Consider a population having a standard deviation equal to 9

Consider a population having a standard deviation equal to 9.93. We wish to estimate the mean of this population.

How large a random sample is needed to construct a 95 percent confidence interval for the mean of this population with a margin of error equal to 1? (Round your answer to the next whole number.)

Suppose that we now take a random sample of the size we have determined in part a. If we obtain a sample mean equal to 337, calculate the 95 percent confidence interval for the population mean. What is the interval’s margin of error? (Round your answers to the nearest whole number.)

Solution

a)

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 =    9.93  
E = margin of error =    1  
      
Thus,      
      
n =    378.7866629  
      
Rounding up,      
      
n =    379   [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    337          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    9.93          
n = sample size =    379          
              
Thus,              
Margin of Error E =    0.999718513   [ANSWER]
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Lower bound =    336.0002815          
Upper bound =    337.9997185          
              
Thus, the confidence interval is              
              
(   336.0002815   ,   337.9997185   ) [ANSWER, CONFIDNECE INTERVAL]

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