1. The average deposit of 20 customers selected at random from the depositors of
ID: 3355795 • Letter: 1
Question
1. The average deposit of 20 customers selected at random from the depositors of a local bank is $83.60 with a standard deviation of $12.41. what would be a 95% confidence epositors Find the maximum error i interval for the mean deposit of al the banks depositors? Find the maximum error in your estimate. 2. The administration at SDSU thinks that up to 75% of the students drink at parties. They would like to form a 95% confidence interval of the true proportion of students who drink at parties. How large of a sample would be necessary, if they want their estimate to be off by less that 5%?Explanation / Answer
Q1.
TRADITIONAL METHOD
given that,
sample mean, x =83.6
standard deviation, s =12.41
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 12.41/ sqrt ( 20) )
= 2.775
II.
maximum of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
maximum of error = 2.093 * 2.775
= 5.808
III.
CI = x ± maximum of error
confidence interval = [ 83.6 ± 5.808 ]
= [ 77.792 , 89.408 ]
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DIRECT METHOD
given that,
sample mean, x =83.6
standard deviation, s =12.41
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 83.6 ± t a/2 ( 12.41/ Sqrt ( 20) ]
= [ 83.6-(2.093 * 2.775) , 83.6+(2.093 * 2.775) ]
= [ 77.792 , 89.408 ]
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interpretations:
1) we are 95% sure that the interval [ 77.792 , 89.408 ] contains the true population mean
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 mean
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