Question

5 6 Problem 214 objections to a new study conducted shows that 50 out of 130 grocery stores in rural area objected the new regulation while 70 out of 120 stores in the city objected the regulation. points): The Municipality department wants to compare the proportion of regulation among grocery stores in a certain city and its rural areas. A a) Is there significant difference between the proportions of objections in rural areas and in city? b) Find a 98% confidence interval for the difference of the two proportions

Answer 1

2.
a.
Given that,
sample one, x1 =50, n1 =130, p1= x1/n1=0.385
sample two, x2 =70, n2 =120, p2= x2/n2=0.583
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.48
q^ Value For Proportion= 1-p^=0.52
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
reject Ho, if zo < -2.326 OR if zo > 2.326
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.385-0.583)/sqrt((0.48*0.52(1/130+1/120))
zo =-3.142
| zo | =3.142
critical value
the value of |z α| at los 0.02% is 2.326
we got |zo| =3.142 & | z α | =2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.142 ) = 0.0017
hence value of p0.02 > 0.0017,here we reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -3.142
critical value: -2.326 , 2.326
decision: reject Ho
p-value: 0.0017
we have enough evidence to support the claim that difference in proportions of objections in rural areas and in city.

b.
TRADITIONAL METHOD
given that,
sample one, x1 =50, n1 =130, p1= x1/n1=0.385
sample two, x2 =70, n2 =120, p2= x2/n2=0.583
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( (0.385*0.615/130) +(0.583 * 0.417/120))
=0.062
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.33
margin of error = 2.33 * 0.062
=0.144
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.385-0.583) ±0.144]
= [ -0.343 , -0.054]
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DIRECT METHOD
given that,
sample one, x1 =50, n1 =130, p1= x1/n1=0.385
sample two, x2 =70, n2 =120, p2= x2/n2=0.583
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 = [ (0.385-0.583) ± 2.33 * 0.062]
= [ -0.343 , -0.054 ]
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interpretations:
1) we are 98% sure that the interval [ -0.343 , -0.054] 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, 98% of these intervals will contains the difference between
true population mean P1-P2