Question Help * To construct a confidence interval for the difference between tw

Question

Question Help * To construct a confidence interval for the difference between two population means 1-2-use the formula shown below, when both population standard de are known, and either both populations are normally distributed or both n 230 and n2 230. Also, the samples must be randomly selected and independent tons n1 n2 n n2 The descriptive statistics for the annual salaries from a random sample of microbiologists from two regions are shown below. Construct a 95% confidence interval for the difference between the mean annual salaries h" $105,720. n1-39, and -S9000, x2-S85.340. n2-35, and 2-$9010 Complete the 95% confidence interval for 1-12 below. (Round to the nearest dollar as needed.)

Explanation / Answer

i)
TRADITIONAL METHOD
given that,
mean(x)=105720
standard deviation , 1 =9000
population size(n1)=39
y(mean)=85340
standard deviation, 2 =9010
population size(n2)=35
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((81000000/39)+(81180100/35))
= 2096.7
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 2096.7
= 4109.6
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (105720-85340) ± 4109.6 ]
= [16270.4 , 24489.6]
= $16270 < u1-u2 < 24489
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=105720
standard deviation , 1 =9000
number(n1)=39
y(mean)=85340
standard deviation, 2 =9010
number(n2)=35
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 = [ ( 105720-85340) ±Z a/2 * Sqrt( 81000000/39+81180100/35)]
= [ (20380) ± Z a/2 * Sqrt( 4396354.5) ]
= [ (20380) ± 1.96 * Sqrt( 4396354.5) ]
= [16270.4 , 24489.6]
= $16270.4 < u1-u2 < 24489.6
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [16270.4 , 24489.6] 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
ii)
TRADITIONAL METHOD
given that,
sample one, x1 =95052, n1 =1068000, p1= x1/n1=0.089
sample two, x2 =135792, n2 =1476000, p2= x2/n2=0.092
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.089*0.911/1068000) +(0.092 * 0.908/1476000))
=0
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table,right tailed z /2 =1.28
margin of error = 1.28 * 0
=0
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.089-0.092) ±0]
= [ -0.003 , -0.003]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =95052, n1 =1068000, p1= x1/n1=0.089
sample two, x2 =135792, n2 =1476000, p2= x2/n2=0.092
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.089-0.092) ± 1.28 * 0]
= [ -0.003 , -0.003 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ -0.003 , -0.003] contains the difference between
true population proportion P1-P2
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 difference between
true population mean P1-P2
iii)
TRADITIONAL METHOD
given that,
mean(x)=135
standard deviation , s.d1=3.77
number(n1)=14
y(mean)=53
standard deviation, s.d2 =2.05
number(n2)=18
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((14.2/14)+(4.2/18))
= 1.1
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 13 d.f is 3
margin of error = 3.012 * 1.1
= 3.4
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (135-53) ± 3.4 ]
= [78.6 , 85.4]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=135
standard deviation , s.d1=3.77
sample size, n1=14
y(mean)=53
standard deviation, s.d2 =2.05
sample size,n2 =18
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 = [( 135-53) ± t a/2 * sqrt((14.2/14)+(4.2/18)]
= [ (82) ± t a/2 * 1.1]
= [78.6 , 85.4]
= $78 < u1-u2 <$85
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [78.6 , 85.4] 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

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