SME
Analytical Chemistry · Problem type
Rejecting a suspect reading (Q test)
Use this for “Four titrations take 24.18, 24.21, 24.22 and 24.39 mL. Qcrit is 0.829 for four readings at 95%. May 24.39 mL be rejected?”
Diagram
3 values: 10.09, 10.18, 10.45.
This is an example. Type your own values to start a new problem.
Q
=
|
−
|
−
xq is the ringed suspect and xₙ the reading nearest it; xₕᵢ and xₗₒ are the highest and lowest readings, the picture’s range.
gap
Gap to the nearest
calculated
w
Range
calculated
Qcrit
Q critical at 95%
example
Formulas
gap = |xq − xₙ|
0.27 = |10.45 − 10.18|
w
=
xₕᵢ
−
xₗₒ
0.36 = 10.45 − 10.09
Q
=
gap
w
0.75
=
0.27
0.36
Assumptions
•
The suspect reading is the highest or the lowest; all readings are in one unit.
•
Qcrit comes from a table for n readings at 95%: 0.970 for 3, 0.829 for 4, 0.710 for 5, 0.625 for 6.
•
Reject the suspect reading only when Q is above Qcrit, and test only one reading per data set.
Step-by-step
WE KNOW
xq = 10.45, xₙ = 10.18, xₕᵢ = 10.45, xₗₒ = 10.09, Qcrit = 0.71
FIND
gap to the nearest (gap), range (w), Q calculated (Q)
Step 1 · Find gap to the nearest (gap)
Use gap = |xq − xₙ|
The gap is how far the suspect reading stands from its nearest neighbor.
gap = |10.45 − 10.18|
gap = |0.27|
gap = 0.27
Step 2 · Find range (w)
Use w = xₕᵢ − xₗₒ
The range is the spread of all the readings, the suspect one included.
w
=
10.45
−
10.09
w
=
0.36
Step 3 · Find Q calculated (Q)
Use Q = gap ÷ w
Q
is
the
gap
as
a
share
of
the
range:
near
1,
the
suspect
reading
stands
far
from
the
rest.
Q
=
0.27
0.36
Q
=
0.75
→
Q
>
Qcrit
=
0.71:
reject
the
suspect
reading
at
95%.
Check
Put the numbers back in. Both sides should match.
0.27 = |10.45 − 10.18| ✓
0.36 = 10.45 − 10.09 ✓
0.75
=
0.27
0.36
✓
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Comparing a mean with a known value (t-test)
Topic check
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Related lessons
Error analysis and statistics
Main lesson
Propagating uncertainty into a concentration
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Comparing a mean with a known value (t-test)
Use this for “A standard certified at 5.00 mg/L lead reads 5.21 mg/L on average over 6 runs, s = 0.18 mg/L. Is the method biased at 95%?”