin{gathered} =1.96\pm 1.96\left(\frac{0.15}{\sqrt{55}}\right) \\ =\left(1.9204,1.9996\right) \end{gathered}$$

Thus, the 95% confidence interval for the mean facial width-to-height ratio (WHR) for all CEOs at publicly traded Fortune 500 firms is (1.9204,1.9996).

03

Step 3:

b.

No, the analyst does not use the value$\mu =2.2$, because the value of $\mu$does not contain in the 95% confidence interval. That is, the value 2.2 does not lie between 1.9204 and 1.9996. Hence, the value $\mu =2.2$is impossible value for the average facial width -to-height ratio (WHR).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!