1. (Use Lynch's Formula) How many sample units must be obtained from 6,500 CSTCians if a 5% margin of error is used?
1. (Use Lynch's Formula) How many sample units must be obtained from 6,500 CSTCians if a 5% margin of error is used?
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.CT: Chapter Test
Problem 24CT: Show the sample space of the experiment: toss a fair coin three times.
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Question
NOTE: PLEASE FOLLOW LYNCH’s FORMULA IN THE PHOTO ATTACHED.
Transcribed Image Text:Å
1. (Use Lynch's Formula) How many sample units must be obtained
from 6,500 CSTCians if a 5% margin of error is used?
Solution
Interpretation
![Lynch's Formula
Where:
n =
n = sample size
N = population size
z = normal variable for reliability level of 95% (usually 1.96)
P = largest possible proportion (usually 0.50)
d = sampling error (usually 0.05)
Example: Compute a sufficient sample size of a target population consisting of 1,524
junior high school students using Lynch's Formula.
Solution:
NZ²-P(1-P)
Nd²+[Z² .P(1-P)]
=
=
=
n =
NZ². P(1-P)
.
Nd² + [Z² · P(1 − P)]
1,524 (1.96)²-(0.50) (1-0.50)
1,524 (0.05)² +[(1.96) ² (0.50)(1-0.50)]
1,463.6496
3.81 +0.9604
1,524(3.8416) (0.50) (0.50)
1,524 (0.0025)+ [(3.8416) (0.50) (0.50)]
5854.5984-0.25
3.81 +[(3.8416) (0.25)]
1,463.6496
4.7704
= 306.8190508
≈ 307](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fbf675e-2739-4cb1-bf6f-4b03cb28814f%2F46cdf47c-d0e7-4ae9-87c8-950b808e3371%2Fffnnbqr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Lynch's Formula
Where:
n =
n = sample size
N = population size
z = normal variable for reliability level of 95% (usually 1.96)
P = largest possible proportion (usually 0.50)
d = sampling error (usually 0.05)
Example: Compute a sufficient sample size of a target population consisting of 1,524
junior high school students using Lynch's Formula.
Solution:
NZ²-P(1-P)
Nd²+[Z² .P(1-P)]
=
=
=
n =
NZ². P(1-P)
.
Nd² + [Z² · P(1 − P)]
1,524 (1.96)²-(0.50) (1-0.50)
1,524 (0.05)² +[(1.96) ² (0.50)(1-0.50)]
1,463.6496
3.81 +0.9604
1,524(3.8416) (0.50) (0.50)
1,524 (0.0025)+ [(3.8416) (0.50) (0.50)]
5854.5984-0.25
3.81 +[(3.8416) (0.25)]
1,463.6496
4.7704
= 306.8190508
≈ 307
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