SME
Analytical Chemistry
Sampling and quality assurance
Use this for “Grab samples of an ore vary with a relative standard deviation of 8%. How many must be averaged to be 95% sure the mean is within 3% of the true value?”
Chart
The curve is centered at μ = 0 with standard error SE = 1.02%.
Interval: 0 ± 2, from −2 to 2, the middle 0.9500 of the curve.
This is an example. Type your own values to start a new problem.
n
=
×
s (%)
e (%)
2
SE
Standard error of the mean
calculated · %
Formulas
n
=
(
z
×
s
e
)
2
24.01
=
(1.96 ×
5
2
)
2
SE
=
s
÷
√
n
1.0204
=
5
÷
√
24.01
Assumptions
•
Sampling error dominates: the spread between samples, s, is much larger than the error of the analysis itself.
•
z is the normal value for the confidence (1.96 at 95%, 2.58 at 99%); for a small n, t is larger, so n would rise.
•
The n samples are averaged, so the mean’s relative standard error is s ÷ √n, and z of them must fit inside e.
Step-by-step
WE KNOW
z = 1.96, s = 5%, e = 2%
FIND
samples needed (n), standard error of the mean (SE)
Step 1 · Find samples needed (n)
Use n = (z × s ÷ e)²
The
mean
of
n
samples
has
the
error
zs
÷
√
n
;
set
it
equal
to
e
and
solve
for
n
.
n
=
(1.96 ×
5
2
)
2
n
=
(
9.8
2
)
2
n
=
4.9
2
n
=
24.01
→
Round
up:
take
25
samples.
Step 2 · Find standard error of the mean (SE)
Use SE = s ÷ √n
The
mean
of
n
samples
scatters
√
n
times
less
than
one
sample
does.
SE
=
5
÷
√
24.01
SE
=
5
4.9
SE = 1.0204%
Check
Put the numbers back in. Both sides should match.
24.01
=
(1.96 ×
5
2
)
2
✓
1.0204
=
5
÷
√
24.01
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Spike recovery
Use this for “A river sample reads 3.6 ppb lead. A spike that adds 5 ppb is put into a second portion, which then reads 8.3 ppb. What is the recovery?”
Detection and quantitation limits
Use this for “Ten blank readings have a standard deviation of 0.0035, and the calibration slope is 0.125 per ppm. Find the detection and quantitation limits.”