SME
Grade 11 Math · Problem type
Margin of error of a sample proportion
Use this for “In a random sample of 400, 60% said yes. What is the margin of error?”
Chart
The curve is centered at p̂ = 0.6 with standard error SE = 0.02449.
Interval: 0.6 ± 0.049, from 0.551 to 0.649, the middle 0.9545 of the curve.
This is an example. Type your own values to start a new problem.
E
=
2
(1
−
)
n
SE
Standard error
worked out (not typed)
L
Lower end
worked out (not typed)
U
Upper end
worked out (not typed)
Formulas
SE
=
√(
p̂
×
1 −
p̂
n
)
0.02449
=
√(0.6
×
1 − 0.6
400
)
E
=
2
×
SE
0.04899 = 2 × 0.02449
L
=
p̂
−
E
0.551 = 0.6 − 0.04899
U
=
p̂
+
E
0.649 = 0.6 + 0.04899
Assumptions
•
The sample is random, and n is large enough that the sample proportions are about normal.
•
The standard error of p̂ is √(p̂(1 − p̂) ÷ n).
•
About 2 standard errors each way cover 95% of samples: that is the margin of error.
Step-by-step
WE KNOW
p̂ = 0.6, n = 400
FIND
standard error (SE), margin of error (E), lower end (L), upper end (U)
Step 1 · Find standard error (SE)
Use SE = √(p̂ × (1 − p̂) ÷ n)
How
much
p̂
varies
from
sample
to
sample
of
size
n
.
SE
=
√(0.6
×
1 − 0.6
400
)
SE
=
√(0.6
×
0.4
400
)
SE
=
√(
0.24
400
)
SE
=
√
0.0006
SE = 0.024495
Step 2 · Find margin of error (E)
Use E = 2 × SE
About 95% of samples land within 2 standard errors.
E
=
2
×
0.024495
E
=
0.04899
Step 3 · Find lower end (L)
Use L = p̂ − E
Go down the margin from the estimate.
L
=
0.6
−
0.04899
L
=
0.551
Step 4 · Find upper end (U)
Use U = p̂ + E
Go up the margin from the estimate.
U
=
0.6
+
0.04899
U
=
0.649
Check
Put the numbers back in. Both sides should match.
0.024495
≈
√(0.6
×
1 − 0.6
400
)
✓
0.04899 = 2 × 0.024495 ✓
0.551 ≈ 0.6 − 0.04899 ✓
0.649 ≈ 0.6 + 0.04899 ✓
Topic check
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