Question

an advertising study interviewed six randomly selected people in each of two cities, recording each person's level of preference for a new product:

milwaukee		green bay
3		4
2		5
1		4
1		3
3		2
2		4

a) is this a paired or unpaired two sample problem?

b) find the average preference level for each city.

c) find the standard error of the difference between these average preference levels (these are small samples)

d) find the 95% two-sided confidence interval for the mean difference in preference between these two cities Green Bay minus Milwaukee).

e) test whether the apparent difference in preference is significant at the 5% test level.

Answer 1

a.
an advertising study interviewed six randomly selected people in each of two cities,so that
given data is un paired data.there are indenpendent to each other.
b.
mean(x)=3.667
standard deviation , s.d1=1.0328
number(n1)=6
y(mean)=2
standard deviation, s.d2 =0.8944
number(n2)=6
c.
the standard error of the difference between these average preference levels = 0.558
d.
TRADITIONAL METHOD
given that,
mean(x)=3.667
standard deviation , s.d1=1.0328
number(n1)=6
y(mean)=2
standard deviation, s.d2 =0.8944
number(n2)=6
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((1.067/6)+(0.8/6))
= 0.558
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 2.571
margin of error = 2.571 * 0.558
= 1.434
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (3.667-2) ± 1.434 ]
= [0.233 , 3.101]
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DIRECT METHOD
given that,
mean(x)=3.667
standard deviation , s.d1=1.0328
sample size, n1=6
y(mean)=2
standard deviation, s.d2 =0.8944
sample size,n2 =6
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 = [( 3.667-2) ± t a/2 * sqrt((1.067/6)+(0.8/6)]
= [ (1.667) ± t a/2 * 0.558]
= [0.233 , 3.101]
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interpretations:
1. we are 95% sure that the interval [0.233 , 3.101] 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
e.
Given that,
mean(x)=3.667
standard deviation , s.d1=1.0328
number(n1)=6
y(mean)=2
standard deviation, s.d2 =0.8944
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3.667-2/sqrt((1.06668/6)+(0.79995/6))
to =2.9887
| to | =2.9887
critical value
the value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 2.571
we got |to| = 2.9887 & | t α | = 2.571
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.9887 ) = 0.03
hence value of p0.05 > 0.03,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.9887
critical value: -2.571 , 2.571
decision: reject Ho
p-value: 0.03
we have enough evidence to support the claim that whether the apparent difference in preference is significant between two cities.