Since in this case we are only using the variance of the sample and not the variance of the real population, therefore we use the t statistic. The formula for the confidence interval is:
CI = X ± t * s / sqrt(n) ---> 1
Where,
X = the sample mean = 84
t = the t score which is obtained in the standard distribution tables at 95% confidence level
s = sample variance = 12.25
n = number of samples = 49
From the table at 95% confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t is around:
t = 1.68
Therefore substituting the given values to equation 1:
CI = 84 ± 1.68 * 12.25 / sqrt(49)
CI = 84 ± 2.94
CI = 81.06, 86.94
Therefore at 95% confidence level, the scores is from 81 to 87.
