SME
Grade 12 Math · Problem type
One-mean t-test
Use this for “A sample of 16 has mean 496 g and s = 8 g. Is the mean different from 500 g?”
Chart
t curve with df = 15 (solid) over the normal (dashed): lower in the middle, heavier in the tails; t⋆ = 2.131 for 95%, where z⋆ = 1.96.
The curve is centered at μ₀ = 500 g with standard error SE = 2 g.
Red outline: the rejection region, area α = 0.05, past t = −2.1314 and t = 2.1314.
Test statistic t = −2.
Amber: the p-value, 0.0639 ≥ α = 0.05: fail to reject H₀.
This is an example. Type your own values to start a new problem.
SE (g)
=
s (g)
n
t
=
x̄ (g)
−
μ₀ (g)
SE (g)
df
Degrees of freedom
worked out (not typed)
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
df
=
n
−
1
15 = 16 − 1
SE
=
s
÷
√
n
2
=
8
÷
√
16
t
=
x̄
−
μ₀
SE
−2
=
496 − 500
2
P
=
2
×
(1
−
tcdf(|
t
|,
df))
0.06395 = 2 × (1 − tcdf(|−2|, 15))
Assumptions
•
H₀: μ = μ₀ and Hₐ: μ ≠ μ₀; σ is unknown, so the sample’s s stands in for it.
•
With s in place of σ the statistic follows a t curve with df = n − 1, wider in the tails than the normal.
•
The sample is random and the population close to normal (or n ≥ 30).
Step-by-step
WE KNOW
μ₀ = 500 g, s = 8 g, n = 16, x̄ = 496 g, α = 0.05
FIND
degrees of freedom (df), standard error (SE), test statistic (t), p-value (P)
Step 1 · Find degrees of freedom (df)
Use df = n − 1
One less than the sample size: the mean uses up one degree of freedom.
df = 16 − 1
df = 15
Step 2 · Find standard error (SE)
Use SE = s ÷ √n
The
sample’
s
s
stands
in
for
σ:
divide
it
by
the
root
of
n
.
SE
=
8
÷
√
16
SE
=
8
4
SE = 2 g
Step 3 · Find test statistic (t)
Use t = (x̄ − μ₀) ÷ SE
How
many
standard
errors
x̄
is
from
the
value
H₀
claims.
t
=
496 − 500
2
t
=
−
4
2
t
=
−2
Step 4 · Find p-value (P)
Use P = 2 × (1 − tcdf(|t|, df))
Hₐ
says
“not
equal”,
so
both
tails
of
the
t
curve
past
|
t
|
count;
tcdf(
t
,
df)
is
the
area
left
of
t
.
P
=
2
×
(1
−
tcdf(|−2|,
15))
P
=
0.06395
→
0.06395
≥
α
=
0.05:
fail
to
reject
H₀
Check
Put the numbers back in. Both sides should match.
15 = 16 − 1 ✓
2
=
8
÷
√
16
✓
−2
=
496 − 500
2
✓
0.06395 = 2 × (1 − tcdf(|−2|, 15)) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Hypothesis tests for one and two samples
Main lesson
One-mean z-test
Use this for “Is the mean less than μ₀?” (or more than, or different from) with σ known.
Matched pairs t-test
Use this for “Twelve students gained 2.5 points on average after a review, s = 3.2. Did the review change scores?”
Two-proportion z-test
Use this for “84 of 150 people sent a reminder voted, against 66 of 150 who were not. Is there a difference?”
Two-sample t-test for means
Use this for “Do two groups have different means?” from two independent samples.
Which alternative hypothesis?
Use this for “Write H₀ and Hₐ for the claim that the new bulbs last longer.”
Type I or Type II error?
Use this for “Name a Type I and a Type II error for this test.”