Question

Neighborhood 1 Neighborhood 2 55 56 34 49

Answer 1

a.
Given that,
mean(x)=55
standard deviation , s.d1=8.8091
number(n1)=6
y(mean)=44
standard deviation, s.d2 =10.6771
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 =2.23
since our test is two-tailed
reject Ho, if to < -2.23 OR if to > 2.23
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*77.6002 + 5*114.0005) / (12- 2 )
s^2 = 95.8004
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=55-44/sqrt((95.8004( 1 /6+ 1/6 ))
to=11/5.651
to=1.9466
| to | =1.9466
critical value
the value of |t α| with (n1+n2-2) i.e 10 d.f is 2.23
we got |to| = 1.9466 & | t α | = 2.23
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 != 1.9466 ) = 0.0776
hence value of p0.05 < 0.0776,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.9466
critical value: -2.23 , 2.23
decision: do not reject Ho
p-value: 0.0776
we do not have enough evidence to support the claim that mean age of houses is same in the two neighbourhood.
b.
TRADITIONAL METHOD
given that,
mean(x)=55
standard deviation , s.d1=8.8091
number(n1)=6
y(mean)=44
standard deviation, s.d2 =10.6771
number(n2)=6
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*77.6 + 5*114) / (12- 2 )
s^2 = 95.8
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 95.8 * (1/6+1/6) )
=5.651
III.
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 (n1+n2-2) i.e 10 d.f is 2.228
margin of error = 2.228 * 5.651
= 12.59
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (55-44) ± 12.59 ]
= [-1.59 , 23.59]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=55
standard deviation , s.d1=8.8091
sample size, n1=6
y(mean)=44
standard deviation, s.d2 =10.6771
sample size,n2 =6
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 = [( 55-44) ± t a/2 * sqrt( 95.8 * (1/6+1/6) ]
= [ (11) ± 12.59 ]
= [-1.59 , 23.59]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.59 , 23.59]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

c.
TRADITIONAL METHOD
given that,
mean(x)=55
standard deviation , s.d1=8.8091
number(n1)=6
y(mean)=44
standard deviation, s.d2 =10.6771
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((77.6/6)+(114/6))
= 5.651
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 * 5.651
= 14.529
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (55-44) ± 14.529 ]
= [-3.529 , 25.529]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=55
standard deviation , s.d1=8.8091
sample size, n1=6
y(mean)=44
standard deviation, s.d2 =10.6771
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 = [( 55-44) ± t a/2 * sqrt((77.6/6)+(114/6)]
= [ (11) ± t a/2 * 5.651]
= [-3.529 , 25.529]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-3.529 , 25.529] 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
option:B
No,
since the standard deviations are fairly close ,the two methods will results in essentially the same confidence intervals and hypothesis tests.