Answer:
Step-by-step explanation:
For private Institutions,
n = 20
Mean, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05
Standard deviation = √(summation(x - mean)²/n
Summation(x - mean)² = (43120 - 34623.05)^2+ (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95
Standard deviation = √(1527829234.95/20
s1 = 8740.22
For public Institutions,
n = 20
Mean, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15
Summation(x - mean)² = (25469 - 25063.15)^2+ (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95
Standard deviation = √(283738188.55/20
s2 = 3766.55
This is a test of 2 independent groups. Let μ1 be the mean out-of-state tuition for private institutions and μ2 be the mean out-of-state tuition for public institutions.
The random variable is μ1 - μ2 = difference in the mean out-of-state tuition for private institutions and the mean out-of-state tuition for public institutions.
We would set up the hypothesis. The correct option is
-B. H0: μ1 = μ2 ; H1: μ1 > μ2
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)
t = 9559.9/2128.12528473889
t = 4.49
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114
df = 26
We would determine the probability value from the t test calculator. It becomes
p value = 0.000065
Since alpha, 0.01 > than the p value, 0.000065, then we would reject the null hypothesis. Therefore, at 1% significance level, the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions.