SME
Grade 12 Math · Statistics
Confidence intervals for a mean and a proportion
Use this for “A sample of 64 has mean 52 and σ = 8. Find a 95% confidence interval for μ.”
Chart
The curve is centered at x̄ = 52 with standard error SE = 1.
Interval: 52 ± 1.96, from 50.04 to 53.96, the middle 0.9500 of the curve.
This is an example. Type your own values to start a new problem.
x̄
±
z⋆
×
σ
n
=
(
L
,
U
)
C
Confidence level
example
SE
Standard error
calculated
E
Margin of error
calculated
Formulas
z⋆
=
invNorm(1
−
1 −
C
2
)
1.96
=
invNorm(1
−
1 − 0.95
2
)
SE
=
σ
÷
√
n
1
=
8
÷
√
64
E
=
z⋆
×
SE
1.96 = 1.96 × 1
L
=
x̄
−
E
50.04 = 52 − 1.96
U
=
x̄
+
E
53.96 = 52 + 1.96
Assumptions
•
A random sample; σ is known (with σ unknown a t interval is used instead).
•
x̄ is close to normal: the population is normal or n ≥ 30.
•
The interval estimates the population mean μ, not where single values fall.
Step-by-step
WE KNOW
x̄ = 52, σ = 8, n = 64, C = 0.95
FIND
critical value (z⋆), standard error (SE), margin of error (E), lower end (L), upper end (U)
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 standard error (SE)
Use SE = σ ÷ √n
Means
of
n
values
spread
less
than
single
values:
divide
σ
by
the
root
of
n
.
SE
=
8
÷
√
64
SE
=
8
8
SE = 1
Step 3 · Find margin of error (E)
Use E = z⋆ × SE
The margin of error is z⋆ standard errors.
E
=
1.96
×
1
E
=
1.96
Step 4 · Find lower end (L)
Use L = x̄ − E
Go down from the estimate by the margin.
L
=
52
−
1.96
L
=
50.04
Step 5 · Find upper end (U)
Use U = x̄ + E
Go up from the estimate by the margin.
U
=
52
+
1.96
U
=
53.96
Check
Put the numbers back in. Both sides should match.
1.96
=
invNorm(1
−
1 − 0.95
2
)
✓
1
=
8
÷
√
64
✓
1.96 = 1.96 × 1 ✓
50.04 = 52 − 1.96 ✓
53.96 = 52 + 1.96 ✓
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Sample means and sums with any σ
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Confidence interval for a mean, σ unknown
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