Question

A developer wants to know if the houses in two different neighborhoods were built at roughly the same time. She takes a random sample of six houses from each neighborhood and finds their ages from local records. The accompanying table shows the data for each sample​ (in years). Assume that the data come from a distribution that is Normally distributed. Complete parts a through c below.

1	2
67	32
55	45
49	37
66	50
54	40
47	60

Find a 95​% confidence interval using the pooled degrees of freedom.

A 95% confidence interval for the mean difference in ages of houses in the two neighborhoods was (__,__).

Is this result different from the result of the​ pooled-t confidence​ interval? Explain why or why not.

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=56.3333
standard deviation , s.d1=8.4301
number(n1)=6
y(mean)=44
standard deviation, s.d2 =10.02
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((71.067/6)+(100.4/6))
= 5.346
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 * 5.346
= 13.744
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (56.3333-44) ± 13.744 ]
= [-1.411 , 26.077]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=56.3333
standard deviation , s.d1=8.4301
sample size, n1=6
y(mean)=44
standard deviation, s.d2 =10.02
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 = [( 56.3333-44) ± t a/2 * sqrt((71.067/6)+(100.4/6)]
= [ (12.333) ± t a/2 * 5.346]
= [-1.411 , 26.077]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.411 , 26.077] 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
b.
TRADITIONAL METHOD
given that,
mean(x)=56.3333
standard deviation , s.d1=8.4301
number(n1)=6
y(mean)=44
standard deviation, s.d2 =10.02
number(n2)=6
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*71.067 + 5*100.4) / (12- 2 )
s^2 = 85.733
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 85.733 * (1/6+1/6) )
=5.346
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 * 5.346
= 11.91
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (56.3333-44) ± 11.91 ]
= [0.423 , 24.244]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=56.3333
standard deviation , s.d1=8.4301
sample size, n1=6
y(mean)=44
standard deviation, s.d2 =10.02
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 = [( 56.3333-44) ± t a/2 * sqrt( 85.733 * (1/6+1/6) ]
= [ (12.333) ± 11.91 ]
= [0.423 , 24.244]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.423 , 24.244]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
from part (a),95% sure that the interval (-1.411 , 26.077)
part (b),95% sure that the interval [0.423 , 24.244]
yes,
result from part (a) different from the result of the​ pooled-t confidence​ interval(part (b))