The monthly incomes for 12 randomly selected people each wit

The monthly incomes for 12 randomly selected people. each with a bachelor\'s degree in economics, are shown on the 4450.08 4596.96 4366.89 4455.61 4151.39 3727.83 Assume the population is normally distributed. 4283.84 4527.88 4407.38 3946.25 4023.17 4221.85 Find the sample mean. x bar= Find the wimple standard deviation. s = Construct a 95% confidence interval for the population mean mu. A 95% confidence interval for the population mean is

Solution

a)
Sample Mean (x) = 4221.6

b)
Sample Standard deviation( sd )=370.4

c)

CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean
Sample Size(n)=12
Confidence Interval = [ 4221.6 ± t a/2 ( 370.4/ Sqrt ( 12) ) ]
= [ 4221.6 - 2.201 * (106.9) , 4221.6 + 2.201 * (106.9) ]
= [ 3986.3,4456.9 ]