Reliance on solid biomass fuel for cooking and heating exposes many children fro
ID: 3313982 • Letter: R
Question
Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. An article presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (LPG). For the 752 children in biomass households, the sample mean peak expiratory flow (a person's maximum speed of expiration) was 3.1 L/s, and the sample standard deviation was 1.2. For the 758 children whose households used liquefied petroleum gas, the sample mean PEF was 4.36 and the sample standard deviation was 1.84 (a) Calculate a confidence interval at the 95% confidence level for the population mean PEF for children in biomass households and then do likewise for children in LPG households. (Round your answers to two decimal places.) biomass households LPG households What is the simultaneous confidence level for the two intervals? (Round your answer to two decimal places.) (b) Carry out a test of hypotheses at significance level 0.01 to decide whether true average PEF is lower for children in biomass households than it is for children in LPG households (the article included a P-value for this es) (Round your test statistic to two decimal places and your P-value to four decimal places.) State the relevant hypotheses. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) P-value =Explanation / Answer
a.
i.
TRADITIONAL METHOD
given that,
sample mean, x =3.1
standard deviation, s =1.2
sample size, n =752
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.2/ sqrt ( 752) )
= 0.04
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 751 d.f is 1.963
margin of error = 1.963 * 0.04
= 0.09
III.
CI = x ± margin of error
confidence interval = [ 3.1 ± 0.09 ]
= [ 3.01 , 3.19 ]
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DIRECT METHOD
given that,
sample mean, x =3.1
standard deviation, s =1.2
sample size, n =752
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 751 d.f is 1.963
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 3.1 ± t a/2 ( 1.2/ Sqrt ( 752) ]
= [ 3.1-(1.963 * 0.04) , 3.1+(1.963 * 0.04) ]
= [ 3.01 , 3.19 ]
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interpretations:
1) we are 95% sure that the interval [ 3.01 , 3.19 ] contains the true population mean
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 mean
ii.
TRADITIONAL METHOD
given that,
sample mean, x =4.36
standard deviation, s =1.84
sample size, n =758
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.84/ sqrt ( 758) )
= 0.07
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 757 d.f is 1.963
margin of error = 1.963 * 0.07
= 0.13
III.
CI = x ± margin of error
confidence interval = [ 4.36 ± 0.13 ]
= [ 4.23 , 4.49 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =4.36
standard deviation, s =1.84
sample size, n =758
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 757 d.f is 1.963
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 4.36 ± t a/2 ( 1.84/ Sqrt ( 758) ]
= [ 4.36-(1.963 * 0.07) , 4.36+(1.963 * 0.07) ]
= [ 4.23 , 4.49 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 4.23 , 4.49 ] contains the true population mean
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 mean
Answers:
biomass households = (3.01 , 3.19 )
LPG households = (4.23 , 4.49)
simultaneous confidence intervals two in percentage
biomass households = (3.0141 , 3.1859) = (301.41%,318.59%)
LPG households = (4.2288 , 4.4912 ) = (422.88%,449.12%)
b.
Given that,
mean(x)=3.1
standard deviation , 1 =1.2
number(n1)=752
y(mean)=4.36
standard deviation, 2 =1.84
number(n2)=758
null, Ho: u1 = u2
alternate, H1: 1 < u2
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.33
since our test is left-tailed
reject Ho, if zo < -2.33
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=3.1-4.36/sqrt((1.44/752)+(3.3856/758))
zo =-15.773
| zo | =15.773
critical value
the value of |z | at los 0.01% is 2.33
we got |zo | =15.773 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -15.773 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 < u2
test statistic: -15.773
critical value: -2.33
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that biomass households is lower than LPG households
c.
TRADITIONAL METHOD
given that,
sample mean, x =2.1
standard deviation, s =0.5
sample size, n =752
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.5/ sqrt ( 752) )
= 0.0182
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 751 d.f is 1.963
margin of error = 1.963 * 0.0182
= 0.0358
III.
CI = x ± margin of error
confidence interval = [ 2.1 ± 0.0358 ]
= [ 2.0642 , 2.1358 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =2.1
standard deviation, s =0.5
sample size, n =752
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 751 d.f is 1.963
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 2.1 ± t a/2 ( 0.5/ Sqrt ( 752) ]
= [ 2.1-(1.963 * 0.0182) , 2.1+(1.963 * 0.0182) ]
= [ 2.0642 , 2.1358 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 2.0642 , 2.1358 ] contains the true population mean
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 mean
biomass households FEV1 = (2.0642 , 2.1358)
simultaneous confidence intervals =(206.42%,213.58%)