Hypothesis Test

(1)

Douglas A Lind, William G Marchal and

Samuel A Wathen. Statistical techniques

in Business and Economics

Chapter 1, 11 Chapter 1, 11

!"pothesis #est

1 1

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What is a !"pothesis$

%



What is a Hypothesis?What is a Hypothesis?



AA !"pothesis!"pothesis is a statement a&out the 'alue o( a is a statement a&out the 'alue o( a populationpopulation

parameter de'eloped (or the purpose o( testing. E)amples parameter de'eloped (or the purpose o( testing. E)amples o( h"potheses made a&out a population parameter are* o( h"potheses made a&out a population parameter are*



 #he mean monthl" income (or s"stems anal"sts is + #he mean monthl" income (or s"stems anal"sts is +,-%.,-%.



What is Hypothesis Testing?What is Hypothesis Testing?



!"pothesis testing is a procedure, &ased on sample!"pothesis testing is a procedure, &ased on sample

e'idence and pro&a&ilit" theor", used to determine /hether e'idence and pro&a&ilit" theor", used to determine /hether the h"pothesis is a reasona&le statement and should not &e the h"pothesis is a reasona&le statement and should not &e re0ected, or is unreasona&le and should &e

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What is a !"pothesis$

%



What is a Hypothesis?What is a Hypothesis?



AA !"pothesis!"pothesis is a statement a&out the 'alue o( a is a statement a&out the 'alue o( a populationpopulation

parameter de'eloped (or the purpose o( testing. E)amples parameter de'eloped (or the purpose o( testing. E)amples o( h"potheses made a&out a population parameter are* o( h"potheses made a&out a population parameter are*



 #he mean monthl" income (or s"stems anal"sts is + #he mean monthl" income (or s"stems anal"sts is +,-%.,-%.



What is Hypothesis Testing?What is Hypothesis Testing?



!"pothesis testing is a procedure, &ased on sample!"pothesis testing is a procedure, &ased on sample

e'idence and pro&a&ilit" theor", used to determine /hether e'idence and pro&a&ilit" theor", used to determine /hether the h"pothesis is a reasona&le statement and should not &e the h"pothesis is a reasona&le statement and should not &e re0ected, or is unreasona&le and should &e

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!"pothesis #esting Steps



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mportant #hings to 2emem&er

a&out !



and !

11

3



!!__* null h"pothesis and !* null h"pothesis and !₁₁* alternate h"pothesis* alternate h"pothesis



!!__ and ! and !₁₁ are mutuall" e)clusi'e and collecti'el" are mutuall" e)clusi'e and collecti'el"

e)hausti'e e)hausti'e



!!__ is al/a"s presumed to &e true is al/a"s presumed to &e true



!!₁₁ has the &urden o( proo( has the &urden o( proo(



A random sample 4A random sample 4nn5 is used to 65 is used to 6reject Hreject H__77



( /e conclude 8do not re0ect !( /e conclude 8do not re0ect !__8, this does not necessaril"8, this does not necessaril"

mean that the null h"pothesis

mean that the null h"pothesis is true, it onl" suggests thatis true, it onl" suggests that there is not su9cient e'idence to re0ect !

there is not su9cient e'idence to re0ect !: re0ecting the: re0ecting the

null h"pothesis then, suggests that the alternati'e null h"pothesis then, suggests that the alternati'e h"pothesis ma" &e true.

h"pothesis ma" &e true.



Equality is always part of HEquality is always part of H₀₀ (e.g. “=” , “≥” , (e.g. “=” , “≥” ,

“”!.



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!o/ to Set ;p a Claim as

!"pothesis





n actual practice, the status quo is set

up as !

__



( the claim is 6&oast(ul7 the claim is set

up as !

_1._1.

2emem&er, !

