Question

4. In a survey conducted to determine, among other things, the cost of vacations taken by single adults in a particular region, 144 individuals were randomly sampled. Each person was asked to assess the total cost of her or his most recent vacation. The average cost was $2386 and the standard deviation was $400. (a) (2) Estimate with a 99% confidence interval the average cost of a trip for all single adults in the region (b) [2] How large a sample would have to be taken to estimate the true average cost to within $60 of the mean with 99% confidence ? Use the sample standard deviation to estimate o.

Answer 1

4.
a.
TRADITIONAL METHOD
given that,
standard deviation, σ =400
sample mean, x =2386
population size (n)=144
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 400/ sqrt ( 144) )
= 33.33
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
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 33.33
= 85.87
III.
CI = x ± margin of error
confidence interval = [ 2386 ± 85.87 ]
= [ 2300.13,2471.87 ]
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DIRECT METHOD
given that,
standard deviation, σ =400
sample mean, x =2386
population size (n)=144
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 2386 ± Z a/2 ( 400/ Sqrt ( 144) ) ]
= [ 2386 - 2.576 * (33.33) , 2386 + 2.576 * (33.33) ]
= [ 2300.13,2471.87 ]
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interpretations:
1. we are 99% sure that the interval [2300.13 , 2471.87 ] contains the true population mean
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 mean

b.
given data,
margin of error= 60 $
confidence level 99%
standard deviation 400$
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.01% LOS is = 2.576 ( From Standard Normal Table )
Standard Deviation ( S.D) = 400
ME =60
n = ( 2.576*400/60) ^2
= (1030.4/60 ) ^2
= 294.92 ~ 295