SME
Analytical Chemistry · Problem type
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%?”
Chart
t curve with df = 4 (solid) over the normal (dashed): lower in the middle, heavier in the tails.
The
middle
0.9500
lies
between
t
=
−2.776
and
2.776:
t⋆
=
2.776
for
95%.
x̄
±
t⋆s
÷
√n
=
10.12
±
2.776
×
0.15
÷
√
5
=
10.12
±
0.1862.
The interval runs from 9.934 to 10.31. t = 1.789 < t⋆ = 2.776: μ = 10 is inside the interval, no significant difference at 95%.
This is an example. Type your own values to start a new problem.
t
=
|
x̄ (mg/L)
−
μ (mg/L)
|
n
s (mg/L)
df
Degrees of freedom
worked out (not typed)
t⋆
Critical t at 95%
worked out (not typed)
Formulas
t
=
|
x̄
−
μ
|
÷
(
s
÷
√
n
)
1.789
=
|10.12
−
10|
÷
(0.15
÷
√
5
)
df
=
n
−
1
4 = 5 − 1
t
⋆
=
invT(0.975,
df)
2.7764 = invT(0.975, 4)
Assumptions
•
Random error only; μ is the known (certified) value.
•
Two-sided at 95%: the difference is significant when t > t⋆, the same as μ lying outside x̄ ± t⋆s ÷ √n.
•
No significant difference does not prove there is no bias; more readings can still find one.
Step-by-step
WE KNOW
x̄ = 10.12 mg/L, μ = 10 mg/L, s = 0.15 mg/L, n = 5
FIND
t calculated (t), degrees of freedom (df), critical t at 95% (t⋆)
Step 1 · Find t calculated (t)
Use t = |x̄ − μ| ÷ (s ÷ √n)
How
many
standard
errors
(
s
÷
√
n
)
the
mean
lies
from
the
known
value.
t
=
|10.12
−
10|
÷
(0.15
÷
√
5
)
t
=
|0.12|
÷
(
0.15
2.236
)
t
=
|0.12|
÷
0.06708
t
=
0.12
0.06708
t
=
1.789
Step 2 · Find degrees of freedom (df)
Use df = n − 1
The
mean
used
up
one
piece
of
the
data,
so
s
has
n
−
1
degrees
of
freedom.
df = 5 − 1
df = 4
Step 3 · Find critical t at 95% (t⋆)
Use t⋆ = invT(0.975, df)
For
95%
two-sided,
2.5%
is
left
in
each
tail,
so
t
⋆
has
0.975
of
the
t
curve
to
its
left.
t
⋆
=
invT(0.975,
4)
t
⋆
=
2.7764
→
t
=
1.789
≤
t
⋆:
no
significant
difference
at
95%.
Check
Put the numbers back in. Both sides should match.
1.789
=
|10.12
−
10|
÷
(0.15
÷
√
5
)
✓
4 = 5 − 1 ✓
2.7764 = invT(0.975, 4) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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