SME
Grade 12 Math · Problem type
One-mean z-test
Use this for “Is the mean less than μ₀?” (or more than, or different from) with σ known.
Chart
The curve is centered at μ₀ = 500 g with standard error SE = 2 g.
Hₐ uses <: the left tail counts.
Red outline: the rejection region, area α = 0.05, past z = −1.6449.
Test statistic z = −2.
Amber: the p-value, 0.0228 < α = 0.05: reject H₀.
This is an example. Type your own values to start a new problem.
Hₐ:
μ
μ₀
SE (g)
=
σ (g)
n
z
=
x̄ (g)
−
μ₀ (g)
SE (g)
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
SE
=
σ
÷
√
n
2
=
12
÷
√
36
z
=
x̄
−
μ₀
SE
−2
=
496 − 500
2
P
=
the
tail
past
z
on
Hₐ’s
side
0.02275 = the tail past −2 on <’s side
Assumptions
•
H₀: μ = μ₀; Hₐ says μ < μ₀, μ > μ₀ or μ ≠ μ₀ (tap the sign in Hₐ).
•
The sample is random, σ is known, and x̄ is close to normal (a normal population or n ≥ 30).
•
Reject H₀ when the p-value is below α.
Step-by-step
WE KNOW
μ₀ = 500 g, Hₐ: μ < μ₀, σ = 12 g, n = 36, x̄ = 496 g, α = 0.05
FIND
standard error (SE), test statistic (z), p-value (P)
Step 1 · 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
=
12
÷
√
36
SE
=
12
6
SE = 2 g
Step 2 · Find test statistic (z)
Use z = (x̄ − μ₀) ÷ SE
How many standard errors the estimate is from the value H₀ claims.
z
=
496 − 500
2
z
=
−
4
2
z
=
−2
Step 3 · Find p-value (P)
Use P = the tail past z on Hₐ’s side
Hₐ
<
takes
the
area
left
of
z
;
Hₐ
>
the
area
right
of
z
;
Hₐ
≠
both
tails
past
|
z
|.
P
=
Φ(
z
)
P
=
Φ(−2)
P
=
0.02275
→
0.02275
<
α
=
0.05:
reject
H₀
Check
Put the numbers back in. Both sides should match.
2
=
12
÷
√
36
✓
−2
=
496 − 500
2
✓
0.02275 = Φ(−2) ✓
Topic check
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