Question

Question 7 of 31 (1 point) 9.2 Section Exercise 9 (table) The data show the heights in feet of waterfalls in Europe and in Asia. Find the 99% confidence for the difference of the means. Source: World Almanac. Round the answers to one decimal places. Europe Asia 830 487 900 1312 345 984 820 614 1137 350 722 722 964 Download data Use i for the mean height of waterfalls in Europe. Assume the variables are normally distributed and the variances are unequal. <H1-H₂ <

Answer 1

7.
TRADITIONAL METHOD
given that,
mean(x)=808
standard deviation , s.d1=349.574
number(n1)=6
y(mean)=762.7143
standard deviation, s.d2 =251.9581
number(n2)=7
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((122202/6)+(63482.9/7))
= 171.6
II.
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 min (n1-1, n2-1) i.e 5 d.f is 4
margin of error = 4.032 * 171.6
= 691.8
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (808-762.7143) ± 691.8 ]
= [-646.5 , 737.1]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=808
standard deviation , s.d1=349.574
sample size, n1=6
y(mean)=762.7143
standard deviation, s.d2 =251.9581
sample size,n2 =7
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 = [( 808-762.7143) ± t a/2 * sqrt((122202/6)+(63482.9/7)]
= [ (45.3) ± t a/2 * 171.6]
= [-646.5 , 737.1]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-646.5 , 737.1] 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
8.
TRADITIONAL METHOD
given that,
mean(x)=16907.8571
standard deviation , s.d1=3470.745
number(n1)=7
y(mean)=8999.8571
standard deviation, s.d2 =3226.6883
number(n2)=7
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((12046070.86/7)+(10411517.39/7))
= 1791.15
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 6 d.f is 1.94
margin of error = 1.943 * 1791.15
= 3480.21
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (16907.8571-8999.8571) ± 3480.21 ]
= [4427.79 , 11388.21]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=16907.8571
standard deviation , s.d1=3470.745
sample size, n1=7
y(mean)=8999.8571
standard deviation, s.d2 =3226.6883
sample size,n2 =7
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 = [( 16907.8571-8999.8571) ± t a/2 * sqrt((12046070.86/7)+(10411517.39/7)]
= [ (7908) ± t a/2 * 1791.15]
= [4427.79 , 11388.21]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [4427.79 , 11388.21] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
23.
TRADITIONAL METHOD
given that,
mean(x)=881
standard deviation , s.d1=375.1586
number(n1)=7
y(mean)=724.1667
standard deviation, s.d2 =267.256
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((140744/7)+(71425.8/6))
= 178.9
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 2
margin of error = 2.015 * 178.9
= 360.5
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (881-724.1667) ± 360.5 ]
= [-203.7 , 517.3]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=881
standard deviation , s.d1=375.1586
sample size, n1=7
y(mean)=724.1667
standard deviation, s.d2 =267.256
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 = [( 881-724.1667) ± t a/2 * sqrt((140744/7)+(71425.8/6)]
= [ (156.8) ± t a/2 * 178.9]
= [-203.7 , 517.3]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-203.7 , 517.3] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
31.
TRADITIONAL METHOD
given that,
mean(x)=17.6154
standard deviation , s.d1=2.6938
number(n1)=13
y(mean)=15.2222
standard deviation, s.d2 =3.993
number(n2)=9
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((7.3/13)+(15.9/9))
= 1.5
II.
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 min (n1-1, n2-1) i.e 8 d.f is 3.4
margin of error = 3.355 * 1.5
= 5.1
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (17.6154-15.2222) ± 5.1 ]
= [-2.7 , 7.5]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=17.6154
standard deviation , s.d1=2.6938
sample size, n1=13
y(mean)=15.2222
standard deviation, s.d2 =3.993
sample size,n2 =9
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 = [( 17.6154-15.2222) ± t a/2 * sqrt((7.3/13)+(15.9/9)]
= [ (2.4) ± t a/2 * 1.5]
= [-2.7 , 7.5]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-2.7 , 7.5] 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