Chapter &, Section 2-CI, Exercise 108 Plastic Miorofiber Polution on Shorelines
ID: 3328916 • Letter: C
Question
Chapter &, Section 2-CI, Exercise 108 Plastic Miorofiber Polution on Shorelines In a stuer, we see that plastic micreparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study also took samples from ocean beaches. five samples were taken fragm each of 18S terent shorenes worldwide, for a total of 90 sampies of size 250 m. The mean number of plastic microparticles founs per 250 mL of sediment was 18.3 with a standard deviation of 8.2 Round your ansiwers to two decimal places Round your answer to two cecmal places the absolute bolerance is fo.osExplanation / Answer
TRADITIONAL METHOD
given that,
sample mean, x =18.3
standard deviation, s =8.2
sample size, n =90
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8.2/ sqrt ( 90) )
= 0.86436
III.
CI = x ± margin of error
confidence interval = [ 18.3 ± 2.27516 ]
= [ 16.02484 , 20.57516 ]
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DIRECT METHOD
given that,
sample mean, x =18.3
standard deviation, s =8.2
sample size, n =90
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 89 d.f is 2.6322
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 = [ 18.3 ± Z a/2 ( 8.2/ Sqrt ( 90) ]
= [ 18.3-(2.6322 * 0.86436) , 18.3+(2.6322 * 0.86436) ]
= [ 16.02484 , 20.57516 ]
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interpretations:
1) we are 99% sure that the interval [ 16.02484 , 20.57516 ] 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
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 89 d.f is 2.6322
margin of error = 2.6322 * 0.86436
= 2.27516 ~ 2.28
III.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.576 ( From Standard Normal Table )
Standard Deviation ( S.D) = 8.2
ME =1
n = ( 2.576*8.2/1) ^2
= (21.123/1 ) ^2
= 446.19 ~ 447
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