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Grade 12 Math · Problem type

Confidence interval for the slope

Use this for “b = 0.8 and SEb = 0.25 from 20 points. Find a 95% confidence interval for the slope.”

Chart

0.050.30.550.81.051.31.55−3−2−10123zSample slope b0.96440.27481.3252
The curve is centered at b = 0.8 with standard error SEb = 0.25.
Interval: 0.8 ± 0.5252, from 0.2748 to 1.3252, the middle 0.9644 of the curve.
This is an example. Type your own values to start a new problem.
b
±
t⋆
×
SEb
=
(
L
,
U
)
n
Number of data points
example
C
Confidence level
example
df
Degrees of freedom
worked out (not typed)
E
Margin of error
worked out (not typed)

Formulas

df
=
n
−
2
18 = 20 − 2
t⋆
=
invT(1
−
1 −
C
2,
df)
2.1009
=
invT(1
−
1 − 0.95
2,
18)
E
=
t⋆
×
SEb
0.5252 = 2.1009 × 0.25
L
=
b
−
E
0.2748 = 0.8 − 0.5252
U
=
b
+
E
1.3252 = 0.8 + 0.5252

Assumptions

•
The interval is b ± t⋆ × SEb, with t⋆ from the t curve with df = n − 2.
•
An interval that doesn’t hold 0 means the slope is not 0 at that level.
•
The same conditions as the test: a straight-line pattern, even scatter, close to normal residuals.

Step-by-step

WE KNOW
b = 0.8, SEb = 0.25, n = 20, C = 0.95
FIND
degrees of freedom (df), critical value (t⋆), margin of error (E), lower end (L), upper end (U)
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 critical value (t⋆)
Use t⋆ = invT(1 − (1 − C) ÷ 2, df)
The
middle
C
of
the
t
curve
lies
within
±t⋆,
so
the
area
left
of
t⋆
is
1
−
1 −
C
2
.
t⋆ = invT(0.975, df)
t⋆ = invT(0.975, 18)
t⋆ = 2.1009
Step 3 · Find margin of error (E)
Use E = t⋆ × SEb
The margin of error is t⋆ standard errors.
E
=
2.1009
×
0.25
E
=
0.5252
Step 4 · Find lower end (L)
Use L = b − E
Go down from the estimate by the margin.
L
=
0.8
−
0.5252
L
=
0.2748
Step 5 · Find upper end (U)
Use U = b + E
Go up from the estimate by the margin.
U
=
0.8
+
0.5252
U
=
1.3252
→
0
is
not
between
0.2748
and
1.3252:
the
slope
differs
from
0
at
the
95%
level
Check
Put the numbers back in. Both sides should match.
18 = 20 − 2 ✓
2.1009 = invT(0.975, 18) ✓
0.5252 ≈ 2.1009 × 0.25 ✓
0.2748 = 0.8 − 0.5252 ✓
1.3252 = 0.8 + 0.5252 ✓
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