A random sample of 600 engineers included 24 Hispanie Americans, 40 African Amer
ID: 3362209 • Letter: A
Question
A random sample of 600 engineers included 24 Hispanie Americans, 40 African Americans, and 102 females. Determine a 90% confidence interval for the proportion of all en- gineers who are female. Determine a 99% confidence interval for the proportion of all en- gineers who are Hispanic American or African American. (b) (c) Determine the minimum. sample size required for a 90% confidence interval for the proportion of engineers who are female to have width no greater than 1%. Calculate both the sample size as determined by the sample proportion, and the sample size as de- termined by the most conservative method.Explanation / Answer
a.
90% confidence interval for the proportion of all engineers who are females
TRADITIONAL METHOD
given that,
possibile chances (x)=102
sample size(n)=600
success rate ( p )= x/n = 0.17
I.
sample proportion = 0.17
standard error = Sqrt ( (0.17*0.83) /600) )
= 0.0153
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.0153
= 0.0252
III.
CI = [ p ± margin of error ]
confidence interval = [0.17 ± 0.0252]
= [ 0.1448 , 0.1952]
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DIRECT METHOD
given that,
possibile chances (x)=102
sample size(n)=600
success rate ( p )= x/n = 0.17
CI = confidence interval
confidence interval = [ 0.17 ± 1.645 * Sqrt ( (0.17*0.83) /600) ) ]
= [0.17 - 1.645 * Sqrt ( (0.17*0.83) /600) , 0.17 + 1.645 * Sqrt ( (0.17*0.83) /600) ]
= [0.1448 , 0.1952]
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interpretations:
1. We are 90% sure that the interval [ 0.1448 , 0.1952] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
99% confidence interval for the proportion of all engineers who are Hispanic americans or African americans
TRADITIONAL METHOD
given that,
possibile chances (x)=64
sample size(n)=600
success rate ( p )= x/n = 0.1067
I.
sample proportion = 0.1067
standard error = Sqrt ( (0.1067*0.8933) /600) )
= 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
III.
CI = [ p ± margin of error ]
confidence interval = [0.1067 ± 0.0325]
= [ 0.0742 , 0.1391]
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DIRECT METHOD
given that,
possibile chances (x)=64
sample size(n)=600
success rate ( p )= x/n = 0.1067
CI = confidence interval
confidence interval = [ 0.1067 ± 2.576 * Sqrt ( (0.1067*0.8933) /600) ) ]
= [0.1067 - 2.576 * Sqrt ( (0.1067*0.8933) /600) , 0.1067 + 2.576 * Sqrt ( (0.1067*0.8933) /600) ]
= [0.0742 , 0.1391]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 99% sure that the interval [ 0.0742 , 0.1391] 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.
90% confidence interval for the proportion of all engineers who are females
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.17
ME = 0.01
n = ( 1.645 / 0.01 )^2 * 0.17*0.83
= 3818.2013 ~ 3819
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