We have studied the effect of the sample size on the margin of error of the conf

Question

We have studied the effect of the sample size on the margin of error of the confidence interval for a single proportion. In this exercise we perform some calculations to observe this effect for the two-sample problem. Suppose that p1 = 0.7 and p2 = 0.5, and n represents the common value of n1 and n2. Compute the 95% margins of error for the difference between the two proportions for n = 40, 50, 60, 80, 380, 480, and 980. Present the results in a table. (Give the large-sample margins of error. Round your answers to three decimal places.)

980

What is your findings.

As sample size increases, margin of error remains constant.

As sample size increases, margin of error increases.    

As sample size increases, margin of error decreases.

There is not enough information.

980

Explanation / Answer

a.
given that,
sample one, n1 =40, p1= x1/n1=0.7
sample two, n2 =40, p2= x2/n2=0.5
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( (0.7*0.3/40) +(0.5 * 0.5/40))
=0.1072
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.1072
=0.2102

b.
given that,
sample one, n1 =50, p1= x1/n1=0.7
sample two, n2 =50, p2= x2/n2=0.5
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( (0.7*0.3/50) +(0.5 * 0.5/50))
=0.0959
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0959
=0.188

c.
sample one, n1 =60, p1= x1/n1=0.7
sample two, n2 =60, p2= x2/n2=0.5
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( (0.7*0.3/60) +(0.5 * 0.5/60))
=0.0876
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0876
=0.1716

d.
given that,
sample one, n1 =80, p1= x1/n1=0.7
sample two, n2 =80, p2= x2/n2=0.5
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( (0.7*0.3/80) +(0.5 * 0.5/80))
=0.0758
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0758
=0.1486

e.
given that,
sample one, n1 =380, p1= x1/n1=0.7
sample two, n2 =380, p2= x2/n2=0.5
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( (0.7*0.3/380) +(0.5 * 0.5/380))
=0.0348
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0348
=0.0682

f.
given that,
sample one, n1 =480, p1= x1/n1=0.7
sample two, n2 =480, p2= x2/n2=0.5
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( (0.7*0.3/480) +(0.5 * 0.5/480))
=0.031
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.031
=0.0607

As sample size increases, margin of error decreases

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