Answer:
(a) PRIVATE COLLEGES
Sample mean is $42.5 thousand
Sample standard deviation is $6.65 thousand
PUBLIC COLLEGES
Sample mean is $22.3 thousand
Sample standard deviation is $4.34 thousand
(b) Point estimate is $20.2 thousand. The mean annual cost to attend private colleges ($42.5 thousand) is more than the mean annual cost to attend public colleges ($22.3 thousand)
(c) 95% confidence interval of the difference between the mean annual cost of attending private and public colleges is $19.2 thousand to $21.2 thousand
Step-by-step explanation:
(a) PRIVATE COLLEGES
Sample mean = Total cost ÷ number of colleges = (51.8+42.2+45+34.3+44+29.6+46.8+36.8+51.5+43) ÷ 10 = 425 ÷ 10 = $42.5 thousand
Sample standard deviation = sqrt[summation (cost - sample mean)^2 ÷ number of colleges] = sqrt([(51.8-42.5)^2 + (42.2-42.5)^2 + (45-42.5)^2 + (34.3-42.5)^2 + (44-42.5)^2 + (29.6-42.5)^2 + (36.8-42.5)^2 + (51.5-42.5)^2 + (43-42.5)^2] ÷ 10) = sqrt (44.24) = $6.65 thousand
PUBLIC COLLEGES
Sample mean = (20.3+22+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) ÷ 12 = 267.6 ÷ 12 = $22.3 thousand
Sample standard deviation = sqrt([(20.3-22.3)^2 + (22-22.3)^2 + (28.2-22.3)^2 + (15.6-22.3)^2 + (24.1-22.3)^2 + (28.5-22.3)^2 + (22.8-22.3)^2 + (25.8-22.3)^2 + (18.5-22.3)^2 + (25.6-22.3)^2 + (14.4-22.3)^2 + (21.8-22.3)^2] ÷ 12) = sqrt (18.83) = $4.34 thousand
(b) Point estimate = mean annual cost of attending private colleges - mean annual cost of attending public colleges = $42.5 thousand - $22.3 thousand = $20.2 thousand.
This implies the the mean annual cost of attending private colleges is greater than the mean annual cost of attending public colleges
(c) Confidence Interval = Mean + or - Margin of error (E)
E = t×sd/√n
Mean = $42.5 - $22.3 = $20.2 thousand
sd = $6.65 - $4.34 = $2.31 thousand
n = 10+12 = 22
degree of freedom = 22-2 = 20
t-value corresponding to 20 degrees of freedom and 95% confidence level is 2.086
E = 2.086×$2.31/√22 = $1.0 thousand
Lower bound = Mean - E = $20.2 thousand - $1.0 thousand = $19.2 thousand
Upper bound = Mean + E = $20.2 thousand + $1.0 thousand = $21.2 thousand
95% confidence interval is $19.2 thousand to $21.2 thousand