₁₁

has the &urden

o( proo(



n pro&lem sol'ing, loo< (or

'ey wor$s

and con'ert them into s"m&ols. Some

<e" /ords include* 6

<e" /ords include* 6improved, better

improved, better

than, as efective as, diferent rom,

than, as efective as, diferent rom, has

has

changed

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=arts o( a Distri&ution in

!"pothesis #esting

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->ne?tail 's. #/o?tail #est

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!"pothesis Setups (or #esting a

Mean 4

µ

5

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!"pothesis Setups (or #esting a

=roportion 4

π

5

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E)ample

1



amesto/n Steel Compan" manu(actures

and assem&les des<s and other o9ce

equipment at se'eral plants in /estern e/

or< State. #he /ee<l" production o( the

Model A% des< at the Fredonia =lant

(ollo/s the normal pro&a&ilit" distri&ution

/ith a mean o( % and a standard de'iation

o( 1-. 2ecentl", &ecause o( mar<et

e)pansion, ne/ production methods ha'e

&een introduced and ne/ emplo"ees hired.

#he 'ice president o( manu(acturing /ould

li<e to in'estigate /hether there has &een a

change in the /ee<l" production o( the

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=ro&lem

11

tep &) tate the null hypothesis an$ the alternate hypothesis.

!_* µ  %

!₁* µ H %

(note) 'eywor$ in the pro*le+ “has hange$”! tep -) elet the leel of signi/ane.

 = 0.0& as state$ in the pro*le+ tep 1) elet the test statisti.

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=ro&lem

1% 58 . 2 not is 55 . 1 50 / 16 200 5 . 203 / 2 / 01 . 2 / 2 / > > − > − > Z Z n X Z Z α α σ µ

tep 2) 3or+ulate the $eision rule.

4e5et H₀ if 676 % 7

Step 5: Make a decision and interpret the result.

Because 1.55 does not fall in the rejection region, H₀ is not rejected. We conclude that the population mean is not different from 00. So ! e !ould report to the "ice president of manufacturing that the sample e"idence does not sho! that the production rate at the #redonia $lant has changed from 00 per !eek.

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8- #"pe o( Errors in !"pothesis #esting

1

 #"pe  Error ?

DeKned as the pro&a&ilit" o( re0ecting the

null h"pothesis /hen it is actuall" true.

 #his is denoted &" the Gree< letter 6_α7

Also <no/n as the signiKcance le'el o( a test  #"pe  Error*

DeKned as the pro&a&ilit" o( 6accepting7 the

null h"pothesis /hen it is actuall" (alse.

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#est Error

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p

?alue in !"pothesis #esting

1

 p9:;<E is the pro&a&ilit" o( o&ser'ing a

sample 'alue as e)treme as, or more e)treme than, the 'alue o&ser'ed, gi'en that the null h"pothesis is true.

n testing a h"pothesis, /e can also compare

the p?'alue to /ith the signiKcance le'el 4α5.

( the p?'alue N signiKcance le'el, H_ is

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p?alue in !"pothesis #esting ?

E)ample

1-2ecall the last pro&lem /here the h"pothesis and decision rules /ere set up as*

!_* µ O %

!₁* µ P %

2e0ect !_ i( I P I_α

/here I  1. and I_α %. 2e0ect !_ i( p?'alue N α

.-- is not N .1 Conclude* Fail to re0ect !_

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What does it mean /hen p?'alue N

α

$

1@

4a5 .1, /e ha'e some e'idence that H_ is not true.

4&5 ., /e ha'e strong e'idence that H_ is not true.

4c5 .1, /e ha'e 'er" strong e'idence that H_ is not true.

4d5 .1, /e ha'e e)tremel" strong e'idence that H_ is not true.

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#esting (or the =opulation Mean* =opulation

Standard De'iation ;n<no/n

1

When the population standard de'iation 4J5 is

un<no/n, the sample standard de'iation 4s5 is used in its place

 #he t ?distri&ution is used as test statistic, /hich

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#esting (or the =opulation Mean* =opulation

Standard De'iation ;n<no/n ? E)ample

1

 #he McFarland nsurance Compan" Claims

Department reports the mean cost to process a claim is +-. An industr" comparison sho/ed this amount to &e larger than most other

insurance companies, so the compan"

instituted cost?cutting measures. #o e'aluate the eQect o( the cost?cutting measures, the

Super'isor o( the Claims Department selected a random sample o( %- claims processed last month. #he sample in(ormation is reported &elo/.

