Families USA a monthly magazine that discusses issues relate

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 21 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,650. The standard deviation of the sample was $1,025. (Use z Distribution Table.)

Based on this sample information, develop a 99% 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 $235 at 90% confidence? (Round up your answer to the next whole number.)

Solution

a)

Note that              

      
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    10650          
t(alpha/2) = critical t for the confidence interval =    2.84533971          
s = sample standard deviation =    1025          
n = sample size =    21          
df = n - 1 =    20          
Thus,              

Lower bound =    10013.57337          
Upper bound =    11286.42663          
              
Thus, the confidence interval is              
              
(   10013.57337   ,   11286.42663   ) [ANSWER]

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B)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    1025  
E = margin of error =    235  
      
Thus,      
      
n =    51.47146386  
      
Rounding up,      
      
n =    52   [ANSWER]