In this chapter, students will: learn concepts ofnull and altemativc h}?othescs, tcst statistic, level ofsignificancc and p-value . conduct tests for a population lrlean based on: o a sample liom a nomal populatiol ofkflown variance o a samplc from a normal population ofunl
.
.
#1.
use
oft-test
Introduction
Olten in scientific enquiry a statement conceming a population palametcr is put forwanJ as a statistical hypotlesis. Its validity is then tcsted, hrsed ofl obsrrvctrorr made from ran.lom samples titken from thc population.
Definirton : A slatistical hypothesis is
a
assertion or (onjccture conccrning onc or more population.
For cxanples, consider the following Ileadlines in Ncwspapers : . Singaporcan students perfonn better tn Mxths and Sclrnec than thusc ftom drc U K. . Stafting school late has reduced thc numbcr ol'latc coners to the Collcgc
.
A\etugc lQ.scorrUl :RJ{ iJn{
I,
aho\(
IUX
ln hpoth€sis testing, we usually takc a random sample lrom the population ulijer invcstigation and use the information contai[ed in lhis sample to deciclc whe{licr a hlpothcsis is likely to bc tnre or false.
#2. Some Definitions
.
Any assumption about the population paraneter (c.g. population mean p which can be ) testcd by using thc observations it a random sample taken liom thc population is a statistical
lrpthcsis
.
Null h)?orhcsrs.
f-l -
ffr.
hl4rothesis on a value of rhe population pa.ameter beforc s.rmplrng is donc It is thc l)ypothcsis lhat we are illterested in
festing and
will .eject if the
sufficient cvidence
c29
I
rlata
in the sample provitlcs
&le : If null hypothesis H0 is known as altemative hypothesis
rejected, the conclusion is that some other hypothesis , is acccpted. There are lhrcc possible Ht that conccm
us.
H6:g=171g,
For example, supposc
Ht
then, depending on the contcxt, the altemalive hlTothesis
Ht:p+17k9 (
could be
)
or ljl:p>l7kg ( or Hl :p
))
|
on"tu'r"a
'"""
Test statistic A quantity calculated from our sample ofdata- Its value is used to decide whethcr or not the null h,?othcsis H0 should be rejectcd in our h)?othesis test.
Critical region The
ofthe test statistic for which Ho is rejected is known critical region. It depends on the t)?e oftest and the signilicance level. The boundaries of thc critical region are called the critical set of values
as tho
values-
Significance lcvel (rr.
x
100%)
o is the probability of rejecting H6 when Hg is in fact true. lt must be chosen, before the tcst is conducted, to aid in deciding whether to accept or rc.iect H0. Note that o is the total probabilify in the critical region.
-
NqEI When the levcl of significance is not given, a 57o level is nolmally uscd. As
examplg the critical regions (shaded) for a z-test lvith a siqdficance levcl of ((l:0.05) for the 3 dillerent types oftests are illustratei below. al1
Two-tailed test
-1.96
\,
1.96
Left-tailed test
5olo
Right-tailed test
,1.645
rcne
Critical valrres
,/ critical value in each case
N9l!9: that the critical values can be obtained tiom the GC or thc tables in MF 15. This be
c29
Z
J
will
.645
dcrDonstratcd in the examplcs.
p - value -. thr: lowest level ofsignificance at which Hg is rcjected. Once the p-r,alue has been dctermincd, wc do not rcjcct H0 at {a x 100% level if p-valuc > c. we reject II0 at (1 x 1007o level ifp-value:, n . Note that thep-v.lue can be calculated by ttre GC.
(]raDhical intcrpretatioI of p-value
What type of test?
Righttailed
Two-tailed
ls the test statistic positive or negative?
P-Value- area to the left of the test statistic
#4. Step 1
P-value = twice the area to the right of the test statistic.
