The mean score, x-bar, on an aptitude test for a random sample of 9 students was 64. Assuming that standard deviation = 16, construct a 95.44% confidence interval for the mean score of all students taking the test.
seeing 95.44 I ASSUME you are assuming the pop st deviation is 16 and that you are sampling from a normal population
Hence use
$\displaystyle \bar X\pm 2\sigma/\sqrt{n} =64\pm 2(16)/\sqrt{9}$
It's weird that someone woudl know the population variance, but not the mean.
If you estimated the pop st deviation with s, you would use a t with 8 degrees of freedom.