HP F2226A - 48GII Graphic Calculator User Manual Page 626

Graphing calculator.

Confidence intervals for the slope (Β) and intercept (A):
First, we obtain t
n-2,
program to solve for t
Next, we calculate the terms
(t
)⋅s
/√S
α
n-2,
/2
e
xx
3.1824...⋅√0.1826...⋅[(1/5)+3
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).
Example 2 --
Suppose that the y-data used in Example 1 represent the
elongation (in hundredths of an inch) of a metal wire when subjected to a
force x (in tens of pounds). The physical phenomenon is such that we expect
the intercept, A, to be zero.
: Α = 0, against the alternative hypothesis, H
null hypothesis, H
0
level of significance α = 0.05.
The test statistic is t
= (a-0)/[(1/n)+x
0
The critical value of t, for ν = n – 2 = 3, and α/2 = 0.025, can
0.44117.
be calculated using the numerical solver for the equation α = UTPT(γ,t)
developed in Chapter 17.
freedom (n-2), and α represents the probability of exceeding a certain value
] = 1 – α.
of t, i.e., Pr[ t>t
α
significance is α = 0.05, g = 3, and t
0.025, t
= t
= 3.18244630528. Because t
α
n-2,
/2
3,0.025
reject the null hypothesis, H
≠ 0, at the level of significance α = 0.05.
This result suggests that taking A = 0 for this linear regression should be
acceptable. After all, the value we found for a, was –0.86, which is relatively
close to zero.
= t
,
= 3.18244630528 (See chapter 17 for a
α
/2
3
0.025
):
ν
,a
= 3.182...⋅(0.1826.../2.5)
2
⋅[(1/n)+x
(t
)⋅s
/S
α
n-2,
/2
e
2
/2.5]
To check if that should be the case, we test the
2
1/2
/S
]
= (-0.86)/ [(1/5)+3
xx
In this program, γ represents the degrees of
For the present example, the value of the level of
= t
α
n-2,
/2
3,0.025
: Α = 0, against the alternative hypothesis, H
0
1/2
= 0.8602...
1/2
]
=
xx
1/2
= 2.65
: Α ≠ 0, at the
1
2
/2.5]
. Also, for γ = 3 and α =
> - t
, we cannot
α
0
n-2,
/2
Page 18-55
½
= -
: Α
1 48gii49g+

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