Chi-square: Exampl-e

Observed Outcome

REGION

Centra] Suburbs Rural_

Cltv

PARTY

Repub l i cans 126 61 38 225 143 .4%)Democra ts 11 , 93 69 233 (45 .0%)

Independen ts 19 14 2 j 60 (11 .6%)

2 1 6 1 6 8 r 3 4 s 1 8 ( 1 0 0 2 )

Expected Percentage when there is no refat ionship between REGIONand PARTY

REGION

Centraf Suburbs Rura.l Rowa i ] . \ r:=_:2PARTY

Repu-bf icans

Democrat s

fndependents

4 3 . 4

4 5 . 0

1 1 . 6

1 0 0

Expected Outcome

REGION

Centraf Suburbs Rura.I

Ci tv

R e p u b l i c a n s 9 3 . 8 7 3 . 0 5 8 - 2 2 2 5 ( 4 3 . 4 2 )

D e m o c r a t s 9 1 . 2 7 5 . 5 6 0 . 3 2 3 3 ( 4 5 . 0 % )

I n d e p e n d e n t s 2 5 . 0 f 9 . 5 1 5 . 5 6 0 ( 1 1 . 6 % )

PARTY

2 L 6 1 6 8 L 3 4 s 1 B ( 1 0 0 % )

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Observed Outcome

PARTYRepubl icansDemoc rat sfndependent s

REGIONSuburbs Rural_Cent. ral-

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ghi*eua1g&S!;_Ex3mple:Question: The following shows the voting pattern by region in one particular year.

REGIONCentral Suburbs .@!eily

PARl]Yl(epublicans 725 61. 38Democrats 71. 93 69Independents 19 74 27

134 518

Concluct a test to find out if voting preference is associatecl with region. Set the level of

2252336 0 '

21e 168

Ho: There is no associationship between the voting Prefernce and region'

,Ha: There is an associationsNp betwrlen the voting prefernce and region'

The statisticai test that can be used to test the above two hypotheses is chi-square test.

We set up "(.

= 0.05 and from Table 5, we reject Ho whenT/'> 9.49, given

significance at 0.05.

Answen We first set up:

d.f.= (c-1) (r-1) = z x2 = 4.t L

cornputarions tor/': ,lf (, +:) _ r I/ L . s , r j .

-to'o uo'1s ' ( hu' U+/'z'*' or'* )

o u,- Q,6lld + ar:tL + -....t?;"t

= --lTT- ?7'L

0 ^ r ' u ,o [ t r l " , , , - r l "

* , "7 " - l

Since 43.665 is >.9.49 we reject Ho.conclusion: There is an ass-ociationship between the voting Patteln region'

Procedures of Statistical Hypothesis Testing Based on Some Observed Data (usually from a

sample)

a) Set up Ho (null hypothesis) to be protected based on the existing wisdom; this also

implies a competing Ha, alternative hypothesis.

b) If Ho is true, what do we expect from the observed data?

c) Choose a risk level, (significance level) -- the risk of making a type I error (that we are

willing to accept)

d) Select an appropriate statistical technique (“test”) for verifying the data

e) Set up a decision rule based on the expected outcome from the sample if Ho is true.

(This involves knowing the characteristics of the sample, etc).

f) Compare the expected outcome under Ho and the observed result.

g) Draw a conclusion: if the expected outcome happens, then retain Ho;

otherwise, reject Ho (indirectly accepting Ha) (this is based on "negative inference")

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of the following area, test the hypotheFis

Example 3

a) The following are ten matchedcollected recently. Find out if there is a

of moisture and temp.

Moisture, X Te4P, Y

47

J 5

z a

z t

4 4

5 5

6 0

, )

J - O

9

z )

, ' 7

? n

J J

4 8

4 9

3 2

b) lf the above data represent a randorn

that there isternp In

iemperature ft soitstween the ranks

Accept Ho

Reject Ho

Ho is true

Conect decision

Ho is false

Type II enor

Conect decision

a)

b)

Set up Fl (null hypothesis) (also impli,rs a

If LIo is tnre, what do we expect from

Choose o (siglificance level), the ri a fype I enor (and we are willing tor

accept)

d) Select an appropriate statistical techniqlue

Set gp,a ileoision rule basqd on the e+p€oted

involves knowing the test statistic, its san

a one-tailecl or two-tailed test, and conring

Do ssmercaloulations to see if what

erqectations.

Draw a conclusion: if the expected outoome

(indirectly accepting H") (this is based on

from the sample if Ho is true. (This

and deciding whether it is

with a critic'al vqlue,)

thp sample confofins to ihe

then retain Ho; otherwise, reject ft

inference")

Example 3B (Rank Correlation)

Question: Given before

Solution:

Set up hypotheses:

We shall test Ho against Ha, where

Ho: There is no linear relationship between the ranks of moisture and temperature (r = 0)

Ha: There is a linear relationship between the ranks ofmoisture and temperature (r "0)

Decision rule: Set the sig. level

The test statistic (from the sample) is r.

Reject Ho when r

When d.f.= no ofpairs ­

Computations

Conclusion: