The online clothing retailer e-Parel is conducting a study to estimate the avera

Question

The online clothing retailer e-Parel is conducting a study to estimate the average size of the orders placed by visitors to its website. The study researcher desires a $50 margin of error at 99% confidence. The population standard deviation is unknown, so a best guess is used as the planning value for a. e-Parel estimates that the smallest order in the population of orders placed at its website is $20 (the price of the least expensive item in its catalog), and the largest order is $720 (three times the price of the most expensive item in its catalog). A rough approximation of the population standard deviation is To satisfy the requirement for the margin of error, a sample size no smaller than is needed.

Explanation / Answer

The range can be assumed to be about 6 standard deviations long.

Hence,

6*sigma = 720-20

sigma = 700/6 = 116.6666667 [ANSWER, ROUGH APPROXIMATION]

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Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    116.6666667  
E = margin of error =    50  
      
Thus,      
      
n =    36.12332596  
      
Rounding up,      
      
n =    37   [ANSWER]

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Hi! If you have a different convention in your class as to how many standard deviation is the range, please resubmit this question together with the convention of your class. That way we can continue helping you! Thanks!

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