A simple random sample with n30 provided a sample mean of 22

A simple random sample with n=30 provided a sample mean of 22.5 and a sample standard deviation of 4.4. Develop a 98% confidence interval for the population mean.
A simple random sample with n=30 provided a sample mean of 22.5 and a sample standard deviation of 4.4. Develop a 98% confidence interval for the population mean.

Solution

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 =    0.01          
X = sample mean =    22.5          
z(alpha/2) = critical z for the confidence interval =    2.326347874          
s = sample standard deviation =    4.4          
n = sample size =    30          
              
Thus,              
Margin of Error E =    1.868816704          
Lower bound =    20.6311833          
Upper bound =    24.3688167          
              
Thus, the confidence interval is              
              
(   20.6311833   ,   24.3688167   ) [answer]

A simple random sample with n=30 provided a sample mean of 22.5 and a sample standard deviation of 4.4. Develop a 98% confidence interval for the population mea

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