Arandom sample of 49 statistics examinations was taken. the average score, in the sample, was 84 with a variance of 12.25. the 95% confidence interval for the average examination score of the population of the examinations is

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.

stops on an even number.