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

A) 83.45 to 84.55.
B) 80.07 to 87.93.
C) 83.29 to 84.71.
D) 80.5 to 87.5.

We can calculate the standard error of the mean as the square root of the variance divided by the sample size, i.e., sqrt(12.25/49) = 0.7. Then, with a 95% confidence level, we can find the t-value for a two-tailed test with 48 degrees of freedom (n-1), which is approximately 2.01. Therefore, the margin of error is 2.01 times the standard error, i.e., 2.01*0.7 = 1.41. Finally, we can construct the confidence interval as the sample mean plus or minus the margin of error, i.e., 84 +/- 1.41. Thus, the interval is 83.29 to 84.71, which corresponds to choice C.