SME
Measurements & Instrumentation · Problem type
Mean, standard deviation and a confidence interval
Use this for “Eight readings average 12.46 mm with s = 0.18 mm. Give the 95% interval for the true value.”
Chart
t curve with df = 7 (solid) over the normal (dashed): lower in the middle, heavier in the tails.
The
middle
0.9500
lies
between
t
=
−2.365
and
2.365:
t⋆
=
2.365
for
95%.
x̄
±
t⋆s
÷
√n
=
12.46
±
2.365
×
0.18
÷
√
8
=
12.46
±
0.1505.
The interval runs from 12.31 to 12.61.
Standard uncertainty s_x̄ = 0.06364 mm
This is an example. Type your own values to start a new problem.
x̄ (mm)
±
t
×
s (mm)
n
s_x̄
Standard uncertainty
calculated · mm
U
Half-width
calculated · mm
xlow
Lower end
worked out (not typed) · mm
xhigh
Upper end
worked out (not typed) · mm
Formulas
t
=
invT(0.975,
n
−
1)
2.3646 = invT(0.975, 8 − 1)
s_x̄
=
s
÷
√
n
0.06364
=
0.18
÷
√
8
U
=
t
×
s_x̄
0.1505 = 2.3646 × 0.06364
xlow
=
x̄
−
U
12.3095 = 12.46 − 0.1505
xhigh
=
x̄
+
U
12.6105 = 12.46 + 0.1505
Assumptions
•
The readings are independent and scatter roughly normally about the true value.
•
s is the sample standard deviation, worked with n − 1 in its denominator.
•
t is the t table’s 95% value at ν = n − 1, worked out from n: 2.365 for 8 readings.
Step-by-step
WE KNOW
n = 8, x̄ = 12.46 mm, s = 0.18 mm
FIND
critical t (t), standard uncertainty (s_x̄), half-width (U), lower end (xlow), upper end (xhigh)
Step 1 · Find critical t (t)
Use t = invT(0.975, n − 1)
The
middle
95%
of
the
t
curve
for
ν
=
n
−
1
lies
within
±
t
:
read
t
from
the
table
at
that
ν.
t
=
invT(0.975,
7)
t
=
2.3646
Step 2 · Find standard uncertainty (s_x̄)
Use s_x̄ = s ÷ √n
Averaging
n
readings
shrinks
the
scatter
by
√
n
,
so
the
mean
is
steadier
than
any
one
reading.
s_x̄
=
0.18
÷
√
8
s_x̄
=
0.18
2.828
s_x̄ = 0.06364 mm
Step 3 · Find half-width (U)
Use U = t × s_x̄
For
a
small
sample
the
half-width
is
t
standard
uncertainties,
not
z
=
1.96
of
them.
U
=
2.3646
×
0.06364
U
=
0.1505
mm
Step 4 · Find lower end (xlow)
Use xlow = x̄ − U
The interval starts one half-width below the mean.
xlow = 12.46 − 0.1505
xlow = 12.3095 mm
Step 5 · Find upper end (xhigh)
Use xhigh = x̄ + U
The interval ends one half-width above the mean.
xhigh = 12.46 + 0.1505
xhigh = 12.6105 mm
Check
Put the numbers back in. Both sides should match.
2.3646 = invT(0.975, 7) ✓
0.06364
=
0.18
÷
√
8
✓
0.1505 = 2.3646 × 0.06364 ✓
12.3095 = 12.46 − 0.1505 ✓
12.6105 = 12.46 + 0.1505 ✓
Topic check
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