Question

The out-of-state tuitions (in dollars) for random samples of both public and private four-year colleges in a New England state are listed. Find the 99% confidence interval for the difference in the means. Assume the variables are normally distributed and the variances are unequal. Use a graphing calculator. Round the answers to two decimal places. Private Public 23,400 16,590 14,291 14,150 17,300 12,500 19,024 13,326 13,600 7,050 6,450 17,7607,050 9,000 9,758 7,871 9,113 15,820 16,100 Source: New York Times Almanac. Download data Use for the mean out-of-state tuition of private colleges. |<H1-H2 <S

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=16020.1111
standard deviation , s.d1=3490.0303
number(n1)=9
y(mean)=10597.2
standard deviation, s.d2 =4270.2979
number(n2)=10
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((12180311.49/9)+(18235444.15/10))
= 1782.39
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.01
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 3.36
margin of error = 3.355 * 1782.39
= 5979.92
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (16020.1111-10597.2) ± 5979.92 ]
= [-557.01 , 11402.83]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=16020.1111
standard deviation , s.d1=3490.0303
sample size, n1=9
y(mean)=10597.2
standard deviation, s.d2 =4270.2979
sample size,n2 =10
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 = [( 16020.1111-10597.2) ± t a/2 * sqrt((12180311.49/9)+(18235444.15/10)]
= [ (5422.91) ± t a/2 * 1782.39]
= [-557.01 , 11402.83]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-557.01 , 11402.83] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
31.
Given that,
mean(x)=0.3233
standard deviation , s.d1=0.0104
number(n1)=6
y(mean)=0.3388
standard deviation, s.d2 =0.0106
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =2.015
since our test is two-tailed
reject Ho, if to < -2.015 OR if to > 2.015
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.3233-0.3388/sqrt((0.00011/6)+(0.00011/6))
to =-2.557
| to | =2.557
critical value
the value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 2.015
we got |to| = 2.55672 & | t α | = 2.015
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.5567 ) = 0.051
hence value of p0.1 > 0.051,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.557
critical value: -2.015 , 2.015
decision: reject Ho
p-value: 0.051
we have enough evidence to support the claim that difference of means between national and american.