Two independent samples have been selected, 5858 observations from population 1

Question

Two independent samples have been selected, 5858 observations from population 1 and 7272observations from population 2. The sample means have been calculated to be x¯¯¯1=9.1x¯1=9.1 and x¯¯¯2=11.5x¯2=11.5. From previous experience with these populations, it is known that the variances are 21=2512=25 and 22=2822=28.

(a)    Determine the rejection region for the test of H0:(12)3.78H0:(12)3.78 and Ha:(12)>3.78Ha:(12)>3.78Use =0.04=0.04. (Hint: Pay attention to the right side of the hypotheses.)
z>z>

(b)    Compute the test statistic.
z=z=

(c)    The final conclusion is

A. There is not sufficient evidence to reject the null hypothesis that (12)3.78(12)3.78.
B. We can reject the null hypothesis that (12)3.78(12)3.78 and support that (12)>3.78(12)>3.78.

(d)    Construct a 9696 % confidence interval for (12)(12).
(12)(12)

Explanation / Answer

Given that,
mean(x)=9.1
standard deviation , 1 =25
number(n1)=58
y(mean)=11.5
standard deviation, 2 =28
number(n2)=72
null, Ho: u1 =u2
alternate, H1: 1 > u2
level of significance, = 0.04
from standard normal table,right tailed z /2 =1.751
since our test is right-tailed
reject Ho, if zo > 1.751
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=9.1-11.5/sqrt((625/58)+(784/72))
zo =-0.5156
| zo | =0.5156
critical value
the value of |z | at los 0.04% is 1.751
we got |zo | =0.516 & | z | =1.751
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: right tail -Ha : ( p > -0.5156 ) = 0.69694
hence value of p0.04 < 0.69694,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: 1 > u2
b.
test statistic: -0.5156
critical value: 1.751
c.
decision: do not reject Ho
p-value: 0.69694
There is not sufficient evidence to reject the null hypothesis that (12)3.78(12)3.78
d.
TRADITIONAL METHOD
given that,
mean(x)=9.1
standard deviation , 1 =25
population size(n1)=58
y(mean)=11.5
standard deviation, 2 =28
population size(n2)=72
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((625/58)+(784/72))
= 4.6545
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.04
from standard normal table, two tailed z /2 =2.054
since our test is two-tailed
value of z table is 2.054
margin of error = 2.054 * 4.6545
= 9.5604
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (9.1-11.5) ± 9.5604 ]
= [-11.9604 , 7.1604]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=9.1
standard deviation , 1 =25
number(n1)=58
y(mean)=11.5
standard deviation, 2 =28
number(n2)=72
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 9.1-11.5) ±Z a/2 * Sqrt( 625/58+784/72)]
= [ (-2.4) ± Z a/2 * Sqrt( 21.6648) ]
= [ (-2.4) ± 2.054 * Sqrt( 21.6648) ]
= [-11.9604 , 7.1604]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 96% sure that the interval [-11.9604 , 7.1604] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 96% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does n't contain a zero we can conclude at 0.04 true mean
difference is not zero

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