the Wall Street Journal reported that automobile crashes cost United States $162

Question

the Wall Street Journal reported that automobile crashes cost United States $162 billion annually. The average cost per person for crashes in the Tampa, Florida area was reported to be $1599. Suppose the average cost was based on a sample of 50 persons who had been involved in crashes and that the population standard deviation is $600.

a.) What is the margin of error for a 95% confidence interval?

b.) what would you recommend if the study required a margin of error of $150 or less?

Show your work and please make it easy to understand. thanks

Explanation / Answer

a) As E = z*sigma / sqrt(n) and for a 95% confidence interval, z = 1.959963985

Then, as sigma = 600, n = 50,

E = 166.3084589

b) Note that n = z(alpha/2)^2 s^2 / E^2   

where  alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also, s = sample standard deviation = 600  
E = margin of error =    150  
      
Thus,          
n =    61.46334113  
      
Rounding up,      
      
n =    62  

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