SME
Grade 12 Math · Problem type
Confidence interval for a proportion
Use this for “240 of 400 people said yes. Find a 95% confidence interval for p.”
Chart
The curve is centered at p̂ = 0.6 with standard error SE = 0.02449.
Interval: 0.6 ± 0.048, from 0.552 to 0.648, the middle 0.9500 of the curve.
This is an example. Type your own values to start a new problem.
p̂
=
k
n
p̂
±
z⋆
×
(1
−
)
n
=
(
L
,
U
)
C
Confidence level
example
SE
Standard error
worked out (not typed)
E
Margin of error
calculated
Formulas
p̂
=
k
n
0.6
=
240
400
z⋆
=
invNorm(1
−
1 −
C
2
)
1.96
=
invNorm(1
−
1 − 0.95
2
)
SE
=
√(
p̂
×
1 −
p̂
n
)
0.02449
=
√(0.6
×
1 − 0.6
400
)
E
=
z⋆
×
SE
0.04801 = 1.96 × 0.02449
L
=
p̂
−
E
0.552 = 0.6 − 0.04801
U
=
p̂
+
E
0.648 = 0.6 + 0.04801
Assumptions
•
A random sample with at least 10 successes and 10 failures, so p̂ is close to normal.
•
The standard error uses p̂, since the true p is what the interval estimates.
•
The interval estimates the population proportion p.
Step-by-step
WE KNOW
k = 240, n = 400, C = 0.95
FIND
sample proportion (p̂), critical value (z⋆), standard error (SE), margin of error (E), lower end (L), upper end (U)
Step 1 · Find sample proportion (p̂)
Use p̂ = k ÷ n
The share of the sample that are successes.
p̂
=
240
400
p̂
=
0.6
Step 2 · 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⋆ = 1.95996
Step 3 · Find standard error (SE)
Use SE = √(p̂ × (1 − p̂) ÷ n)
The
spread
of
p̂
from
sample
to
sample,
estimated
with
p̂
itself.
SE
=
√(0.6
×
1 − 0.6
400
)
SE
=
√(0.6
×
0.4
400
)
SE
=
√(
0.24
400
)
SE
=
√
0.0006
SE = 0.024495
Step 4 · Find margin of error (E)
Use E = z⋆ × SE
The margin of error is z⋆ standard errors.
E
=
1.95996
×
0.024495
E
=
0.04801
Step 5 · Find lower end (L)
Use L = p̂ − E
Go down from the estimate by the margin.
L
=
0.6
−
0.04801
L
=
0.552
Step 6 · Find upper end (U)
Use U = p̂ + E
Go up from the estimate by the margin.
U
=
0.6
+
0.04801
U
=
0.648
Check
Put the numbers back in. Both sides should match.
0.6
=
240
400
✓
1.95996
=
invNorm(1
−
1 − 0.95
2
)
✓
0.024495
≈
√(0.6
×
1 − 0.6
400
)
✓
0.04801 ≈ 1.95996 × 0.024495 ✓
0.552 ≈ 0.6 − 0.04801 ✓
0.648 ≈ 0.6 + 0.04801 ✓
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