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1: Covariance: 2.025
These results are interpreted as a = -0.86, b = 3.24, r
xy
= 0.989720229749,
and s
xy
= 2.025. The correlation coefficient is close enough to 1.0 to confirm
the linear trend observed in the graph.
From the
Single-var… option of the ‚Ù menu we find: ⎯x = 3, s
x
=
0.790569415042,⎯y = 8.86, s
y
= 2.58804945857.
Next, with n = 5, calculate
Confidence intervals for the slope (Β) and intercept (A):
Θ First, we obtain t
n-2,α/2
= t
3
,
0.025
= 3.18244630528 (See chapter 17 for
a program to solve for t
ν,a
):
Θ Next, we calculate the terms
(t
n-2,α/2
)⋅s
e
/√S
xx
= 3.182…⋅(0.1826…/2.5)
1/2
= 0.8602…
(t
n-2,α/2
)⋅s
e
⋅[(1/n)+⎯x
2
/S
xx
]
1/2
=
3.1824…⋅√0.1826…⋅[(1/5)+3
2
/2.5]
1/2
= 2.65
Θ Finally, for the slope B, the 95% confidence interval is
(-0.86-0.860242, -0.86+0.860242) = (-1.72, -0.00024217)
For the intercept A, the 95% confidence interval is (3.24-2.6514,
3.24+2.6514) = (0.58855,5.8914).
5.2427905694150.0)15()1(
22
=⋅−=⋅−=
xxx
snS
=−⋅⋅
−
−
= )1(
2
1
222
xyye
rs
n
n
s
...1826.0)...9897.01(...5880.2
25
15
22
=−⋅⋅
−
−