Answer:
99% confidence interval for the population mean is [19.891 , 24.909].
Step-by-step explanation:
We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.
Assuming the population has a normal distribution.
Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;
          P.Q. = 
 ~ 
where, 
 = sample mean age of selected students = 22.4 years
              s = sample standard deviation = 3.8 years
              n = sample of students = 19
              
 = population mean
<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>
So, 99% confidence interval for the population mean, 
 is ;
P(-2.878 < 
 < 2.878) = 0.99  {As the critical value of t at 18 degree of 
                                                 freedom are -2.878 & 2.878 with P = 0.5%}
P(-2.878 < 
 < 2.878) = 0.99
P( 
 < 
 < 
 ) = 0.99
P( 
 < 
 < 
 ) = 0.99
<u>99% confidence interval for</u> 
 = [ 
 , 
 ]
                                                  = [ 
 , 
 ]
                                                  = [19.891 , 24.909]
Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].