The annual incomes for 14 randomly selected people each with

The annual incomes for 14 randomly selected people, each with a bachelor\'s degree. Assume that the population of the monthly incomes in normally distributed.

What is the sample mean, x-bar?

What is the sample standard deviation, s?

Construct a 95% confidence interval for the population of mean  µ.

Solution

a)

By technology,

X = sample mean =    55006.21429   [ANSWER]

***************

b)

By technology,

s = sample standard deviation =    12674.73377 [ANSWER]

****************

c)          

Note that              
              
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 =    55006.21429          
t(alpha/2) = critical t for the confidence interval =    2.160368656          
s = sample standard deviation =    12674.73377          
n = sample size =    14          
df = n - 1 =    13          
Thus,              
              
Lower bound =    47688.04088          
Upper bound =    62324.38769          
              
Thus, the confidence interval is              
              
(   47688.04088   ,   62324.38769   ) [ANSWER]