Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 ofx¯=1.96 and a standard deviation of s=0.15.

a. Find and interpret a 95% confidence interval for μ, 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 μ=2.2. Do you recommend he use this value?

Short Answer

Expert verified

a.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).

b.No, the analyst does not use the valueμ=2.2

Step by step solution

01

Given information

x¯=1.96,n=55ands=0.15

02

Calculating 95% confidence interval for μ

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

1-α=0.95α=0.05α2=0.025

From z score table, the value of zα2is given below:

zα2=z0.025=1.96

Thus, the value ofzα2 is 1.96.

The confidence interval is obtained below:

x¯±zα2σx¯=x¯±zα2σn

=1.96±1.960.1555=1.9204,1.9996

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 valueμ=2.2, because the value of μdoes 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 μ=2.2is 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.

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

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In each case, find the approximate sample size required to construct a 95% confidence interval for p that has sampling error of SE = .08.

a. Assume p is near .2.

b. Assume you have no prior knowledge about p, but you wish to be certain that your sample is large enough to achieve the specified accuracy for the estimate.

Water pollution testing. The EPA wants to test a randomlyselected sample of n water specimens and estimate themean daily rate of pollution produced by a mining operation.If the EPA wants a 95% confidence interval estimatewith a sampling error of 1 milligram per liter (mg/L),how many water specimens are required in the sample?Assume prior knowledge indicates that pollution readingsin water samples taken during a day are approximately

normally distributed with a standard deviation equal to5 mg/L.

Question: Is Starbucks coffee overpriced? The Minneapolis Star Tribune (August 12, 2008) reported that 73% of Americans say that Starbucks coffee is overpriced. The source of this information was a national telephone survey of 1,000 American adults conducted by Rasmussen Reports.

a. Identify the population of interest in this study.

b. Identify the sample for the study.

c. Identify the parameter of interest in the study.

d. Find and interpret a 95% confidence interval for the parameter of interest.

Suppose you have selected a random sample of n = 5 measurements from a normal distribution. Compare the standard normal z-values with the corresponding t-values if you were forming the following confidence intervals.

a. 80% confidence interval

b. 90% confidence interval

c. 95% confidence interval

d. 98% confidence interval

e. 99% confidence interval

f. Use the table values you obtained in parts a–e to sketch the z- and t-distributions. What are the similarities and differences?

Do social robots walk or roll? Refer to the International Conference on Social Robotics (Vol. 6414, 2010) study of the trend in the design of social robots, Exercise 5.44 (p. 320). The researchers obtained a random sample of 106 social robots through a Web search and determined that 63 were designed with legs, but no wheels.

a. Find a 99% confidence interval for the proportion of all social robots designed with legs but no wheels. Interpret the result.

b. In Exercise 5.42, you assumed that 40% of all social robots are designed with legs but no wheels. Comment on the validity of this assumption.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free