Suppose a population has a mean of 44 and a standard deviati

Suppose a population has a mean of 44 and a standard deviation equal to 12. Using a sample size of 36, calculate the symmetrical interval that includes 95% of the sample means for the population.

A. What is the lower bound of this interval?

B. What is the upper bound of this interval?

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)=44
Standard deviation( sd )=12
Sample Size(n)=36
Confidence Interval = [ 44 ± Z a/2 ( 12/ Sqrt ( 36) ) ]
= [ 44 - 1.96 * (2) , 44 + 1.96 * (2) ]
= [ 40.08,47.92 ]

A. Lower = 40.08
B. Upper = 47.92