Construct a 90 confidence interval for the population mean

Construct a 90% confidence interval for the population mean, . Assume the population has a normal distribution. A sample of 15 randomly selected math majors has a grade point average of 2.86 with a standard deviation of 0.78. Round to the nearest hundredth.
Question 7 options:
  
a)    (2.41, 3.42)
  
b)    (2.28, 3.66)
  
c)    (2.37, 3.56)
  
d)    (2.51, 3.21)

Solution

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 =    0.05          
X = sample mean =    2.86          
t(alpha/2) = critical t for the confidence interval =    1.761310136          
s = sample standard deviation =    0.78          
n = sample size =    15          
df = n - 1 =    14          
Thus,              
Margin of Error E =    0.354719291          
Lower bound =    2.505280709          
Upper bound =    3.214719291          
              
Thus, the confidence interval is              
              
(   2.51   ,   3.21   ) [ANSWER, D]