SME
Grade 12 Math · Problem type
Is the correlation significant?
Use this for “r = 0.6 for 18 pairs of data. Is the correlation significant at 0.05?”
Chart
t = 3 is 3 standard deviations above the mean: z = 3.
This is an example. Type your own values to start a new problem.
r
Correlation coefficient
example
n
Number of data points
example
df
Degrees of freedom
worked out (not typed)
t
Test statistic
worked out (not typed)
Hₐ
Alternative hypothesis
example
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
df
=
n
−
2
16 = 18 − 2
t
=
r
×
√df
÷
√
1 −
r
2
3
=
0.6
×
√
16
÷
√
1 −
0.6
2
P
=
the
t
tail
past
t
on
Hₐ’s
side,
with
df
degrees
of
freedom
0.008479
=
the
t
tail
past
3
on
ρ
≠
0’s
side,
with
16
degrees
of
freedom
Assumptions
•
H₀: ρ = 0 (no linear relationship in the population); Hₐ is ρ ≠ 0, ρ > 0 or ρ < 0.
•
t = r√(n − 2) ÷ √(1 − r²), with df = n − 2: the same t as the slope test.
•
The pairs are a random sample and the scatter follows a straight-line pattern.
Step-by-step
WE KNOW
r = 0.6, n = 18, Hₐ: ρ ≠ 0, α = 0.05
FIND
degrees of freedom (df), test statistic (t), p-value (P)
Step 1 · Find degrees of freedom (df)
Use df = n − 2
The fitted line uses up two degrees of freedom, its slope and its intercept.
df = 18 − 2
df = 16
Step 2 · Find test statistic (t)
Use t = r × √df ÷ √(1 − r²)
A
strong
r
or
many
points
make
t
large;
√
1 −
r
2
shrinks
as
r
nears
±1.
t
=
0.6
×
√
16
÷
√
1 −
0.6
2
t
=
0.6
×
√
16
÷
√
1 − 0.36
t
=
0.6
×
√
16
÷
√
0.64
t
=
0.6
×
4
0.8
t
=
2.4
0.8
t
=
3
Step 3 · Find p-value (P)
Use P = the t tail past t on Hₐ’s side, with df degrees of freedom
Hₐ
≠
counts
both
tails
past
|
t
|;
Hₐ
>
the
area
right
of
t
;
Hₐ
<
the
area
left
of
t
(tcdf
is
the
area
left).
P
=
2
×
(1
−
tcdf(|
t
|,
df))
P
=
2
×
(1
−
tcdf(|3|,
16))
P
=
2
×
(1
−
0.99576)
P
=
2
×
0.00424
P
=
0.008479
→
Two
tails:
below
α
=
0.05,
so
reject
H₀:
convincing
evidence
of
a
linear
relationship
between
x
and
y
Check
Put the numbers back in. Both sides should match.
16 = 18 − 2 ✓
3
=
0.6
×
√
16
÷
√
1 −
0.6
2
✓
0.008479 = 2 × (1 − tcdf(|3|, 16)) ✓
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