Question

We have two normal populations with unknown means μ₁ and μ₂ and a common variance σ². Two independent samples are randomly drawn from the populations with the following data: n₁ = 26   xത₁ = 8.35 s₁ = 1.7 n₂ = 34   xത₂ = 7.46 s₂ = 2.2

(a)[5] At the level of significance α = 0.01, test H₀: μ₁ = μ₂ versus H₁: μ₁ ≠ μ₂. Sketch the test.

(b)[2] Sketch and find the p‐value of the test in part (a). Would you reject H₀ if α = 0.10?

(c)[3] Construct the 99% CI for the sample mean difference (xത₁ ‐ xത₂). Based on the CI, can the two means be the same? Do you see the conclusion similarity between part (a) and part (c)? Hint: Use 5 decimal places. Use some Excel lookups for values and probabilities.

Answer 1

a.
Given that,
mean(x)=8.35
standard deviation , s.d1=1.7
number(n1)=26
y(mean)=7.46
standard deviation, s.d2 =2.2
number(n2)=34
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.663
since our test is two-tailed
reject Ho, if to < -2.663 OR if to > 2.663
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (25*2.89 + 33*4.84) / (60- 2 )
s^2 = 3.999
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=8.35-7.46/sqrt((3.999( 1 /26+ 1/34 ))
to=0.89/0.521
to=1.708
| to | =1.708
critical value
the value of |t α| with (n1+n2-2) i.e 58 d.f is 2.663
we got |to| = 1.708 & | t α | = 2.663
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 1.7083 ) = 0.0928
hence value of p0.01 < 0.0928,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.708
critical value: -2.663 , 2.663
decision: do not reject Ho
p-value: 0.0928
we do not have enough evidence to support the claim that difference of means.
b.
Given that,
mean(x)=8.35
standard deviation , s.d1=1.7
number(n1)=26
y(mean)=7.46
standard deviation, s.d2 =2.2
number(n2)=34
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.672
since our test is two-tailed
reject Ho, if to < -1.672 OR if to > 1.672
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (25*2.89 + 33*4.84) / (60- 2 )
s^2 = 3.999
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=8.35-7.46/sqrt((3.999( 1 /26+ 1/34 ))
to=0.89/0.521
to=1.708
| to | =1.708
critical value
the value of |t α| with (n1+n2-2) i.e 58 d.f is 1.672
we got |to| = 1.708 & | t α | = 1.672
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 1.7083 ) = 0.0928
hence value of p0.1 > 0.0928,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.708
critical value: -1.672 , 1.672
decision: reject Ho
p-value: 0.0928
we have enough evidence to support the claim that difference of means.
c.
TRADITIONAL METHOD
given that,
mean(x)=8.35
standard deviation , s.d1=1.7
number(n1)=26
y(mean)=7.46
standard deviation, s.d2 =2.2
number(n2)=34
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (25*2.89 + 33*4.84) / (60- 2 )
s^2 = 3.99948
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 3.99948 * (1/26+1/34) )
=0.52102
III.
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 (n1+n2-2) i.e 58 d.f is 2.663
margin of error = 2.663 * 0.52102
= 1.38747
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (8.35-7.46) ± 1.38747 ]
= [-0.49747 , 2.27747]
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DIRECT METHOD
given that,
mean(x)=8.35
standard deviation , s.d1=1.7
sample size, n1=26
y(mean)=7.46
standard deviation, s.d2 =2.2
sample size,n2 =34
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 8.35-7.46) ± t a/2 * sqrt( 3.99948 * (1/26+1/34) ]
= [ (0.89) ± 1.38747 ]
= [-0.49747 , 2.27747]
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interpretations:
1. we are 99% sure that the interval [-0.49747 , 2.27747]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
from part (a),(b) are both conclusions are different.
yes,
the conclusion similarity between part (a) and part (c)