Q:

A random sample of 25 statistics examinations was selected. The average score in the sample was 76 with a variance of 144. Assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is _____. a. 70.02 to 81.98 b. 69.82 to 82.18 c. 69.29 to 82.71 d. 70.06 to 81.94

Accepted Solution

A:
Answer:99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.Step-by-step explanation:Confidence interval = mean +/- margin of error (E)mean = 76variance = 144sd = sqrt(variance) = sqrt(144) = 12n = 25degree of freedom (df) = n - 1 = 25 - 1 = 24confidence level (C) = 99% = 0.99significance level = 1 - C = 1 - 0.99 = 0.01 = 1%t-value corresponding to 24 df and 1% significance level is 2.797E = t×sd/√n = 2.797×12/√25 = 6.7128Lower limit = mean - E = 76 - 6.7128 = 69.2872Upper limit = mean + E = 76 + 6.7128 = 82.712899% confidence interval is (69.2872, 82.7128)