The average annual cost (including tuition, room, board, books and fees) to attend a public college takes nearly a third of the annual income of a typical family with college-age children (Money, April 2012). At private colleges, the average annual cost is equal to about 60% of the typical family's income. The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars. Private Colleges 52.8 30.6 43.2 45.8 45.0 37.8 33.3 50.5 44.0 42.0 Public Colleges 20.3 22.8 22.0 25.8 28.2 18.5 15.6 25.6 24.1 14.4 28.5 21.8 a. Compute the sample mean and sample standard deviation for private and public colleges. Round your answers to two decimal places. S1 =
S2 =
b. What is the point estimate of the difference between the two population means? Round your answer to one decimal place.
Interpret this value in terms of the annual cost of attending private and public colleges.$
c. Develop a 95% confidence interval of the difference between the annual cost of attending private and pubic colleges.
95% confidence interval, private colleges have a population mean annual cost $ to $ more expensive than public colleges.

Step-by-step explanation:

a) For private colleges,

Mean = (52.8 + 30.6 + 43.2 + 45.8 + 45.0 + 37.8 + 33.3 + 50.5 + 44.0 + 42.0)/10 = 42.5

x1 = 42.5 × 1000 = 42500

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (52.8 - 42.5)^2 + (30.6 - 42.5)^2 + (43.2 - 42.5)^2 + (45.8 - 42.5)^2 + (45.0 - 52.5)^2 + (37.8 - 42.5)^2 + (33.3 - 42.5)^2 + (50.5 - 42.5)^2 + (44.0 - 42.5)^2 + (42.0 - 52.5)^2 = 598.56

standard deviation = √(598.56/10 = 7.74

s1 = 7.74 × 1000 = 7740

For public colleges,

Mean = (20.3 + 22.8 + 22.0 + 25.8 + 28.2 + 18.5 + 15.6 + 25.6 + 24.1 + 14.4 28.5 + 21.8)/12 = 22.3

x2 = 22.3 × 1000 = 22300

n2 = 12

Summation(x - mean)² = (20.3 - 22.3)^2 + (22.8 - 22.3)^2 + (22 - 22.3)^2 + (25.8 - 22.3)^2 + (28.2 - 22.3)^2 + (18.5 - 22.3)^2 + (15.6 - 22.3)^2 + (25.6 - 22.3)^2 + (24.1 - 22.3)^2 + (14.4 - 22.3)^2 + (28.5 - 22.3)^2 + (21.8 - 22.3)^2 = 225.96

standard deviation = √(225.96/12 = 4.34

s2 = 4.34 × 1000 = 4340

b) The point estimate is the difference between the sample means

Point estimate = 42500 - 22300 = 20200

The best guess for the difference in population mean annual cost of attending private and public colleges is $20200. The range of the value is determined by the margin of error.

c) Confidence interval = point estimate ± margin of error

Margin of error = z√(s²/n1 + s2²/n2)

Where z is the test score for 95% confidence level from the t distribution table. To find the test score, we would first find degree of freedom, df

df = (n1 - 1) + (n2 - 1) = (10 - 1) + (12 - 1) = 20

From the t distribution table,

z = 2.086

Margin of error = 2.086√(7.74²/10 + 4.34²/12) = 4.84

4.84 × 1000 = 4840

Confidence interval = 20200 ± 4840

it should be a and d. lmk if i'm wrong xd

x + 4/7

step-by-step explanation:

use cymath