Question

Randomly selected 10 student cars have ages with a mean of 7.2 years and a standard deviation of 3.4 years, while randomly selected 31 faculty cars have ages with a mean of 5.9 years and a standard deviation of 3.5 years.

1. Use a 0.01 significance level to test the claim that student cars are older than faculty cars.

(a) The test statistic is

(b) The critical value is 2.326

(c) Is there sufficient evidence to support the claim that student cars are older than faculty cars? A. No B. Yes

2. Construct a 99% confidence interval estimate of the difference μs−μf, where μs is the mean age of student cars and μf is the mean age of faculty cars. <(μs−μf)<

Answer 1

1.

Given that,
mean(x)=7.2
standard deviation , s.d1=3.4
number(n1)=10
y(mean)=5.9
standard deviation, s.d2 =3.5
number(n2)=31
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.821
since our test is left-tailed
reject Ho, if to < -2.821
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =7.2-5.9/sqrt((11.56/10)+(12.25/31))
to =1.04
| to | =1.04
critical value
the value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 2.821
we got |to| = 1.04379 & | t α | = 2.821
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:left tail - Ha : ( p < 1.0438 ) = 0.8381
hence value of p0.01 < 0.8381,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
a.
test statistic: 1.04
b.
critical value: -2.821
decision: do not reject Ho
p-value: 0.8381
c.
No,
we do not have enough evidence to support the claim that student cars are older than faculty cars.
d.
TRADITIONAL METHOD
given that,
mean(x)=7.2
standard deviation , s.d1=3.4
number(n1)=10
y(mean)=5.9
standard deviation, s.d2 =3.5
number(n2)=31
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((11.56/10)+(12.25/31))
= 1.25
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 9 d.f is 3.25
margin of error = 3.25 * 1.25
= 4.05
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (7.2-5.9) ± 4.05 ]
= [-2.75 , 5.35]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=7.2
standard deviation , s.d1=3.4
sample size, n1=10
y(mean)=5.9
standard deviation, s.d2 =3.5
sample size,n2 =31
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 = [( 7.2-5.9) ± t a/2 * sqrt((11.56/10)+(12.25/31)]
= [ (1.3) ± t a/2 * 1.25]
= [-2.75 , 5.35]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-2.75 , 5.35] 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