SME
Grade 12 Math · Problem type
The standard error of the slope
Use this for “The residuals have s = 2, the x values have sₓ = 1.5 and n = 10. Is b = 1.2 significant?”
Chart
The curve is centered at μ = 0 with standard error SEb = 0.4444.
b = 1.2 is 2.7 standard deviations above the mean: z = 2.7.
This is an example. Type your own values to start a new problem.
SEb
=
s
×
−
1
t
=
b
SEb
df
Degrees of freedom
worked out (not typed)
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
SEb
=
s
sₓ ×
√
n
− 1
0.4444
=
2
1.5 ×
√
10 − 1
t
=
b
SEb
2.7
=
1.2
0.4444
df
=
n
−
2
8 = 10 − 2
P
=
2
×
(1
−
tcdf(|
t
|,
df))
0.02707 = 2 × (1 − tcdf(|2.7|, 8))
Assumptions
•
SEb = s ÷ (sₓ√(n − 1)): s is the spread of the residuals, sₓ the spread of the x values.
•
Points spread widely in x pin the slope down well, so SEb is small.
•
Then t = b ÷ SEb with df = n − 2, as in the slope test.
Step-by-step
WE KNOW
s = 2, sₓ = 1.5, n = 10, b = 1.2, α = 0.05
FIND
standard error of the slope (SEb), test statistic (t), degrees of freedom (df), p-value (P)
Step 1 · Find standard error of the slope (SEb)
Use SEb = s ÷ (sₓ × √(n − 1))
The
residuals’
spread
over
the
x
values’
spread,
shrinking
as
n
grows.
SEb
=
2
1.5 ×
√
10 − 1
SEb
=
2
1.5 ×
√
9
SEb
=
2
1.5 × 3
SEb
=
2
4.5
SEb = 0.4444
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
=
1.2
0.4444
t
=
2.7
Step 3 · Find degrees of freedom (df)
Use df = n − 2
The fitted line uses up two degrees of freedom, its slope and its intercept.
df = 10 − 2
df = 8
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(|2.7|,
8))
P
=
2
×
(1
−
0.98646)
P
=
2
×
0.01354
P
=
0.02707
→
Hₐ:
β
≠
0,
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.
0.4444
≈
2
1.5 ×
√
10 − 1
✓
2.7
≈
1.2
0.4444
✓
8 = 10 − 2 ✓
0.02707 = 2 × (1 − tcdf(|2.7|, 8)) ✓
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Analysis of variance (ANOVA) and the F distribution
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