Question

From a random sample of 66 students in an introductory finance class that uses​ group-learning techniques, the mean examination score was found to be 79.79 and the sample standard deviation was 2.7. For an independent random sample of 99 students in another introductory finance class that does not use​ group-learning techniques, the sample mean and standard deviation of exam scores were 72.14 and 8.8 respectively. Estimate with 90​% confidence the difference between the two population mean​ scores; do not assume equal population variances.

The 90% confidence interval is from a lower limit of __ to an upper limit of __

​(Round to two decimal places as​ needed.)

Answer 1

We have, n1 66, x1 = 79.79, s1 2.7 n2 99, 272.14, s2 = 8.8 The confidence interval for the difference of two population means is given by 1 1 -+ 1 Х — Х2 — 2 |s \N1 Where (п1 — 1)s{ + (nz - 1)s3 = п + пz — 2 (66 1)2.72 + (99 - 1)8.82 66 99 2 = 49.4661 Since, population variances are unequal we need to use pooled variance. Let's find out z for 90% confidence %06 . 5% 90% confidence is nothing but area between -z and z is 90% (middle 90%), so area to the left of z would be 5% 90% 95% Using NORMS.INV(probability) function in excel we can get the z value for which probability to the left is 0.95 =NORM.S.INV(0.95) which is 1.6449 z Now, plug the values in the formula above 1 79.79 72.14 - 1.644949.4661 66 79.79 72.14 +1.6449 49.4661 66 (7.65 1.8384, 7.65 1.8384) (5.8116, 9.4884) The 90% confidence interval is from a lower limit of 5.81 to an upper limit of 9.49