At the .1 signiKcance le'el is it reasona&le a

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#esting (or a =opulation Mean /ith a

Rno/n =opulation Standard De'iation? E)ample

%

!_* µ  +-

!₁* µ N +-

(note) 'eywor$ in the pro*le+ “now less than”!

tep -) elet the leel of signi/ane.  = 0.0& as state$ in the pro*le+ tep 1) elet the test statisti.

(22)

#esting (or a =opulation Mean /ith a

Rno/n =opulation Standard De'iation? E)ample

%1

2e0ect !_ i( t N ?t_α_,n?1

Because -1.818 does not fall in the rejection region, H₀ is not rejected at the . 01 significance level. We have not demonstrated that the cost-cutting

measures reduced the mean cost per claim to less than $60. he difference of $!."8 #$"6.% - $60& 'et(een the sample mean and the population mean could 'e due to sampling error.

(23)

#esting (or a =opulation Mean /ith an ;n<no/n =opulation Standard De'iation? E)ample

%%

 #he current rate (or producing  amp (uses at

ear" Electric Co. is % per hour. A ne/ machine has &een purchased and installed

that, according to the supplier, /ill increase the production rate. A sample o( 1 randoml"

selected hours (rom last month re'ealed the mean hourl" production on the ne/ machine /as %- units, /ith a sample standard

de'iation o( - per hour.

At the . signiKcance le'el can ear"

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#esting (or a =opulation Mean /ith a

Rno/n =opulation Standard De'iation?

E)ample continued

%

Step 1* State the null and the alternate h"pothesis.

H_* T O %: H₁* T P %

Step %* Select the le'el o( signiKcance. t is ..

Step * Find a test statistic. ;se the t distri&ution &ecause the population standard de'iation is

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#esting (or a =opulation Mean /ith a

Rno/n =opulation Standard De'iation?

E)ample continued

%3

Step 3* State the decision rule.

#here are 1 V 1   degrees o( (reedom. #he null h"pothesis is re0ected i( t P 1..

Step * Ma<e a decision and interpret the results. #he null h"pothesis is re0ected. #he mean num&er

produced is more than % per hour.

162 . 3 10 6 250 256 = − = − = n s X t µ

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#ests Concerning =roportion

%

A =roportion is the (raction or percentage that

indicates the part o( the population or sample ha'ing a particular trait o( interest.

 #he sample proportion is denoted &" p and is

(ound &" x/n

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Assumptions in #esting a =opulation

=roportion using the U?Distri&ution

%-A random sample is chosen (rom the population.

t is assumed that the &inomial assumptions discussed

in Chapter - are met*

1. the sample data collected are the result o( counts:

%. the outcome o( an e)periment is classiKed into one o(

t/o mutuall" e)clusi'e categoriesa 6success7 or a 6(ailure7:

. the pro&a&ilit" o( a success is the same (or each trial:

and 435 the trials are independent

 #he test /e /ill conduct shortl" is appropriate /hen

&oth nπ and n41? π 5 are at least .

When the a&o'e conditions are met, the normal

distri&ution can &e used as an appro)imation to the &inomial distri&ution

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#est Statistic (or #esting a Single

=opulation =roportion

%@ n p z

)

1 (

π π π − − = )ample proportion H*pothesi+ed population proportion )ample si+e

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#est Statistic (or #esting a Single

=opulation =roportion ? E)ample

%

Suppose prior elections in a certain state

indicated it is necessar" (or a candidate (or go'ernor to recei'e at least  percent o( the 'ote in the northern section o( the state to &e elected. #he incum&ent go'ernor is interested in assessing his chances o( returning to o9ce and plans to conduct a sur'e" o( %,

registered 'oters in the northern section o( the state. ;sing the h"pothesis?testing procedure, assess the go'ernorXs chances o( reelection.

