© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:

Conclusion: We have found ________________ evidence at the _____ level of significance that

© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:

Conclusion: We have found ________________ evidence at the _____ level of significance that

.05

.05

© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:

Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches.

.05

.05

μ ≠ .25

μ = .25

© Buddy Freeman, 2015

level of data?

Sign Test*pp. 631-634

ordinal

Parameter ?

meanor

median

proportion

varianceor

standard deviation Normal

population?

yes

no?

chi-square (df = n-1)pp. 336-344

t with df = n-1pp. 288-294

Wilcoxon Signed-Ranks*pp. 610-614(assumes population is symmetric)

known?

yes

Normalpopulation

?

no

Normalpopulation

?

yes

noyesn > 30

?

yes

no

Z using σpp. 277-284

1 Group Flowchart

at leastinterval

np > 5and

n(1- p) > 5?

Binomial/Hypergeometric

yes

no

Zpp. 294-298

Critical Value(s)Table

Z-table

WSRTable

t-table

Sign Table

Z-table

BinomialTable

Chi-squareTable

Wald-WolfowitzOne-Sample RunsTest for Randomnesspp. 634-638

2

3

4

5

6

7

1

* means coverage is different from text.

non > 30

?

yes

no

Jaggia and Kelly(1st edition)

Default case

© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches.

.05

.05

μ ≠ .25

μ = .25

n

S

Xtn 0

1

© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches.

t = 2.00030

.05

.05

μ ≠ .25

μ = .25

Do not reject H0 Reject H0Reject H0 .025.025

df = n – 1 = 60

t = -2.0003

n

S

Xtn 0

1

© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If -2.0003 < tcomputed < 2.0003

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches.

.05

.05

μ ≠ .25

μ = .25

t = 2.00030

Do not reject H0 Reject H0Reject H0 .025.025

df = n – 1 = 60

t = -2.0003

n

S

Xtn 0

1

© Buddy Freeman, 2015

3 Steps to standard deviation

1. Calculate the variation of the sample, SS.

2. Calculate the variance of the sample, S2.

3. Calculate the standard deviation of the sample, S.

2130009130567.

61

3101.1584352227.3

)( 222

n

xxSS

1202170000152176.

60

2130009130567.

1

2

n

SSS

003900976.1202170000152176.2 SS

© Buddy Freeman, 2015

H0:

H1:

α =

Decision Rule:

If -2.0003 < tcomputed < 2.0003

then do not reject H0, otherwise reject H0.

Test Statistic:

Decision:Conclusion: We have found ________________ evidence at the _____ level of significance that the true average diameter of the ball bearings produced differs from .25 inches.

.05

.05

µ ≠ .25

μ = .25

Do not reject H0.insufficient

t = 2.00030

Reject H0Reject H0 .025.025

df = n – 1 = 60

t = -2.0003

Do not reject H0Do not reject H0

9726.1

61

003900976.

25.613101.15

01

n

S

Xtn