Find the interval within which 95 percent of the sample me

Find the interval [ ] within which 95 percent of the sample means would be expected to fall, assuming that each sample is from a normal population. (a) Mu = 159, sigma = 20, n = 44. (Round your answers to 2 decimal places.) The 95% range s from to (b) Mu = 1,036, sigma = 25, n = 6. (Round your answers to 2 decimal places.) The 95% range is from . to (c) Mu = 44, sigma = 25, n = 20. (Round your answers to 3 decimal places.) The 95% range s from to

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 =    0.025          
X = sample mean =    159          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    20          
n = sample size =    44          
              
Thus,              
Margin of Error E =    5.909513763          
Lower bound =    153.0904862          
Upper bound =    164.9095138          
              
Thus, the confidence interval is              
              
(   153.0904862   ,   164.9095138   ) [ANSWER]

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

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 =    1036          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    25          
n = sample size =    6          
              
Thus,              
Margin of Error E =    20.00379865          
Lower bound =    1015.996201          
Upper bound =    1056.003799          
              
Thus, the confidence interval is              
              
(   1015.996201   ,   1056.003799   ) [ANSWER]

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

c)

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 =    44          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    3          
n = sample size =    20          
              
Thus,              
Margin of Error E =    1.314783811          
Lower bound =    42.68521619          
Upper bound =    45.31478381          
              
Thus, the confidence interval is              
              
(   42.68521619   ,   45.31478381   ) [ANSWER]