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#est Statistic (or #esting a Single

=opulation =roportion ? E)ample

%

!_* π  .

!₁* π N .

(note) 'eywor$ in the pro*le+ “at least ”!

tep -) elet the leel of signi/ane.  = 0.0& as state$ in the pro*le+ tep 1) elet the test statisti.

;se I?distri&ution since the assumptions are met and nπ and n41?π5  

(31)

#esting (or a =opulation =roportion ? E)ample



4e5et H₀ if Z #97

he computed value of z#%.80& is in the rejection region, so the null h*pothesis is rejected at the .0" level. he difference of %." percentage points 'et(een the sample percent #." percent& and the h*pothesi+ed population percent #80& is statisticall* significant. he

evidence at this point does not support the claim that the incum'ent governor (ill return to the governors mansion for another four *ears.

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=ro&lem Y 

1



According to a recent sur'e", Americans

get a mean o( @ hours o( sleep per night. A

random sample o(  students at West

irginia ;ni'ersit" re'ealed the mean

num&er o( hours slept last night /as

-hours and 3 minutes. #he standard

de'iation o( the sample /as . hours. s it

reasona&le to conclude that students at

West irginia sleep less than the t"pical

American$ Compute the p?'alue.

(33)

=ro&lem Y

3-%



n the "ear %- the mean (are to Z" (rom

Charlotte, orth Carolina, to Seattle,

Washington, on a discount tic<et /as +%-@.

A random sample o( round?trip (ares on this

route last month gi'es in the (ollo/ing

ta&le. At the .1 signiKcance le'el can /e

conclude that the mean (are has increased$

What is the p?'alue$

(34)

=ro&lem Y @





#he ;.S. presidentXs call (or designing and

&uilding a missile de(ense s"stem that

ignores restrictions o( the Anti?Ballistic

Missile De(ense S"stem treat" 4ABM5 is

supported &" 3 o( the respondents in a

nation/ide poll o( 1% adults. s it

reasona&le to conclude that the nation is

e'enl" di'ided on the issue$ ;se the .

signiKcance le'el.

(35)

Testing Hypothesis a*out >ierene *etween Two @opulation Aeans (Two a+ple Tests of Hypothesis! •ull !"pothesis* !_* 4T₁ ? T_%5D or T₁  T_% •Alternate !"pothesis !₁* 4T₁ ? T_%5PD 4T₁ ? T_%5ND [T₁ P T_% T₁ N T_%\ ..one?tailed test Alternate !"pothesis !₁* 4T₁ ? T_%5 HD [T_{1 H} T_% \ .. t/o?tailed test •_{#est Statistics*} 2 2 2 1 2 1 2 1 2 2 2 1 2 1 0 2 1 ) ( ) ( n n x x z or n n D x x z σ σ σ σ + − = + − − =

4replace s /hen J not a'aila&le5

•2e0ection 2egion U P U_] or U N ?U_] ₄one tailed test5

U P U_]^% or U N ?U_]^% ₄t/o tailed test5

(36)

Mar" o FitUpatric is the ice =resident (or ursing Ser'ices at St. Lu<eXs Memorial !ospital. 2ecentl" she noticed in the 0o& posting (or nurses that those that are unionised seem to oQer higher /ages. She decided to in'estigate and gathered the (ollo/ing sample in(ormation. Group Mean Wage Sample Standard De'iation Sampl e SiUe ;nion +%.@ +%.% 3 onunion +1. +1. 3

Would it &e reasona&le (or her to conclude that there is signiKcant diQerence in earning &et/een union and non?union nurses$ ;se the . 1 signiKcance le'el.

