# HP 50g User Manual Page 623

Graphing calculator.

Example 2 --
elongation (in hundredths of an inch) of a metal wire when subjected to a force
x (in tens of pounds).
intercept, A, to be zero.
hypothesis, H
:
0
of significance
The test statistic is t
-0.44117. The critical value of t, for
calculated using the numerical solver for the equation
in Chapter 17. In this program, represents the degrees of freedom (n-2), and
represents the probability of exceeding a certain value of t, i.e., Pr[ t>t ] = 1
– . For the present example, the value of the level of significance is
g = 3, and t
n-2, /2
= 3.18244630528. Because t
hypothesis, H
:
0
of significance
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.
Example 3 – Test of significance for the linear regression. Test the null
hypothesis for the slope H
0, at the level of significance
The test statistic is t
18.95. The critical value of t, for
obtained in Example 2, as t
t
, we must reject the null hypothesis H
/2
= 0.05, for the linear fitting of Example 1.
Suppose that the y-data used in Example 1 represent the
The physical phenomenon is such that we expect the
To check if that should be the case, we test the null
= , against the alternative hypothesis, H
= 0.05.
= (a- )/[(1/n)+ x
0
= t
. Also, for = 3 and
3,0.025
0
= , against the alternative hypothesis, H
= 0.05.
:
= , against the alternative hypothesis, H
0
= 0.05, for the linear fitting of Example 1.
= (b -
)/(s
0
0
n-2, /2
2
1/2
/S
]
= (-0.86)/ [(1/5)+3
xx
= n – 2 = 3, and /2 = 0.025, can be
> - t
, we cannot reject the null
n-2, /2
/ S
) = (3.24-0)/(
e
xx
= n – 2 = 3, and /2 = 0.025, was
= t
= 3.18244630528. Because, t
3,0.025
:
0, at the level of significance
1
:
0, at the level
1
2
= UTPT( ,t) developed
= 0.025, t
n-2, /2
:
0, at the level
1
.18266666667/2.5) =
½
/2.5]
=
= 0.05,
= t
3,0.025
:
1
>
0
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