A statistics professor has 115 students in a statistics clas

A statistics professor has 115 students in a statistics class and would like to estimate the number of hours each student studied for the last exam. A random sample of 41 students was found to study an average of 7.3 hours with a standard deviation of 1.9 hours. The 98% confidence interval to estimate the average number of hours studying for the exam is ________.

(5.18, 9.42)

(5.82, 8.79)

(6.11, 8.49)

(6.72, 7.88)

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 =    0.01          
X = sample mean =    7.3          
t(alpha/2) = critical t for the confidence interval =    2.423256779          
s = sample standard deviation = s*sqrt[(N-n)/(N-1)] =   1.530795          
n = sample size =    41          
df = n - 1 =    40          
Thus,              
Margin of Error E =    0.579328032          
Lower bound =    6.720671968          
Upper bound =    7.879328032          
              
Thus, the confidence interval is              
              
(   6.72   ,   7.88   ) [ANSWER, D]