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E)ample

_{A manpower-development statistician is asked to}

determine whether the horl! wa"es o# semiskilled workers are the same in two cities. $he reslts o# the srve! are presented in the #ollowin" ta%le&

'orl! a"e ate

(38)

-Solution

Step 1:

$his is a two-tailed test. $he h!pothesis is stated %elow. $he si"ni#icance level is 0.05 ("iven)

'₀& *₁ + *₂ verss '₁& *₁ , *₂

Step 2:

ince this is a test o# the means and the de"rees o# #reedom (n₁  n₂ - 2) is in e/cess o# 30 a  test is appropriate. $he critical

vales are  1.6 (#rom a  ta%le).

Step 3:

(39)

Solution

_{tep 4& ketch the distri%tion locate the critical vales}

and the test statistic.

_{tep 5& ecide ince the test statistic vales lies within}

the rejection re"ion then there is s##icient statistical evidence %ased on this sample to re7ect '₀.

_{$he test #or a di##erence %etween parameters does not have}

to %e ero it can %e non-ero. 8or e/ample&

'₀& *₁ - *₂ 9 0.10 verss '₁& *₁ - *₂ : 0.10

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@ro*le+) B91



#/o research la&oratories ha'e

independentl" produced drugs that

pro'ide relie( to arthritis suQerers. #he

Krst drug /as tested on a group o( 

arthritis suQerers and produced an

a'erage o( . hours o( relie(, and a

sample standard de'iation o( 1. hours.

#he second drug /as tested on 

arthritis suQerers, producing an a'erage

o( @. hours o( relie(, and a sample

standard de'iation o( %.1 hours. At the

. le'el o( signiKcance, does the

second drug pro'ide a signiKcantl"

(41)

@ro*le+) B9C



ot/ithstanding the Equal =a" Act o(

1-, in 1 it still appeared that men

earned more than /omen in similar 0o&s.

A random sample o(  male machine

tool operators (ound a mean hourl"

/age o( +11., and the sample

standard de'iation /as +1.3. A random

sample o( 3 (emale machine?tool

operators (ound their mean /age to &e

+.3%,

and

the

sample

standard

de'iation /as +1.1. >n the &asis o(

these samples, is it reasona&le to

conclude 4at a  .15 that the male

operators are earning o'er +%. more

per hour than the (emale operators$

(42)

$estin" '!pothesis a%ot i##erence

%etween $wo ;oplation ;roportion

1.ull !"pothesis* !_* 4p₁ ? p_%5D or p₁  p_% %.Alternate !"pothesis !₁* 4p₁ ? p_%5PD 4p₁ ? p_%5ND [p₁ P p_% p₁ N p_%\ ..one?tailed test Alternate !"pothesis!₁* 4p₁ ? p_%5 HD [p_{1 H} p_% \ .. t/o?tailed test . #est Statistics*

2e0ection 2egion U P U_] or U N ?U_] ₄one tailed test5

U P U_]^% or U N ?U_]^% ₄t/o tailed test5

(43)

$estin" '!pothesis a%ot i##erence %etween

$wo ;oplation ;roportion& ;ooled

When population parameters ‘p’ and ‘q’ are unknown: .Dull !"pothesis* !_* p₁Gp_% [p₁?

p%G\

.Alternate !"pothesis !₁* p₁Pp_% p₁Np_% ..one?tailed test

Alternate !"pothesis !1* p1 Hp% t/o?tailed test

.=ooled proportion* 2 1 2 1 < n n x x p + + = . #est Statistics* = 1 1 > < < ) < < ( 2 1 2 1 n n q p p p z + − =

.2e0ection 2egion U PU]or U N?U] 4one tailed test5

U PU]^% or U N?U]^%4t/o tailed test5

(44)

E)ample*

1. Accordin" to a report %! the American ?ancer ociet! more men

than women smoke and twice as man! smokers die prematrel!

than nonsmokers. @n random samples o# 200 males and 200 #emales 62 o# the males and 54 o# the #emales were smokers. @s there

s##icient evidence to conclde that the proportion o# male smokers hi"her #rom the proportion o# #emale smokers when  + .01B

