HP 50G, Руководство пользователя

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Page 18-55
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 . 2 42 7905694150 . 0 ) 1 5 ( ) 1 (
2 2
= ⋅ − = ⋅ − =
x xx
s n S
= − ⋅ ⋅
−
−
= ) 1 (
2
1
2 2 2
xy y e
r s
n
n
s
... 1826 . 0 ) ... 9897 . 0 1 ( ... 5880 . 2
2 5
1 5
2 2
= − ⋅ ⋅
−
−