1. In an experiment to investigate the effect of color of paper (blue=1, green=2
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Question
1. In an experiment to investigate the effect of color of paper (blue=1, green=2, orange=3) on response rates for questionnaires distributed by the “windshield method” in supermarket parking lots, 12 representative supermarket parking lots were selected in an area and each color was assigned at random to four of the lots. The response rates (in percentage) and the size of parking lots are shown below.
Response Rate Color
27 1
26 1
31 1
27 1
34 2
29 2
25 2
31 2
31 3
25 3
27 3
29 3
(e) Compute 99% C. I. for contrast L=u2-.5(u1+u3), and complete the hypothesis test: H0= L=0 vs Ha: not H0. given alpha = .05 When you run the test, please try both t and F tests
Explanation / Answer
1.
e.
i.
TRADITIONAL METHOD
given that,
mean(x)=28.5
standard deviation , s.d1=2.8123
number(n1)=12
y(mean)=2
standard deviation, s.d2 =0.8528
number(n2)=12
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((7.909/12)+(0.727/12))
= 0.848
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 11 d.f is 3.106
margin of error = 3.106 * 0.848
= 2.635
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (28.5-2) ± 2.635 ]
= [23.865 , 29.135]
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DIRECT METHOD
given that,
mean(x)=28.5
standard deviation , s.d1=2.8123
sample size, n1=12
y(mean)=2
standard deviation, s.d2 =0.8528
sample size,n2 =12
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 = [( 28.5-2) ± t a/2 * sqrt((7.909/12)+(0.727/12)]
= [ (26.5) ± t a/2 * 0.848]
= [23.865 , 29.135]
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interpretations:
1. we are 99% sure that the interval [23.865 , 29.135] 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
ii.
Given that,
mean(x)=28.5
standard deviation , s.d1=2.8123
number(n1)=12
y(mean)=2
standard deviation, s.d2 =0.8528
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =28.5-2/sqrt((7.90903/12)+(0.72727/12))
to =31.237
| to | =31.237
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 2.201
we got |to| = 31.23724 & | t | = 2.201
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 31.2372 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 31.237
critical value: -2.201 , 2.201
decision: reject Ho
p-value: 0
iii.
Given that,
sample 1
s1^2=7.909, n1 =12
sample 2
s2^2 =0.7272, n2 =12
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.05
from standard normal table, two tailed f /2 =3.474
since our test is two-tailed
reject Ho, if F o < -3.474 OR if F o > 3.474
we use test statistic fo = s1^1/ s2^2 =7.909/0.7272 = 10.88
| fo | =10.88
critical value
the value of |f | at los 0.05 with d.f f(n1-1,n2-1)=f(11,11) is 3.474
we got |fo| =10.876 & | f | =3.474
make decision
hence value of | fo | > | f | and here we reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 10.88
critical value: -3.474 , 3.474
decision: reject Ho
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