Question

Q’s # 3 and 4

Answer 1

3.
a.
Given that,
mean(x)=42340
standard deviation , sigma1 =9639.19
number(n1)=158
y(mean)=33865.98
standard deviation, sigma2 =8298.82
number(n2)=148
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=42340-33865.98/sqrt((92913983.8561/158)+(68870413.3924/148))
zo =8.256
| zo | =8.256
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =8.256 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 8.256 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: μ1 != u2
test statistic: 8.256
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference of men and women

b.
Given that,
mean(x)=42340
standard deviation , sigma1 =9639.19
number(n1)=158
y(mean)=33865.98
standard deviation, sigma2 =8298.82
number(n2)=148
null, Ho: u1 = u2
alternate, H1: μ1 > u2
level of significance, alpha = 0.05
from standard normal table,right tailed z alpha/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=42340-33865.98/sqrt((92913983.8561/158)+(68870413.3924/148))
zo =8.256
| zo | =8.256
critical value
the value of |z alpha| at los 0.05% is 1.645
we got |zo | =8.256 & | z alpha | =1.645
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: right tail -Ha : ( p > 8.256 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: μ1 > u2
test statistic: 8.256
critical value: 1.645
decision: reject Ho
p-value: 0

c.
TRADITIONAL METHOD
given that,
mean(x)=42340
standard deviation , σ1 =9639.19
population size(n1)=158
y(mean)=33865.98
standard deviation, σ2 =8298.82
population size(n2)=148
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((92913983.8561/158)+(68870413.3924/148))
= 1026.3546
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1026.3546
= 2011.6551
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (42340-33865.98) ± 2011.6551 ]
= [6462.3649 , 10485.6751]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=42340
standard deviation , σ1 =9639.19
number(n1)=158
y(mean)=33865.98
standard deviation, σ2 =8298.82
number(n2)=148
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 = [ ( 42340-33865.98) ±Z a/2 * Sqrt( 92913983.8561/158+68870413.3924/148)]
= [ (8474.02) ± Z a/2 * Sqrt( 1053403.82) ]
= [ (8474.02) ± 1.96 * Sqrt( 1053403.82) ]
= [6462.3649 , 10485.6751]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [6462.3649 , 10485.6751] 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, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero

4.
sample of 52 statistics elementary students 13 women are included
a.
percentage of women in the elementary statistic = 13/52 =1/4 =25%
Given that,
possibile chances (x)=13
sample size(n)=52
success rate ( p )= x/n = 0.25
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5  
alternate, H1: p!=0.5
level of significance, alpha = 0.02
from standard normal table, two tailed z alpha/2 =2.326
since our test is two-tailed
reject Ho, if zo < -2.326 OR if zo > 2.326
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.25-0.5/(sqrt(0.25)/52)
zo =-3.606
| zo | =3.606
critical value
the value of |z alpha| at los 0.02% is 2.326
we got |zo| =3.606 & | z alpha | =2.326
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.60555 ) = 0.00031
hence value of p0.02 > 0.0003,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: -3.606
critical value: -2.326 , 2.326
decision: reject Ho
p-value: 0.00031

b.
Given that,
possibile chances (x)=13
sample size(n)=52
success rate ( p )= x/n = 0.25
success probability,( po )=0.4
failure probability,( qo) = 0.6
null, Ho:p=0.4  
alternate, H1: p<0.4
level of significance, alpha = 0.1
from standard normal table,left tailed z alpha/2 =1.282
since our test is left-tailed
reject Ho, if zo < -1.282
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.25-0.4/(sqrt(0.24)/52)
zo =-2.208
| zo | =2.208
critical value
the value of |z alpha| at los 0.1% is 1.282
we got |zo| =2.208 & | z alpha | =1.282
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: left tail - Ha : ( p < -2.20794 ) = 0.01362
hence value of p0.1 > 0.01362,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.4
alternate, H1: p<0.4
test statistic: -2.208
critical value: -1.282
decision: reject Ho
p-value: 0.01362

c.
power of the test when the proportion is 46%
power = 1- beta
beta = p(Zalpha< p-po/sqrt(po(1-po)/n)
beta = p(Zalpha<(0.25-0.46)/sqrt(0.46(1-0.46)/52)
beta = p(Zalpha<-3.038)
beta = 0.001191
power = 1-0.001191
power = 0.9988