Assumes that monthly returns are normally distributed with a

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.

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]

Assumes that monthly returns are normally distributed with a

Solution

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