Question

An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.1 years of the population mean. Assume the population of ages is normally distributed (a) Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.7 years (b) The sample mean is 21 years of age. Using the minimum sample size with a 90% level of confidence, does it seem likely that the population mean could be within 8% of the sample mean? within 9% of the sample mean? Explain Click here to view page 1 of the Standard NormalTable. Click here to view. Page 2 of the Standard NormalTable

Answer 1

Solution :

a.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.3
ME =1.1
n = ( 1.645*1.3/1.1) ^2
= (2.14/1.1 ) ^2
= 3.78 ~ 4          
b.
i.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.3
mean estimate is 8% of sample mean
when sample mean is 20 years
so that mean estimate is 1.6
ME =1.6
n = ( 1.645*1.3/1.6) ^2
= (2.14/1.6 ) ^2
= 1.79 ~ 2      
ii.  
when sample mean is 20 years
mean estimate is 9% of sample mean
mean estimate =20*0.09=1.8
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.3
ME =1.8
n = ( 1.645*1.3/1.8) ^2
= (2.14/1.8 ) ^2
= 1.41 ~ 2          

TRADITIONAL METHOD
given that,
standard deviation, σ =1.3
sample mean, x =20
population size (n)=4
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 1.3/ sqrt ( 4) )
= 0.65
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.65
= 1.07
III.
CI = x ± margin of error
confidence interval = [ 20 ± 1.07 ]
= [ 18.93,21.07 ]
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DIRECT METHOD
given that,
standard deviation, σ =1.3
sample mean, x =20
population size (n)=4
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 20 ± Z a/2 ( 1.3/ Sqrt ( 4) ) ]
= [ 20 - 1.645 * (0.65) , 20 + 1.645 * (0.65) ]
= [ 18.93,21.07 ]
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interpretations :
1. we are 90% sure that the interval [18.93 , 21.07 ] contains the true population mean
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 mean