help D Question 17 15 pts Problem 17 Confidence Interval Sa

help

D Question 17 15 pts Problem 17 (Confidence Interval & Sample Size) In a sample of 50 retirees ages 60-7 was 7, and the standard deviation was 2 1 number of jobs for this age group. To how large of a sample is needed? 0, the average number of jobs they had during their lifetimes Find the 99% confidence interval or the mean estimate the true mean within 1/2job with 99% confidence e -T 1, Font Sizes Paragraph

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 =    0.005          
X = sample mean =    7          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    2.1          
n = sample size =    50          
              
Thus,              
Margin of Error E =    0.764982274          
Lower bound =    6.235017726          
Upper bound =    7.764982274          
              
Thus, the confidence interval is              
              
(   6.235017726   ,   7.764982274   ) [ANSWER]

************************

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) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    2.1  
E = margin of error =    0.5  
      
Thus,      
      
n =    117.039576  
      
Rounding up,      
      
n =    118   [ANSWER]