HP 50g Graphing Calculator User Manual

Page 623

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Page 18-56

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. To check if that should be the case, we test the null
hypothesis, H

0

:

Α = 0, against the alternative hypothesis, H

1

:

Α ≠ 0, at the level

of significance

α = 0.05.

The test statistic is t

0

= (a-

0)/[(1/n)+⎯x

2

/S

xx

]

1/2

= (-0.86)/ [(1/5)+3

2

/2.5]

½

=

-0.44117. The critical value of t, for

ν = n – 2 = 3, and α/2 = 0.025, can be

calculated using the numerical solver for the equation

α = UTPT(γ,t) developed

in Chapter 17. In this program,

γ represents the degrees of freedom (n-2), and

α represents the probability of exceeding a certain value of t, i.e., Pr[ t>t

α

] = 1

α. For the present example, the value of the level of significance is α = 0.05,

g = 3, and t

n-2,

α/2

= t

3,0.025

. Also, for

γ = 3 and α = 0.025, t

n-2,

α/2

= t

3,0.025

= 3.18244630528. Because t

0

> - t

n-2,

α/2

, we cannot reject the null

hypothesis, H

0

:

Α = 0, against the alternative hypothesis, H

1

:

Α ≠ 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.

Example 3 – Test of significance for the linear regression. Test the null
hypothesis for the slope H

0

:

Β = 0, against the alternative hypothesis, H

1

:

Β ≠

0, at the level of significance

α = 0.05, for the linear fitting of Example 1.

The test statistic is t

0

= (b -

Β

0

)/(s

e

/

√S

xx

) = (3.24-0)/(

√0.18266666667/2.5) =

18.95. The critical value of t, for

ν = n – 2 = 3, and α/2 = 0.025, was

obtained in Example 2, as t

n-2,

α/2

= t

3,0.025

= 3.18244630528. Because, t

0

>

t

α/2

, we must reject the null hypothesis H

1

:

Β ≠ 0, at the level of significance α

= 0.05, for the linear fitting of Example 1.

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