Families USA a monthly magazine that discusses issues relate

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 19 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,500. The standard deviation of the sample was $1,060. (Use z Distribution Table.)

Based on this sample information, develop a 90% confidence interval for the population mean yearly premium. (Round your answers to the nearest whole number.)

How large a sample is needed to find the population mean within $250 at 99% confidence? (Round up your answer to the next whole number.)

Solution

A)

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 =    0.05          
X = sample mean =    10500          
z(alpha/2) = critical z for the confidence interval =    1.64          
s = sample standard deviation =    1060          
n = sample size =    19          
              
Thus,              
Margin of Error E =    398.8163118          
Lower bound =    10101.18369          
Upper bound =    10898.81631          
              
Thus, the confidence interval is              
              
(   10101.18369   ,   10898.81631   ) [ANSWER]

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B)

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) =    2.58  
      
Also,      
      
s = sample standard deviation =    1060  
E = margin of error =    250  
      
Thus,      
      
n =    119.6660966  
      
Rounding up,      
      
n =    120   [ANSWER]