SME
Grade 12 Math · Problem type
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?”
Chart
t curve with df = 11 (solid) over the normal (dashed): lower in the middle, heavier in the tails; t⋆ = 2.201 for 95%, where z⋆ = 1.96.
The curve is centered at μd = 0 with standard error SE = 0.9238.
Red outline: the rejection region, area α = 0.05, past t = −2.201 and t = 2.201.
Test statistic t = 2.7063.
Amber: the p-value, 0.0204 < α = 0.05: reject H₀.
This is an example. Type your own values to start a new problem.
SE
=
s
n
t
=
d̄
SE
df
Degrees of freedom
worked out (not typed)
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
df
=
n
−
1
11 = 12 − 1
SE
=
s
÷
√
n
0.9238
=
3.2
÷
√
12
t
=
d̄
SE
2.7063
=
2.5
0.9238
P
=
2
×
(1
−
tcdf(|
t
|,
df))
0.02042 = 2 × (1 − tcdf(|2.7063|, 11))
Assumptions
•
Each subject is measured twice (before and after): work with the differences d, one per pair.
•
H₀: μd = 0 and Hₐ: μd ≠ 0; the differences are a one-mean t-test with df = n − 1.
•
The pairs are random and the differences close to normal (or n ≥ 30).
Step-by-step
WE KNOW
d̄ = 2.5, s = 3.2, n = 12, 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 = 12 − 1
df = 11
Step 2 · Find standard error (SE)
Use SE = s ÷ √n
The
differences’
s
over
the
root
of
the
number
of
pairs.
SE
=
3.2
÷
√
12
SE
=
3.2
3.464
SE = 0.92376
Step 3 · Find test statistic (t)
Use t = d̄ ÷ SE
H₀
says
the
mean
difference
is
0:
count
how
many
standard
errors
d̄
is
from
0.
t
=
2.5
0.92376
t
=
2.7063
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(|2.7063|,
11))
P
=
0.02042
→
0.02042
<
α
=
0.05:
reject
H₀
Check
Put the numbers back in. Both sides should match.
11 = 12 − 1 ✓
0.92376
≈
3.2
÷
√
12
✓
2.7063
≈
2.5
0.92376
✓
0.02042 = 2 × (1 − tcdf(|2.7063|, 11)) ✓
Topic check
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