Question

The stocks included in the S and P 500 are those of large publicity held companies that trade on either the New York Stock exchange or the NASDAQ. In 2008, the Sand P 500 was down 38.5%, but what about financial compensation (salary, bonuses, stock options, etc.) to the 500 CEOs that run the companies? To learn more about the mean CEO compensation, an alphabetical list of the 500 companies was obtained and ordered from 1 (3M) to 500 (Zions Bancorp). Next, the random number table was used to select a random number from 1 to 50. The number was selected was 10. then, the companies numbered 10,60,110,160,210,260,310,360,410 and 460 were investigated and the total CEO compensation recorded. The data stored in are as follows:

Number Company Compensation

10 Alfac 10,783,232

60 Big Lota 9,862, 262

110 Comerica 4,108,245

160 EMC 13,874, 262

210 Harley Davidson 6,048,027

260 Kohl's 11,638,049

310 Molson Coors Brewing 5,558,499

360 Pfizer 6,629,955

410 Sigma Aldrich 3,983,596

460 United Parcel Service 5,168, 664

a). Construct a 95% confidence interval estimate for the mean 2008 compensation for CEOs of s and p 500 companies

b). Construct a 99% confidence interval estimate for the mean 2008 compensation for CEOs of s and p 500 companies'

c). comment on the effect that changing the level of confidence had on your answers in a) and b).

Answer 1

We calculate the average and standard deviation (s) compensation(sample mean) of the CEOs using the given data       
X̅ = 7765479  Sample Mean   (Using Excel function average)  
s = 3485663.1  Sample Standard Deviation   (Using Excel function stdev.s)  
        
a) 95% confidence interval for population mean       
n = 10  Sample Size     
Since sample size is less than 30 and population standard deviation is unknown       
we use t-distribution       
Confidence interval for mean is given by       
        
        
Degrees of freedom = df = n - 1 = 10 - 1 = 9       
For 95%, α = 0.05       
From Excel function T.INV.2T(0.05, 9)        
t = T.INV.2T(0.05, 9) = 2.262       
                
= (5271985.83, 10258972.17)       
95% confidence interval is (5271985.83, 10258972.17)       
        
b) 99% confidence interval for population mean       
n = 10  Sample Size     
Since sample size is less than 30 and population standard deviation is unknown       
we use t-distribution       
Confidence interval for mean is given by       
        
Degrees of freedom = df = n - 1 = 10 - 1 = 9       
For 99%, α = 0.01       
From Excel function T.INV.2T(0.01, 9)        
t = T.INV.2T(0.01, 9) = 2.262       
          
= (4183304.05, 11347653.95)       
99% confidence interval is (4183304.05, 11347653.95)       
        
c) From the confidence intervals calculated in (a) and (b), we can see that        
as the confidence level percentage increases the width (difference between lower level and upper level)

of the interval increases