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Inferences concerning one variance, Inferences concerning one variance ,18-47 – HP 50g Graphing Calculator User Manual

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Page 18-47

Thus, we accept (more accurately, we do not reject) the hypothesis: H

0

:

μ

1

−μ

2

=

0, or H

0

:

μ

1

2

, against the alternative hypothesis H

1

:

μ

1

−μ

2

<

0, or H

1

:

μ

1

2

.

The test t value is t

0

= -1.341776, with a P-value = 0.09130961, and critical t is

–t

α

= -1.659782. The graphical results are:

These three examples should be enough to understand the operation of the
hypothesis testing pre-programmed feature in the calculator.

Inferences concerning one variance

The null hypothesis to be tested is , H

o

:

σ

2

=

σ

o

2

, at a level of confidence (1-

α)100%, or significance level α, using a sample of size n, and variance s

2

. The

test statistic to be used is a chi-squared test statistic defined as

Depending on the alternative hypothesis chosen, the P-value is calculated as
follows:

Θ H

1

:

σ

2

<

σ

o

2

,

P-value = P(

χ

2

<

χ

o

2

) = 1-UTPC(

ν,χ

o

2

)

Θ H

1

:

σ

2

>

σ

o

2

,

P-value = P(

χ

2

>

χ

o

2

) = UTPC(

ν,χ

o

2

)

Θ H

1

:

σ

2

≠ σ

o

2

,

P-value =2

⋅min[P(χ

2

<

χ

o

2

), P(

χ

2

>

χ

o

2

)] =

2

⋅min[1-UTPC(ν,χ

o

2

), UTPC(

ν,χ

o

2

)]

where the function min[x,y] produces the minimum value of x or y (similarly,
max[x,y] produces the maximum value of x or y). UTPC(

ν,x) represents the

calculator’s upper-tail probabilities for

ν = n - 1 degrees of freedom.

2

0

2

2

)

1

(

σ

χ

s

n

o

=