Question

The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time.com). The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars.

Private Colleges

53.8	44.2	45.0	32.3	45.0
30.6	44.8	38.8	51.5	43.0

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) Compute the sample mean (in thousand dollars) and sample standard deviation (in thousand dollars) for private colleges. (Round the standard deviation to two decimal places.)

sample mean$_________ thousand

sample standard deviation$_________ thousand

Compute the sample mean (in thousand dollars) and sample standard deviation (in thousand dollars) for public colleges. (Round the standard deviation to two decimal places.)

sample mean$____________ thousand

sample standard deviation$____________ thousand

(b)What is the point estimate (in thousand dollars) of the difference between the two population means? (Use Private − Public.)

$__________ thousand

Interpret this value in terms of the annual cost (in dollars) of attending private and public colleges.

We estimate that the mean annual cost to attend private colleges is $___________ more than the mean annual cost to attend public college

(c) Develop a 95% confidence interval (in thousand dollars) of the difference between the mean annual cost of attending private and public colleges. (Use Private − Public. Round your answers to one decimal place.)

$________ thousand to $__________ thousand

Answer 1

solution:-

(a) private colleges

sample mean x1 = 42.9

sample standard deviation s1 = 7.36

public colleges

sample mean x2 = 22.3

sample standard deviation s2 = 4.53

b) the point estimate = (x1-x2) = (42.9-22.3)

= 20.6

We estimate that the mean annual cost to attend private colleges is $ 20.6 more than the mean annual cost to attend public college

(c) confidence interval

degree of freedom df = (n1+n2)-2 = (10+12)-2 = 20

the valu of 95% confidence from t table with df is t = 2.086

confidence interval formula

=> point estimate +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> 20.6 +/- 2.086*sqrt((7.36^2/10)+(4.53^2/12))

=> 15.0 thousand to $ 26.2 thousand