Question

From a random sample of 77 students in an introductory finance class that uses​ group-learning techniques, the mean examination score was found to be 76.9376.93 and the sample standard deviation was 2.52.5. For an independent random sample of 88 students in another introductory finance class that does not use​ group-learning techniques, the sample mean and standard deviation of exam scores were 70.8870.88 and 8.58.5​, respectively. Estimate with 9999​% confidence the difference between the two population mean​ scores; do not assume equal population variances.

Answer 1

Mean examination score of group-learning techniques, m1 = 76.93

Mean examination score of without group-learning techniques, m2 = 70.88

Standard deviation of score of group-learning techniques, s1 = 2.5

Standard deviation of score of no group-learning techniques, s2 = 8.5

Standard error of difference in means =

= 3.150255

Since, we do not assume equal population variances,

Degree of freedom, df =  (s₁²/n₁ + s₂²/n₂)² / { [ (s₁² / n₁)² / (n₁ - 1) ] + [ (s₂² / n₂)² / (n₂ - 1) ] }

= (2.5²/7 + 8.5²/8)² / { [ (2.5² / 7)² / (7 - 1) ] + [ (8.5² / 8)² / (8 - 1) ] }

= 8 (Rounding to nearest integer)

Critical value of t at 99% confidence interval and df = 8 is  3.355

Margin of error = Std error * t

= 3.150255 * 3.355

= 10.56911

Difference in means = m1 - m2 = 76.93 - 70.88 = 6.05

99% confidence interval of the difference between the two population mean​ scores is,

(6.05 - 10.56911, 6.05 + 10.56911)

(-4.51911,  16.61911)