Question

A random sample of 100 San Diego adults reveals 11 smokers. To test whether the prevalence of smoking in San Diego is lower than national prevalence value of 0.21, compute the 95% C.I. of the population p using the San Diego random sample.

Answer 1

a.
Given that,
possibile chances (x)=11
sample size(n)=100
success rate ( p )= x/n = 0.11
success probability,( po )=0.21
failure probability,( qo) = 0.79
null, Ho:p=0.21
alternate, H1: p<0.21
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.11-0.21/(sqrt(0.1659)/100)
zo =-2.455
| zo | =2.455
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.455 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -2.45514 ) = 0.00704
hence value of p0.05 > 0.00704,here we reject Ho
ANSWERS
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null, Ho:p=0.21
alternate, H1: p<0.21
test statistic: -2.455
critical value: -1.645
decision: reject Ho
p-value: 0.00704
we have enough evidence to support the claim that whether the prevalence of smoking in San Diego is lower than national prevalence value of 0.21.
b.
TRADITIONAL METHOD
given that,
possible chances (x)=11
sample size(n)=100
success rate ( p )= x/n = 0.11
I.
sample proportion = 0.11
standard error = Sqrt ( (0.11*0.89) /100) )
= 0.031
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.031
= 0.061
III.
CI = [ p ± margin of error ]
confidence interval = [0.11 ± 0.061]
= [ 0.049 , 0.171]
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DIRECT METHOD
given that,
possible chances (x)=11
sample size(n)=100
success rate ( p )= x/n = 0.11
CI = confidence interval
confidence interval = [ 0.11 ± 1.96 * Sqrt ( (0.11*0.89) /100) ) ]
= [0.11 - 1.96 * Sqrt ( (0.11*0.89) /100) , 0.11 + 1.96 * Sqrt ( (0.11*0.89) /100) ]
= [0.049 , 0.171]
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interpretations:
1. We are 95% sure that the interval [ 0.049 , 0.171] 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