Question

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Chapter 8 Reflection:

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 colleges students was used in this study.

1. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses.

2. Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses.
3. What happens to the margin of error as the confidence is increased from 95% to 99%? Explain why this makes sense.

4. Suppose the NCES would like to re-do their study. If they want a 95% confidence interval with a margin of error for the population proportion within ±0.03, determine the sample size they should use. Use p*=0.47, the sample proportion from their previous study.

Answer 1

8.

a.
TRADITIONAL METHOD
given that,
possible chances (x)=211.5
sample size(n)=450
success rate ( p )= x/n = 0.47
I.
sample proportion = 0.47
standard error = Sqrt ( (0.47*0.53) /450) )
= 0.0235
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 * 0.0235
= 0.0461
III.
CI = [ p ± margin of error ]
confidence interval = [0.47 ± 0.0461]
= [ 0.4239 , 0.5161]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=211.5
sample size(n)=450
success rate ( p )= x/n = 0.47
CI = confidence interval
confidence interval = [ 0.47 ± 1.96 * Sqrt ( (0.47*0.53) /450) ) ]
= [0.47 - 1.96 * Sqrt ( (0.47*0.53) /450) , 0.47 + 1.96 * Sqrt ( (0.47*0.53) /450) ]
= [0.4239 , 0.5161]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.4239 , 0.5161] 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
b.
TRADITIONAL METHOD
given that,
possible chances (x)=211.5
sample size(n)=450
success rate ( p )= x/n = 0.47
I.
sample proportion = 0.47
standard error = Sqrt ( (0.47*0.53) /450) )
= 0.0235
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 * 0.0235
= 0.0606
III.
CI = [ p ± margin of error ]
confidence interval = [0.47 ± 0.0606]
= [ 0.4094 , 0.5306]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=211.5
sample size(n)=450
success rate ( p )= x/n = 0.47
CI = confidence interval
confidence interval = [ 0.47 ± 2.576 * Sqrt ( (0.47*0.53) /450) ) ]
= [0.47 - 2.576 * Sqrt ( (0.47*0.53) /450) , 0.47 + 2.576 * Sqrt ( (0.47*0.53) /450) ]
= [0.4094 , 0.5306]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.4094 , 0.5306] 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.
yes,
the margin of error as the confidence is increased from 95% to 99%
margin of error is 0.0461 for confidence interval 95%
margin of error is 0.0606 for confidence interval 99%
Because confidence level is increases then margin of error is also increases.
d.
Given data,
sample proportion =0.47
confidence level is 95%
margin of error =0.03
sample size = n
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.47
ME = 0.03
n = ( 1.96 / 0.03 )^2 * 0.47*0.53
= 1063.2695 ~ 1064