SME
Grade 12 Math · Problem type
Sample size for a margin of error
Use this for “How many people must be asked for a 95% margin of 3 points?”
Chart
The curve is centered at p = 0.5 with standard error SE = 0.0153.
Interval: 0.5 ± 0.03, from 0.47 to 0.53, the middle 0.9501 of the curve.
This is an example. Type your own values to start a new problem.
n
≥
z⋆
E
2
×
p
(1
−
p
)
C
Confidence level
example
SE
Standard error with that n
worked out (not typed)
Formulas
z⋆
=
invNorm(1
−
1 −
C
2
)
1.96
=
invNorm(1
−
1 − 0.95
2
)
n
=
⌈z⋆²
×
p
×
1 −
p
E
2
⌉
1,068
=
⌈
1.96
2
×
0.5
×
1 − 0.5
0.03
2
⌉
SE
=
√(
p
×
1 −
p
n
)
0.0153
=
√(0.5
×
1 − 0.5
1,068
)
Assumptions
•
The margin is E = z⋆√(p(1 − p) ÷ n); solve it for n.
•
With no earlier estimate, use p = 0.5: it gives the largest n, so the margin is met whatever p is.
•
Round n up: rounding down would make the margin a little too wide.
Step-by-step
WE KNOW
E = 0.03, C = 0.95, p = 0.5
FIND
critical value (z⋆), sample size needed (n), standard error with that n (SE)
Step 1 · Find critical value (z⋆)
Use z⋆ = invNorm(1 − (1 − C) ÷ 2)
The
middle
C
of
the
curve
is
inside
±z⋆,
so
half
of
the
rest,
1
−
C
,
sits
in
each
tail.
z⋆
=
invNorm(1
−
1 − 0.95
2
)
z⋆ = 1.96
Step 2 · Find sample size needed (n)
Use n = ⌈z⋆² × p × (1 − p) ÷ E²⌉
Square
E
=
z⋆√(
p
(1 −
p
)
n
)
and
solve
for
n
,
then
round
up
to
a
whole
person.
n
=
⌈
1.96
2
×
0.5
×
1 − 0.5
0.03
2
⌉
1.96
2
×
0.5
×
0.5
0.03
2
=
1,067.0719,
rounded
up
to
1,068
n
=
1,068
Step 3 · Find standard error with that n (SE)
Use SE = √(p × (1 − p) ÷ n)
The standard error with that many people: z⋆ standard errors fit within the margin.
SE
=
√(0.5
×
1 − 0.5
1,068
)
SE
=
√(0.5
×
0.5
1,068
)
SE
=
√(
0.25
1,068
)
SE
=
√
0.0002341
SE = 0.0153
Check
Put the numbers back in. Both sides should match.
1.96
=
invNorm(1
−
1 − 0.95
2
)
✓
1,068
=
⌈
1.96
2
×
0.5
×
1 − 0.5
0.03
2
⌉
✓
0.0153
≈
√(0.5
×
1 − 0.5
1,068
)
✓
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