Assumes that monthly returns are normally distributed with a mean of 1% and a la

Question

Assumes that monthly returns are normally distributed with a mean of 1% and a large sample standard d deviation of 4%. The polulation standard deviation is unknown. Construct a 95% confidence interval for the sample mean of monthly returns of the sample size is 24. Round to nearest hundredth. Assumes that monthly returns are normally distributed with a mean of 1% and a large sample standard d deviation of 4%. The polulation standard deviation is unknown. Construct a 95% confidence interval for the sample mean of monthly returns of the sample size is 24. Round to nearest hundredth.

Explanation / Answer

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 =    0.025          
X = sample mean =    1          
t(alpha/2) = critical t for the confidence interval =    2.06865761          
s = sample standard deviation =    4          
n = sample size =    24          
df = n - 1 =    23          
Thus,              
Margin of Error E =    1.689051866          
Lower bound =    -0.689051866          
Upper bound =    2.689051866          
              
Thus, the confidence interval is              
              
(   -0.689051866%   ,   2.689051866%   ) [ANSWER]

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