Chapter 8.1 Problem 43, page 363: A college admissions officer takes a simple ra
ID: 3359724 • Letter: C
Question
Chapter 8.1 Problem 43, page 363: A college admissions officer takes a simple random sample of 1o0 entermng tresnmen and computes their mean SAT score to be 458. Assume the population standard deviation is 116: a) Construct a 99% confidence interval for the mean SAT scores. b) If the sample size is 75. What would the margin of error be for a 99% confidence interva c) Construct a 95% confidence interval for the mean SAT scores of these students? d) If the administrator wanted a margin of error of (+/-)10, what size of sample would be necessaryExplanation / Answer
PART A.
TRADITIONAL METHOD
given that,
standard deviation, =116
sample mean, x =458
population size (n)=100
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 116/ sqrt ( 100) )
= 11.6
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
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 11.6
= 29.882
III.
CI = x ± margin of error
confidence interval = [ 458 ± 29.882 ]
= [ 428.118,487.882 ]
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DIRECT METHOD
given that,
standard deviation, =116
sample mean, x =458
population size (n)=100
level of significance, = 0.01
from standard normal table, two tailed z /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 = [ 458 ± Z a/2 ( 116/ Sqrt ( 100) ) ]
= [ 458 - 2.576 * (11.6) , 458 + 2.576 * (11.6) ]
= [ 428.118,487.882 ]
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interpretations:
1. we are 99% sure that the interval [428.118 , 487.882 ] 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
PART B.
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 116/ sqrt ( 75) )
= 13.395
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
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 13.395
= 34.504
PART C.
given that,
standard deviation, =116
sample mean, x =458
population size (n)=75
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 458 ± Z a/2 ( 116/ Sqrt ( 75) ) ]
= [ 458 - 1.96 * (13.395) , 458 + 1.96 * (13.395) ]
= [ 431.747,484.253 ]
PART D.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 116
ME =10
n = ( 1.96*116/10) ^2
= (227.36/10 ) ^2
= 516.926 ~ 517