Question

A certain test preparation course is designed to help students improve their scores on the MCAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 7 students' scores on the exam after completing the course: 21.21, 23, 20, 28, 22, 15 Using these data, construct a 90 % confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Copy Data Step 3 of 4: Find the critical value that should be used in constructing the confidence interval, Round your answer to three decimal places. Tables Keypad Answer How to enter your answer Previous step answers

Answer 1

Solution: dr - A = x - 21 dr2 0 0 0 0 4 20 -1 28 49 22 1 1 15 -6 36 Σ (dz)2 91 Σ ( da)-3 150 Σε Mean n 21 21 23 20 28 22 15 7 150 7 = 21.4286 (Σά Σde Sample Standard deviation S - 1 (3)3 91 7 91 1.2857 6 89.7143 6 14.9524 = 3.8668 222

Critical value at 90 % CI is,

Solution: Given That- CI 90.0% X= 21.429 df n-1 df 6 7 n = S = 3.8668 90.0% Confidence Level't' is, At a 1 90.0 % a 1 0.9 a 0.1 a/2 0.05 ta/2, df 1.9432 90.0% Confidence Interval for Mean is, At n

CI for Mean using t CI for 90.0% 7 n Mean (X bar) t-value of 90% CI and df 6 std. Dev. (S) SE std.dev./sqrt(n) ME = t*SE Lower Limit Mean ME 21.429 1.9432 3.8668 1,46151 2.83998 18.58902 Upper Limit Mean ME 90% CI 24.26898 (18.589,24.269

At 90 %, CI for mean is ( 18.589 , 24.269 ).