2. A #inancial anal!st wants to compare the trnover rates in percent

#or shares o# oil related stocks verss other stocks. he selected 32 oil-related stocks and 4 other stocks. $he mean trnover o# oil

related stocks is 31.4 percent and the standard deviation 5.1 percent. 8or the other stocks the mean rate was compted to %e 34. percent and the standard deviation 6.C percent. @s there a si"ni#icant

di##erence in the trnover rates o# the two t!pes o# stockB

(45)

@ro*le+)

B9--A coal?Kred po/er plant is considering t/o

diQerent s"stems (or pollution a&atement. #he Krst s"stem has reduced the emission o(

pollutants to accepta&le le'els - percent o( the time, as determined (rom % air samples. #he second, more e)pensi'e s"stem has

reduced the emission o( pollutants to ac cepta&le le'els @- percent o( the time, as determined (rom % air samples. ( the e)pensi'e s"stem is signiKcantl" more eQecti'e than the ine)pensi'e s"stem in

reducing pollutants to accepta&le le'els, then the management o( the po/er plant /ill install the e)pensi'e s"stem. Which s"stem /ill &e installed i( management uses a signiKcance

(46)

@ro*le+) B9-1

A group o( clinical ph"sicians is per(orming tests on patients to

determine the eQecti'eness o( a ne/ antih"pertensi'e drug.

=atients /ith high &lood pressure /ere randoml" chosen and then randoml" assigned to either the control group 4/hich recei'ed a /ell?esta&lished antih"pertensi'e5 or the treatment group 4/hich recei'ed the ne/ drug5. #he doctors noted the percentage o(

patients /hose &lood pressure /as reduced to a normal le'el /ithin 1 "ear. At the .1 le'el o( signiKcance, test appropriate h"potheses to determine /hether the ne/ drug is signiKcantl"

more eQecti'e than the older drug in reducing high &lood pressure.

Droup @roportion That

+proe$ Fu+*er of @atients #reatment .3 1% Control .- 1 3

(47)

Test for $ierene *etween

Aeans) s+all sa+ple siGe



For small samples siUes, /e must estimate

a 8pooled8 estimate 4a.<.a. a /eighted

a'erage5 o( the 'ariances (or the t/o

groups. #his estimate is*



and then, the estimated standard error is*

(48)

3- #ests (or DiQerence Bet/een #/o Means* Small Sample SiUe

.Dull !"pothesis !_* T₁GT_%

.Alternati'e !"pothesis !1* T1 PT% or T1 NT%

4one tailed test5

!1* T1 H T% 4t/o?tailed test5 .=ooled Estimator* ₍ ₁₎ ₍ ₁₎ ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 2 − + − − + − = n n s n s n s_p . #est Statistics* 2 1 < 2 1 x x x x t − − = σ 2 1 2 1 1 1 n n s x x t p + − =

)

2 (

₁ + ₂ − =

n

df

.2e0ection 2egion t Pt] or t N?t]4one tailed test5

t Pt]^% or t N?t]^%4t/o tailed test5

(49)

E)ample*

_{A compan! wishes to test when the sensitivit! achieved %!}

a new pro"ram is si"ni#icantl! hi"her than achieved nder the le"ac! pro"ram. $he #ollowin" in#ormation is

availa%le #rom test reslts.

ensitivit!

ean

)tandard

/eviation

)ample

)i+e

roposed

%

1"

1%

..2.

8

1

1"

3

(50)

Solution

Step 1:

This is a one-tailed test. The hypothesis is stated below. The significance level is 0.05 (given) H₀: μ₁ ≤ μ_ ve!s"s H₁: μ₁ # μ_

$tep :

$ince this is a test of the %eans and neithe! n₁ o! n_ is in e&cess of '0 a t test is app!op!iate. The c!itical val"e is 1.0* (f!o% a t table with 5 deg!ees of f!eedo%). $tep ':

(51)

Solution



tep 4&

_{ketch the distri%tion locate the critical vales and the}

test statistic.



tep 5&

_{ecide ince the test statistic vales lies within the}

retention re"ion then there is no s##icient statistical evidence %ased on this sample to re7ect '₀.

(52)

@ro*le+) B9

