Question

In a sample of 4500 first year college students, a national survey reported that 39% participated in community service or volunteering work.

a)v Find the margin of error for99% confidence level. Explain the findings of the study to someone who does not know statistics.

b) Suppose a similar study collects a new sample of 4,500 first year students. If a 99% confidence interval is computed from the new sample, will the confidence interval will be exactly the same as the results in a)? Explain your answer.

c) The same survey also asked about academic life. In response to one of the questions, 42% of first students plan to study abroad. Give the 95% confidence interval for the population proportion.

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=1755
sample size(n)=4500
success rate ( p )= x/n = 0.39
I.
sample proportion = 0.39
standard error = Sqrt ( (0.39*0.61) /4500) )
= 0.007
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.007
= 0.019
b.
given that,
possible chances (x)=1755
sample size(n)=4500
success rate ( p )= x/n = 0.39
I.
sample proportion = 0.39
standard error = Sqrt ( (0.39*0.61) /4500) )
= 0.007
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
margin of error = 2.576 * 0.007
= 0.019
III.
CI = [ p ± margin of error ]
confidence interval = [0.39 ± 0.019]
= [ 0.371 , 0.409]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=1755
sample size(n)=4500
success rate ( p )= x/n = 0.39
CI = confidence interval
confidence interval = [ 0.39 ± 2.576 * Sqrt ( (0.39*0.61) /4500) ) ]
= [0.39 - 2.576 * Sqrt ( (0.39*0.61) /4500) , 0.39 + 2.576 * Sqrt ( (0.39*0.61) /4500) ]
= [0.371 , 0.409]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.371 , 0.409] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
c.
TRADITIONAL METHOD
given that,
possible chances (x)=1890
sample size(n)=4500
success rate ( p )= x/n = 0.42
I.
sample proportion = 0.42
standard error = Sqrt ( (0.42*0.58) /4500) )
= 0.007
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 * 0.007
= 0.014
III.
CI = [ p ± margin of error ]
confidence interval = [0.42 ± 0.014]
= [ 0.406 , 0.434]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=1890
sample size(n)=4500
success rate ( p )= x/n = 0.42
CI = confidence interval
confidence interval = [ 0.42 ± 1.96 * Sqrt ( (0.42*0.58) /4500) ) ]
= [0.42 - 1.96 * Sqrt ( (0.42*0.58) /4500) , 0.42 + 1.96 * Sqrt ( (0.42*0.58) /4500) ]
= [0.406 , 0.434]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.406 , 0.434] 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