Question

What if we found that the sample statistic and the corresponding margin of error based on the ENTIRE American Community Survey(ACS) sample turned out to be a sample proportion is 0.155 with a margin of error of 0.0001 (for 90% confidence level) again for the entire sample. (important we stay with the results written here).

Answer 1

a.
Given that,
sample one, x1 =861, n1 =1000, p1= x1/n1=0.861
sample two, x2 =139, n2 =1000, p2= x2/n2=0.139
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.861-0.139)/sqrt((0.5*0.5(1/1000+1/1000))
zo =32.289
| zo | =32.289
critical value
the value of |z α| at los 0.1% is 1.645
we got |zo| =32.289 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 32.2888 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 32.289
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference of proportion of ENTIRE American Community Survey(ACS).

b.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.155
ME = 0.0001
n = ( 1.645 / 0.0001 )^2 * 0.155*0.845
= 35442162.438 ~ 35442163