Question

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 income. The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars. Click on the webfile logo to reference the data. a. Compute the sample mean and sample standard deviation for private and public colleges. Round your answers to two decimal places. S1 = S2 =

Answer 1

Answer:

a) Private Colleges

Sample mean = 42.5 thousand dollars

Standard deviation = S1 = 6.62 thousand dollars.

Public colleges

Sample mean = 22.3 thousand dollars

Standard deviation = 4.34 thousand dollars

b) The difference in sample mean for both cases = 42.5 - 22.3 = 20.2 thousand dollars

The average amount of going to a Private college is 20.2 thousand dollars more than the average cost of going to public colleges

c) 95% confidence interval for a sampling distribution of the difference of the cost of private and public colleges is given as

(15.0, 25.4) thousand dollars.

Step-by-step explanation:

Private colleges.

52.8 43.2 45.0 33.3 44.0 30.6 45.8 37.8 50.5 42.0

Public colleges.

20.3 22.0 28.2 15.6 24.1 28.5 22.8 25.8 18.5 25.6 14.4 21.8

a) Calculate sample mean and standard deviation for both data set.

Mean = (Σx)/N

where N = Sample size

Σx = sum of all variables

Private colleges

Σx = (52.8+43.2+45.0+33.3+44.0+30.6+45.8+37.8+50.5+42.0) = 425

N = 10

Mean = 425/10 = 42.5 thousand dollars

Standard deviation = S1 = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = (52.8-42.5)² + (43.2-42.5)²

+ (45.0-42.5)² + (33.3-42.5)²

+ (44.0-42.5)² + (30.6-42.5)² + (45.8-42.5)² + (37.8-42.5)² + (50.5-42.5)² + (42.0-42.5)² = 438.56

N = 10

Standard deviation = √[Σ(x - xbar)²/N]

Standard deviation = √(438.56/10) = 6.62 thousand dollars

Public colleges

Mean = (Σx)/N

Σx =

(20.3+22.0+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) = 267.6

N = 12

Mean = (267.6/12) = 22.3 thousand dollars

Standard deviation = √[Σ(x - xbar)²/N]

[Σ(x - xbar)²

(20.3-22.3)² + (22.0-22.3)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)+(21.8-22.3) = 225.96

N = 12

standard deviation = s2 = √(225.96/12) = 4.34 thousand dollars

b) The difference in sample mean for both cases = 42.5 - 22.3 = 20.2 thousand dollars

The average amount of going to a Private college is 20.2 thousand dollars more than the average cost of going to 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.

To combine the distribution in this manner,

Sample mean of difference = 20.2 thousand dollars

Combined standard deviation of the sampling distribution = √[(S1²/n1) + (S2²/n2)]

= √[(6.62²/10) + (4.34²/12)] = 2.44 thousand dollars

Confidence interval = (Sample mean) ± (Margin of error)

Sample mean = 20.2

Margin of error = (critical value) × (standard deviation of the sampling distribution)

standard deviation of the sampling distribution = 2.44

To obtain the critical value, we need the t-score at a significance level of 5%; α/2 = 0.025

we obtain the degree of freedom too

The degree of freedom, df, is calculated in the attached image.

df = 15

t (0.025, 15) = 2.13145 from the tables

Margin of error = 2.13145 × 2.44 = 5.20

Confidence interval = (Sample mean) ± (Margin of error)

= (20.2 ± 5.2) = (15.0, 25.4)

Hope this Helps!!!