A sample of 225 elements from a population with a standard d

A sample of 225 elements from a population with a standard deviation of 90 is selected. The sample mean is 200 and the margin of error at a 99% confidence level is 15.455. What is the upper value for a 99% confidence interval for the population mean? Enter your answer rounded to three decimal places.

Solution

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=225
Standard deviation( sd )=90
Sample Size(n)=200
Confidence Interval = [ 225 ± Z a/2 ( 90/ Sqrt ( 200) ) ]
= [ 225 - 2.58 * (6.364) , 225 + 2.58 * (6.364) ]
= [ 208.581,241.419 ]