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 $11,100. The standard deviation of the sample was $920. (Use z Distribution Table.)

a. Based on this sample information, develop a 98% confidence interval for the population mean yearly premium.

b.How large a sample is needed to find the population mean within $250 at 95% confidence?

Solution

a)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
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 =    0.01          
X = sample mean =    11000          
t(alpha/2) = critical t for the confidence interval =    2.55237963          
s = sample standard deviation =    920          
n = sample size =    19          
df = n - 1 =    18          
Thus,              
Margin of Error E =    538.7115623          
Lower bound =    10461.28844          
Upper bound =    11538.71156          
              
Thus, the confidence interval is              
              
(   10461.28844   ,   11538.71156   ) [ANSWER]

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b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    920  
E = margin of error =    250  
      
Thus,      
      
n =    52.02257193  
      
Rounding up,      
      
n =    53   [ANSWER]