Question

1. Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time. The bank takes a random sample of 200 recently approved loans.

   1. What values of sample proportions of clients who will not make timely payments would be unusual? Explain.

   2. Construct and interpret a 95% confidence interval for the true proportion of clients who will not make timely payments.

   3. Since the U.S. economy has changed, bank officials would like to do a new study to estimate the true proportion of clients who will not make timely payments. Assume you have no preconceived idea of what that proportion would be. What sample size is needed if you wish to be 99% confident that your estimate is within 2% of the true proportion.   

   4. Based on previous research, you assume the proportion of clients who will not make timely payments is 7%. What sample size is needed if you wish to be 99% confident that your estimate is within 2% of the true proportion.

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=14
sample size(n)=200
success rate ( p )= x/n = 0.07
I.
sample proportion = 0.07
standard error = Sqrt ( (0.07*0.93) /200) )
= 0.018
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 * 0.018
= 0.0354
III.
CI = [ p ± margin of error ]
confidence interval = [0.07 ± 0.0354]
= [ 0.0346 , 0.1054]
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DIRECT METHOD
given that,
possible chances (x)=14
sample size(n)=200
success rate ( p )= x/n = 0.07
CI = confidence interval
confidence interval = [ 0.07 ± 1.96 * Sqrt ( (0.07*0.93) /200) ) ]
= [0.07 - 1.96 * Sqrt ( (0.07*0.93) /200) , 0.07 + 1.96 * Sqrt ( (0.07*0.93) /200) ]
= [0.0346 , 0.1054]
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interpretations:
1. We are 95% sure that the interval [ 0.0346 , 0.1054] 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.
given data,
margin of error =2%
Assume sample proportion =50%
confidencelevel is 99%
sample size =n
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.5
ME = 0.02
n = ( 2.576 / 0.02 )^2 * 0.5*0.5
= 4147.36 ~ 4148          
c.
given data,
margin of error =2% ,   the proportion of clients who will not make timely payments is 7%.
sample proportion =7%
confidencelevel is 99%
sample size =n
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.07
ME = 0.02
n = ( 2.576 / 0.02 )^2 * 0.07*0.93
= 1079.9725 ~ 1080