P-value = h,,r'ice the area to the left of the test statistic
-
P-value area to the right of the test statistic
Procedure for Hypothesis Testing State I I0 and Hl
Step 2
Detcrminc the tlpe oftest (one-tailed or two-tailed) and selcct the signilicance levcl not given)_
Step 3
Determine the distribution under I{o
Step 4
Caiculatc the value oflhc test statistics and be done with the GC_
c29
3
tlicp value,
(iI
basod on sampLc data. This calr
.tt,
5 : (Two methods to decide) (1) R'jject Ho ifthep-value :: o(at significancc level Cr x 100o/o). Do not reject Ho if the p-value > o (at significance level d x 1009/0)-
(2)
Reject H6 if the test statistic lies in lhe critical region (at significance level 100%). Otherwise, do dot reject Ho (at significance level
(4.1) Use of Z-Test To
in
cx
Sisnificance testine concerning pezu!4lie!Uqe4!,!
tcst Ilo: p=p0
against Hl :
!L
t
lto
It
at
u
r
1007o level
>Po
ofsignificance.
Useful Formulae 1)Sample mean
t,
- It,'
,t
n
2) In the lbllowing cases, ifpopulation varirnc. ol is unklo*n, replcce o2 with .-' (unbiascd variance estimate). [Jse ary of the following formulae according to given irfonl1ation.
. ', - , ' tl' 1t," ," (lt' n.-1)'I where cis a constant. I
c29-4
(ii) A largc sample fron
a nonnal population, population variance .' 2 is u[known (conlpare this to the usc of l-test whcre samplc size is small.)
(iii) A large samplc (,
I
50 as a guide)
from any populatiot (not Normal)
-r-N( whcre s2
)
is the
unbiascd
variance estimate.
(r)
By CLT,
(l)
,:i - _i
t-rv(
when o2 krown, (2)
- - 1/(
when o2 unhown, where s' is lhe unbiased var-iance esfimate-
Erarnplc
'Ihc
I:
lengths of mctal bars produccd by a particular machire arc nonnally dislributed with mean length 420 cl}I and staodard deviation l2 cnr Thc rn:chrne Ls servrcetl, ritcr wlrich a sar'plc of lr){r b,lrs gr\e\ I tcJn lcnAth ol 4/ r cm Is therc cviclencc, at thc 5o4 lcvcl , ofa chnngc io the m(jan lenglh ofthc bam producerl by lhe machine, assuming tltat the standard dcviation rcmains thc same ,l
JSolnl:
Let
Step 1:
F cm denote population mean length-
:
To test
116
against
lli: llr
7z
=
Stcp 2:
Two-tailed tcsl af
Stcp 3:
uracr Ho,
Step .l:
Tq qqt Ge ttq calculale tesr rt41!!!!
. 0."..
.
Filund
setecr <1> o. prc..
F-
level. (i.c. (l
Lr[
:
l
q!d? f4!!r.
use righr arrow key ro sclecr
.
I=NIFR
I
EIIT
CHLC
IEZ-Teqi.
larFltr
2: T-Test... 2-Sar.rt'zTest... 4: 2-SaneTTesL-.. 3!
c29
5
5: 1*PrspzTeet... 6! z-Fr'rFZTeEt... 7J.Z Inter,,ra 1...
Select and cnter the given values for Then select + .
l0
to indicate a two{ailed test followed by
Z-TesL I nPt: nEt s 1rr:428
s!
Z-Test, $+428 F=.8124193597
I
12
rr=
n: 169
po,o,X,n.
E:EM {rrD Calculele
)uo
1e0
I
Eraur
Retum to the Z-Tcst screen above and select instead of to obtain a drawing ofthe arca associated with thep-value. Step 5:
Sincep-value:
, w€
and conclude that therc is that the mean lcngth ofmetal bars has changed
itt
Altcrnntivelv
F=-012i|
(use
critical reeion)
whv
Stcp 4: (additioftll step ifusirg this method)
o.975? Use InvNonn to detellnine thc critical valucs.
inuHorn(8. 975)
ffi*3TEH*. 2: noFPia I cdf ( 3: i nuNorm ( 4: tpdft 5: Lcdf( 6: )tu Fdf (
TiHzcdf(
Thus Cntrcal Rcgion: Rrjcct H"
if 2.,
_
or
zdn >
Step 5: Since test statistic
:
and conclude that there is ,we thal lhe mean length ofmctal ban has changed at
216r
Do it yourself Re-do Example I on your own but now using a significant level of 37o. Which pafts of the steps arc the same and which parts need changcs? Did you get the samc conclusion? Why?
