Question

A sample consisting of 922 men and 503 women was asked various questions pertaining to the nation's economy. For a 99% level of confidence, find the margin of error associated with the following samples. (Round your answers to one decimal place.)

(a) the male sample
±  %

(b) the female sample
±  %

(c) the combined sample
±  %

Answer 1

Solution:

Given data:

Number of mens=922

Number of womens=503

Level of confidence=99%=0.99

(a) the male sample
possibile chances (x)=922
sample size(n)= 922 + 503 = 1425
success rate ( p )= x/n = 0.6470
I.
sample proportion = 0.6470

standard error = Sqrt ( (p*(1-p)) /n) )
standard error = Sqrt ( (0.6470*0.353) /1425) )
= 0.0126
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.01
from standard normal table, two tailed z ?/2 =2.576
margin of error = 2.576 * 0.0126
= 0.0325

(b) the female sample
possibile chances (x)=503
sample size(n)=1425
success rate ( p )= x/n = 0.3529
I.
sample proportion = 0.3529
standard error = Sqrt ( (0.3529*0.6471) /1475) ) =sqrt(1.5*10-4)
= 0.0124
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.01
from standard normal table, two tailed z ?/2 =2.576
margin of error = 2.576 * 0.0124
= 0.0319

(c) the combined sample
sample one, x1 =922, n1 =1425, p1= x1/n1=0.647
sample two, x2 =503, n2 =1425, p2= x2/n2=0.352
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( (0.647*0.352/1425) +(0.352 * 0.647/1425))
=0.0178
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.01
from standard normal table, two tailed z ?/2 =2.58
margin of error = 2.58 * 0.0178
=0.046