enoVO HLAD3104-2017 9.1 Estimating a Population Proportion Objective: Construct
ID: 3319399 • Letter: E
Question
enoVO HLAD3104-2017 9.1 Estimating a Population Proportion Objective: Construct and in 9.1.27 in a poll, 103 of 1030 randomly selected adults aged 18 or older stated that they belleve there is too ile spending on national delense Use this information to complete parts (a) through (e) below pr (Sinngary your answer ) lb) Very that the requroments for constructing a conlidance interval about p are saltisfed Ave the regsérements for consatucting a cortédence interval abpsatd sample is not satisfed C No, the that the sample size is no more han 5% of population is not satisfied (c) Conwua a 90% corience interva The 90% confidence interval is ( 0085 0115) Round to three decimal places as needed ) (d) 1sR possible that more than 15% of aduts aged 18 al for the proportion of adults aged 18 or older who believe there is too lile spending on national defense older believe there is too iew spending on naional defense? Is it A It is possible, but not Ikely R is possible and lkely X t is not possble sul, a pan (c) to construct .90% codimee intera ter me popdason proporSon of aans aged 16 o' old" who do not for he pepulation proportion of aduats aged 18 or oider who do not beleve there is soo litle spending on national defense The 90% confidence interval is Round to three decimal places as needed ) All parts showing Ask meExplanation / Answer
9.1.27.
a.
given that,
possibile chances (x)=103
sample size(n)=1030
success rate ( p )= x/n = 1/10
b.
Yes, the requirements for constructing the confidence interval is satisfied
c.
TRADITIONAL METHOD
given that,
possibile chances (x)=103
sample size(n)=1030
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /1030) )
= 0.009
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.009
= 0.015
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.015]
= [ 0.085 , 0.115]
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DIRECT METHOD
given that,
possibile chances (x)=103
sample size(n)=1030
success rate ( p )= x/n = 0.1
CI = confidence interval
confidence interval = [ 0.1 ± 1.645 * Sqrt ( (0.1*0.9) /1030) ) ]
= [0.1 - 1.645 * Sqrt ( (0.1*0.9) /1030) , 0.1 + 1.645 * Sqrt ( (0.1*0.9) /1030) ]
= [0.085 , 0.115]
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interpretations:
1. We are 90% sure that the interval [ 0.085 , 0.115] 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
We are 90% sure that the interval [ 0.085 , 0.115] proportion of adults of aged 18 years or older believe that there is too life spending on defense
d.
Yes,it is possible but not likely
more than 15% of adults aged 18 or older believe that there is too life spending on national defense
e.
as the part.C 90% sure that confidence interval [ 0.085 , 0.115],
now proportion of adults of aged 18 years or older do not believe that there is too life spending on defense so that,
90% confidence interval is (1-0.085,1-0.115) = (0.915,0.885)
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