Transcript Nature of Air Contaminants

Occupational Air Sampling
Strategies – who, when, how….
Lecture Notes
1
Components of a Sampling
Strategy
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
Characterization and information gathering

Risk assessment and sampling priorities

Air sampling strategy and analysis

Data interpretation

Recommendation and reporting

Re-evaluation
Air sampling strategy
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
Which employee or employees should be
sampled?

How many samples should be taken on each
workday sampled to define the employee’s
exposure?

How long should the sampling interval be for
a measurement sample?
Air sampling strategy
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
What periods during the workday should the
employee’s exposure be sampled?

How many workdays during the year should
be sampled and when?

Time to result – acute vs. chronic and direct
reading real time vs. sampling media and
two-week lab time.
Which employee or employees should
be sampled?
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
OSHA regulation requires the sampling of the
“employee believed to have the greatest
exposure” or the “maximum risk employee” principle extended to include groups of
employees

Use the exposure risk/health risk priority
matrix
Which employee or employees should
be sampled?

If the maximum risk employee or group can’t
be identified then do random sampling of the
group of workers.

Objective is to select a subgroup of adequate
size so that there is a high probability that the
random sample will contain at least one
worker with high exposure if one exists.

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Want to be careful about use of group statistics
NIOSH’s Occupational Exposure
Sampling Strategies Manual

Gives one sample size
model


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Set up to ensure with
90% confidence that at
least one person from the
highest 10% exposure
group is contained in the
sample
Conversely 10% chance
of missing someone in the
highest 10% exposure
group
Sampling
periods
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
Various types of
sampling periods
possible

As you increase
the # of sample
periods in a shift
the analysis
becomes more
sophisticated
How many workdays during the year
should be sampled and when?

OSHA regulation can differ

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Substance specific standards specify sample
interval dependent on the exposure level relative
to the PEL
1910.1025 Lead

1910.1025(c)(1) The employer shall assure that no
employee is exposed to lead at concentrations greater
than fifty micrograms per cubic meter of air (50 mg/m3)
averaged over an 8-hour period.

1910.1025(c)(2) If an employee is exposed to lead for
more than 8 hours in any work day, the permissible
exposure limit, as a time weighted average (TWA) for
that day, shall be reduced according to the following
formula:

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Maximum permissible limit (in mg/m3)=400 divided by hours
worked in the day.
1910.1025 Lead
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
1910.1025 (d) Exposure monitoring -

1910.1025(d)(1)(ii) With the exception of monitoring
under paragraph (d)(3), the employer shall collect full
shift (for at least 7 continuous hours) personal
samples including at least one sample for each shift
for each job classification in each work area.

1910.1025(d)(1)(iii) Full shift personal samples shall
be representative of the monitored employee's
regular, daily exposure to lead.
1910.1025 Lead
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
1910.1025(d)(2) Initial determination. Each employer
who has a workplace or work operation covered by this
standard shall determine if any employee may be
exposed to lead at or above the action level.

1910.1025(d)(4)(i) Where a determination conducted
under paragraphs (d)(2) and (3) of this section shows
the possibility of any employee exposure at or above
the action level, the employer shall conduct monitoring
which is representative of the exposure for each
employee in the workplace who is exposed to lead.
1910.1025 Lead


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1910.1025(d)(6) Frequency.
1910.1025(d)(6)(i) If the initial monitoring reveals employee
exposure to be below the action level the measurements need
not be repeated except as otherwise provided in paragraph (d)(7)
of this section.
1910.1025(d)(6)(ii) If the initial determination or subsequent
monitoring reveals employee exposure to be at or above the
action level but below the permissible exposure limit the employer
shall repeat monitoring in accordance with this paragraph at least
every 6 months. The employer shall continue monitoring at the
required frequency until at least two consecutive measurements,
taken at least 7 days apart, are below the action level at which
time the employer may discontinue monitoring for that employee
except as otherwise provided in paragraph (d)(7) of this section.
1910.1025 Lead

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1910.1025(d)(6)(iii) If the initial monitoring reveals that employee
exposure is above the permissible exposure limit the employer shall
repeat monitoring quarterly. The employer shall continue monitoring at the
required frequency until at least two consecutive measurements, taken at
least 7 days apart, are below the PEL but at or above the action level at
which time the employer shall repeat monitoring for that employee at the
frequency specified in paragraph (d)(6)(ii), except as otherwise provided in
paragraph (d)(7) of this section.
1910.1025(d)(7) Additional monitoring. Whenever there has been a
production, process, control or personnel change which may result in new
or additional exposure to lead, or whenever the employer has any other
reason to suspect a change which may result in new or additional
exposures to lead, additional monitoring in accordance with this paragraph
shall be conducted.
Limited basic statistics review

What is a population?

