A random sample of 90 observations produced a mean of x293 f

A random sample of 90 observations produced a mean of x=29.3 from a population with a normal distribution and a standard deviation =2.75.
a)Find a 99% confidence interval for
b)Find a 90% confidence interval for
c)Find a 95% confidence interval for

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 =    29.3          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    2.75          
n = sample size =    90          
              
Thus,              
Margin of Error E =    0.746669684          
Lower bound =    28.55333032          
Upper bound =    30.04666968          
              
Thus, the confidence interval is              
              
(   28.55333032   ,   30.04666968   ) [ANSWER]

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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 =    0.05          
X = sample mean =    29.3          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    2.75          
n = sample size =    90          
              
Thus,              
Margin of Error E =    0.476802689          
Lower bound =    28.82319731          
Upper bound =    29.77680269          
              
Thus, the confidence interval is              
              
(   28.82319731   ,   29.77680269   ) [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 =    29.3          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2.75          
n = sample size =    90          
              
Thus,              
Margin of Error E =    0.568145446          
Lower bound =    28.73185455          
Upper bound =    29.86814545          
              
Thus, the confidence interval is              
              
(   28.73185455   ,   29.86814545   ) [ANSWER]

A random sample of 90 observations produced a mean of x=29.3 from a population with a normal distribution and a standard deviation =2.75. a)Find a 99% confidenc
A random sample of 90 observations produced a mean of x=29.3 from a population with a normal distribution and a standard deviation =2.75. a)Find a 99% confidenc

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