Question

A random 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

Answer 1

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.