All of something, e.g.



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What is a sample?

Observations selected from a larger population
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A lot of silicon wafers produced in a semiconductor factory
The 2003 semester students in IH&S 725
All lead exposures to “metal pourers”
A sample of the silicon wafers from a lot
A sample of students from the 2003 IH&S 725 class
A sample of lead exposures from all possible exposures
Population and sample means
P opulat ion
Sam ple
n
N
m
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x
i 1
N
i
x
x
i 1
n
i
Variance and standard deviation
Population
N
Variance :  2 
 x
i 1
Standard deviation:  
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i
 m
N
2
Sample
n
2
s2 
s 
 x
i
i 1
 m
n 1
s2
2
Point estimate

A point estimate of some population
parameter is a single numerical value of a
statistic
Sample mean is a point estimate of an unknown
population mean
 Sample variance is a point estimate of an
unknown population variance

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2-sided confidence interval on the
mean
CI1-


Two sided CI
P(L ≤ µ ≤ U) = 1-α where 0 ≤ α ≤ 1




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 s 
 x  t1 , n1   
2
 n
We have a probability of 1-α of selecting a sample that will
produce a interval containing the true value of µ
The interval l ≤ µ ≤ u is called a 100(1-α) percent confidence
interval
I and u are the lower- and upper-confidence limits
(1-α) is the confidence coefficient
Confidence Interval on sample mean:
One-sided Confidence Interval
CI1-

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 s 
 x  t1 , n1  

 n
α is assigned to either the upper or lower tail of the tdistribution
Where does t come from

Definition

Let X1 X2 ….Xn be a random sample for a normal
distribution with unknown mean and unknown
variance σ2. The quantity
X m
T
S
n
has a t-distribution with n-1 degrees of freedom.
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t-distribution and α



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Probability density
function i.e.. area under
the curve = 1.
α represents the area
assigned to the tail(s) of
the distribution –
probability of T > t1-α
&/or -tα
Note that t1-α = -tα
Example #1

Using the t-table determine the t-value with 14 df that
leaves an area of 0.025 to the left (and therefore an
area of 0.975 to the right).
t1  t  t0.975  t0.025
t0.975  2.145
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Example #2

What is the P(-t0.025 < T < t0.025)?

Since t0.025 leaves an area of 0.025 to the
right and -t0.025 leaves and area of 0.025 to
the left the total area (probability) between t0.025 and t0.025 is?
1- 0.025 – 0.025 = 0.95
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Influence of degrees of freedom on the
t-distribution


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The area to the left
or right of tα and t1-α
decreases with
increasing df
Note from the ttable that as df gets
large it the t-value
approaches the z
value from the SND
table
Standard normal distribution, SND

Definition

If is the mean of a random sample of size n taken
from a population with mean µ and variance σ2,
then the limiting form of the distribution of:
Z
X m

n
as n
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∞, is the standard normal distribution snd(z;0,1).
Using the standard normal
distribution

Using the z-table, find the area under the
curve that lies:


to the right of z = 1.84

1- area to the left of z = 1.84

1-0.9671= 0.0329 or 3.29%
between z = -1.97 and z = 0.86

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0.8051 - 0.0244 = 0.7807
Example #3

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Given a normal distribution with µ = .05mg/m3
and σ = 0.01, find the probability that a sample
point estimate of µ would fall between 0.04 and
0.06mg/m3.
Example #3 solution

First calculate the z-value corresponding to
x1 = .04mg/m3 and x2 = .06mg/m3
0.04  0.05
z1 
 1
0.01
0.06  0.05
z2 
1
0.01
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Example #3 solution con’t

Second determine the probability

P(0.04mg/m3 < x < 0.06mg/m3) = P(-1< z < 1)
P( 1  z  1)  P( z  1)  P( z  1)
P  0.8413  0.1587
P  0.6826 or 68.3%
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