SME
Analytical Chemistry
Error analysis and statistics
Use this for “Six readings of nitrate in a well-water sample average 8.42 mg/L with s = 0.21 mg/L. Find the 95% confidence interval for the mean.”
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.
This is an example. Type your own values to start a new problem.
x̄ (mg/L)
±
t⋆
×
s (mg/L)
n
df
Degrees of freedom
worked out (not typed)
Δ
Half-width
calculated · mg/L
L
Lower end
calculated · mg/L
U
Upper end
calculated · mg/L
Formulas
df
=
n
−
1
4 = 5 − 1
t⋆ = invT(0.975, df)
2.7764 = invT(0.975, 4)
Δ
=
t⋆
×
s
÷
√
n
0.1862
=
2.7764
×
0.15
÷
√
5
L
=
x̄
−
Δ
9.9338 = 10.12 − 0.1862
U
=
x̄
+
Δ
10.3062 = 10.12 + 0.1862
Assumptions
•
Random error only: the readings scatter normally about the true mean, with no bias.
•
The interval is where the true mean lies with 95% confidence, not where 95% of the readings lie.
•
s comes from the same n readings, so Student’s t with n − 1 degrees of freedom replaces z.
Step-by-step
WE KNOW
n = 5, x̄ = 10.12 mg/L, s = 0.15 mg/L
FIND
degrees of freedom (df), critical t at 95% (t⋆), half-width (Δ), lower end (L), upper end (U)
Step 1 · 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 2 · 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
Step 3 · Find half-width (Δ)
Use Δ = t⋆ × s ÷ √n
The
standard
error
of
a
mean
is
s
÷
√
n
,
and
the
half-width
is
t⋆
of
them.
Δ
=
2.7764
×
0.15
÷
√
5
Δ
=
2.7764
×
0.15
2.2361
Δ
=
0.41646
2.2361
Δ
=
0.1862
mg/
L
Step 4 · Find lower end (L)
Use L = x̄ − Δ
The interval starts one half-width below the mean.
L
=
10.12
−
0.1862
L
=
9.9338
mg/
L
Step 5 · Find upper end (U)
Use U = x̄ + Δ
The interval ends one half-width above the mean.
U
=
10.12
+
0.1862
U
=
10.3062
mg/
L
Check
Put the numbers back in. Both sides should match.
4 = 5 − 1 ✓
2.7764 = invT(0.975, 4) ✓
0.1862
=
2.7764
×
0.15
÷
√
5
✓
9.9338 = 10.12 − 0.1862 ✓
10.3062 = 10.12 + 0.1862 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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