Question

2. In a large class, an in-person exam, where the lazy students couldn't do cheating, in a random sample of 8 lazy students, average exam score was 42.6 with sample standard deviation 4.4, while in a random sample of 10 non-lazy students, the average exam score was 49.8 with sample standard deviation 3.1. Assume that the two populations (Exam scores for the students who do not cheat) are normally distributed. Exam was out of 60points. (a) Assume that the two population variances are equal. Construct a 95% confidence interval for the difference between the overall average exam scores for the non-lazy students and for the lazy students. (b) Since the large sample variance is more than twice as big as the smaller one, the assumption of equal variances is questionable here. Construct a 95% confidence interval for the difference between the overall average exam scores for the non-lazy students and lazy students without assuming that two population variances are equal. Use Welch's T.

Answer 1

(a) The 95% confidence interval is between -10.9457 and -3.4543.

Lazy students	Non-Lazy students
42.6	49.8	mean
4.4	3.1	std. dev.
8	10	n
	16	df
	-7.2000	difference (Lazy students - Non-Lazy students)
	13.8756	pooled variance
	3.7250	pooled std. dev.
	1.7669	standard error of difference
	-10.9457	confidence interval 95.% lower
	-3.4543	confidence interval 95.% upper
	3.7457	margin of error

(b) The 95% confidence interval is between -11.2063 and -3.1937.

Lazy students	Non-Lazy students
42.6	49.8	mean
4.4	3.1	std. dev.
8	10	n
	12	df
	-7.2000	difference (Lazy students - Non-Lazy students)
	1.8387	standard error of difference
	-11.2063	confidence interval 95.% lower
	-3.1937	confidence interval 95.% upper
	4.0063	margin of error