SME
Grade 12 Math · Problem type
Sampling distribution of a proportion
Use this for “40% of voters agree. In a sample of 150, what is P(p̂ > 0.46)?”
Chart
The curve is centered at p = 0.4 with standard error SE = 0.04.
P(X ≥ 0.46) = 0.0668.
This is an example. Type your own values to start a new problem.
SE
=
(1
−
)
n
z
=
−
SE
P
Chance the sample proportion is more than p̂
calculated
Formulas
SE
=
√(
p
×
1 −
p
n
)
0.04
=
√(0.4
×
1 − 0.4
150
)
z
=
p̂
−
p
SE
1.5
=
0.46 − 0.4
0.04
P
=
1
−
Φ(
z
)
0.06681 = 1 − Φ(1.5)
Assumptions
•
The sample proportions p̂ center on the population proportion p.
•
They spread by √(p(1 − p) ÷ n) and are close to normal when np ≥ 10 and n(1 − p) ≥ 10.
•
Samples are random and less than 10% of the population.
Step-by-step
WE KNOW
p = 0.4, n = 150, p̂ = 0.46
FIND
standard error of p̂ (SE), z-score of p̂ (z), chance the sample proportion is more than p̂ (P)
Step 1 · Find standard error of p̂ (SE)
Use SE = √(p × (1 − p) ÷ n)
Proportions
of
n
people
spread
by
√(
p
(1 −
p
)
n
).
SE
=
√(0.4
×
1 − 0.4
150
)
SE
=
√(0.4
×
0.6
150
)
SE
=
√(
0.24
150
)
SE
=
√
0.0016
SE = 0.04
Step 2 · Find z-score of p̂ (z)
Use z = (p̂ − p) ÷ SE
How
many
standard
errors
p̂
is
from
p
.
z
=
0.46 − 0.4
0.04
z
=
0.06
0.04
z
=
1.5
Step 3 · Find chance the sample proportion is more than p̂ (P)
Use P = 1 − Φ(z)
The
whole
area
is
1;
take
away
the
area
left
of
z
.
P
=
1
−
Φ(1.5)
P
=
0.06681
→
at
most
p̂
:
1
−
0.06681
=
0.93319
Check
Put the numbers back in. Both sides should match.
0.04
=
√(0.4
×
1 − 0.4
150
)
✓
1.5
=
0.46 − 0.4
0.04
✓
0.06681 = 1 − Φ(1.5) ✓
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Sampling distributions and the central limit theorem
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Mean and spread of a binomial count
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