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Grade 12 Math · Statistics

Inference for the slope of a regression line

Use this for “A slope of 0.8 has a standard error of 0.25 from 20 points. Is the slope significant?”

Chart

−0.75−0.5−0.2500.250.50.75−3−2−10123zSample slope b if β = 0
The curve is centered at μ = 0 with standard error SEb = 0.25.
b = 0.8 is 3.2 standard deviations above the mean: z = 3.2.
This is an example. Type your own values to start a new problem.
df
=
n
−
2
t
=
b
SEb
Hₐ
Alternative hypothesis
example
P
p-value
worked out (not typed)
α
Significance level
example

Formulas

df
=
n
−
2
18 = 20 − 2
t
=
b
SEb
3.2
=
0.8
0.25
P
=
the
t
tail
past
t
on
Hₐ’s
side,
with
df
degrees
of
freedom
0.004962
=
the
t
tail
past
3.2
on
β
≠
0’s
side,
with
18
degrees
of
freedom

Assumptions

•
H₀: β = 0 (no linear relationship); Hₐ is β ≠ 0, or β > 0 (β < 0) for a positive (negative) one.
•
t = b ÷ SEb follows a t curve with df = n − 2.
•
The points scatter evenly about a straight line, with residuals close to normal and independent.

Step-by-step

WE KNOW
b = 0.8, SEb = 0.25, n = 20, Hₐ: β ≠ 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 = 20 − 2
df = 18
Step 2 · Find test statistic (t)
Use t = b ÷ SEb
How many standard errors the sample slope is from 0, the slope H₀ claims.
t
=
0.8
0.25
t
=
3.2
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(|3.2|,
18))
P
=
2
×
(1
−
0.997519)
P
=
2
×
0.002481
P
=
0.004962
→
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.
18 = 20 − 2 ✓
3.2
=
0.8
0.25
✓
0.004962 = 2 × (1 − tcdf(|3.2|, 18)) ✓
Topic check
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