Question

In: Statistics and Probability

The average annual cost (including tuition, room, board, books, and fees) to attend a public college...

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. At private colleges, the average annual cost is equal to about 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 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 Using manual calculation to answer the following questions:

a. Compute the sample mean and the sample standard deviation for private and public collages

b. What is the point estimate of the difference between the population mean?

c. Interpret this value in terms of the annual cost of attending private and public colleges.

d. Develop a 95% confidence interval of the difference between the mean annual cost of attending private and public colleges.

Solutions

Expert Solution

(a)

Private colleges: x-bar = 42.5, s = 6.9806

Public colleges: x-bar = 22.3, s = 4.5323

(b)

Point estimate = 42.5 - 22.3 = 20.2

(c)

On average, there is a difference of about $20200 dollars in cost between attending a public college and a private college

(d)

n1 = 10

n2 = 12

x1-bar = 42.5

x2-bar = 22.3

s1 = 6.9806

s2 = 4.5323

% = 95

Degrees of freedom = n1 + n2 - 2 = 10 + 12 -2 = 20

Pooled s = √(((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = √(((10 - 1) * 6.9806^2 + ( 12 - 1) * 4.5323^2)/(10 + 12 -2)) = 5.764191892

SE = Pooled s * √((1/n1) + (1/n2)) = 5.76419189232107 * √((1/10) + (1/12)) = 2.468079516

t- score = 2.085963441

Width of the confidence interval = t * SE = 2.08596344129554 * 2.468079516164 = 5.148323641

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = 20.2 - 5.1483236409285 = 15.05167636

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = 20.2 + 5.1483236409285 = 25.34832364

The confidence interval is [15.05, 25.35], that is [$15052, $25348]

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