6. The average annual cost (including tuition, room, board, books and fees) to a
ID: 3314483 • Letter: 6
Question
6.
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.
Click on the datafile 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.
x1=
S1 =
x2=
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. Round your answer to hundreds. Use z-table.
From 95% confidence interval, private colleges have a population mean annual cost $ to $ which is more expensive than public colleges.
Explanation / Answer
a.
x1= 42.5
S1 = 6.9806
x2= 22.3
S2 =4.5323
b.
Given that,
mean(x)=42.5
standard deviation , s.d1=6.9806
number(n1)=10
y(mean)=22.3
standard deviation, s.d2 =4.5323
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =42.5-22.3/sqrt((48.72878/10)+(20.54174/12))
to =7.872
| to | =7.872
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 7.87198 & | t | = 2.262
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 7.872 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 7.872
critical value: -2.262 , 2.262
decision: reject Ho
p-value: 0
c.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 42.5-22.3) ± t a/2 * sqrt((48.729/10)+(20.542/12)]
= [ (20.2) ± t a/2 * 2.566]
= [14.396 , 26.004]
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interpretations:
1. we are 95% sure that the interval [14.396 , 26.004] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
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