c29
6
Example 2 In a population of chickcns fiom a particular farmer, thc distribution of t_he mass ot a chrcken has mean p kg and standard deviation o kg. Tho farmer claims that the mean mass is l.?5 kg. A random sample of 100 chickens is taken fron thc population_ The mean mass for the sampie is denoted
hv
,Y . Srarc thc apprnximate
distnbution
ol X
The sarnple values are summarised by l,,c = lS0- l, lr. - 401.7,1 , where,r kg is the mass of a 'rest, chicken. at 17o levcl ofsignificancc, whether the farmer has underostimated the mean mass of the chickens, stating whether you are using a one-tailed or a two-tailed test and statc your conclusion clear-ly-
lSolnl:
IIo: / = 1.75 against Hl: ll >
Stcp 1:
To test
Step 2:
One-tailed test at Sample mean
I
Jr
t,
=4n
level (u
=
nbiascd \arinrce (srimarc
(: ' f r ,'I rl" ft , l )
+ Stcp 3:
:
s = 0.88401t99
lirJ,r Hu.sIrrrsJDrllcsr,/(ut tU0rjlrrBc, br ( LT, ) Thus
Vtt,.-
l0{)
r.
)r .' N(
Step 4:
IffitiPg'. *=
Z-Tesf,
o: .88498952755... R: 1. BrAl n! l0B
Z-Test-
u)l.75
z=.576S643937 P=. ?82S154885 *=1. 801 n=168
f;t[,:rltgffi" Stcp 5:
Sincep-value: at
we
and conclude that there is that the farmer has underestimate.d the mean maLss of the chickcns
c29
7
Alternatively (usc critical reqion) Step 4: (additional step ifusing this method)
Thus Critical Region: Re1ecl H.
if
2..,
-
Step 5: Sinc€ the test statistic ztd there is chickens at
:
and cllnchrde thrr that the farmer has underestimated the mean nass ofthe
I)o it yourself? What if the farmer claims that the mean mass is l.85kg? Repeat Example 2 on your own but this time test at 27o level ofsignificarce, whether the fanner has unde.estimated the mean mass ofthe chickens
c29
{t
Example 3 : A cedain ricc company packs rice in bags which the company claims contain l0 kg ofrice each. on average. A random sample of80 bags is examined and thc mass, r kg, ofthe contents of each bag is deternrined. Ir is found thar tOy =,Zl.Z, l0)' =85.1. Tr*r at 5% level of
l{t
I("
significance whethcr the manufacturcr is overstating the avcrage nass of the contents ofa bag_
ISolnl: Step
1:
Let p derote the average To test H0: l= against H1
Step
2:
mass
ofthe contorts ofa bag.
: p<
One-tailed test at 5% level. (u ..0.05) SarnDle data:
.
samPlc mean
fr, ,) r z=-n
Urrhir.e,t r anrnce csrirrrrrc
|.
s n' tl" [1,, ., lI('
)) i I
-.> s = 0.9'/98'.7214
Step
3:
Sincc sanrple sizc of80 is large, by
'r)x Nf
u,,d.r He.
)
Step 4i
Z-Test
nPt i DEt a FtFlEt t-r0: lLt n: .97987?44867... x: 9. 66 n: 88 Lt: +1I 0 I )r'.r I
UElculaLe
DrsLJ
Z-Test
s( 1g z= -3. 18f,518786 F=9.5623865E -4 x=9. 66 n=80
I
Sincep value =
and coicludc that therc is
,et
overstating tlie avcrage lnitss ofthe cottents
, that the
ola
c)9
blrg.
9
ma ul'acturcr is
Alternatively (use criaical region) Step 4: (additional step ifusing this
whv
inuHorF ( 8. 65
Thus Critical Region: R€ject FL
o.os?
if zd
<
-
Step 5: Since the test statistic z,*, = there is I ,at overstating the average mass of the contents of a bag.
and conclude that that the manufacfurer is
Do it yours€lf What if in Example 3, you are asked to test at 5% level ofsignificance whether the manufacturer is understating the average mass of the contents of a bag? Which parts of the test will change? What $.ill be your conclusion then?
c29
r0
[x:tmple 4 : Thc random variable ,l1 has a normal distribution with mean p (unknown) and variance o 2 (known). To test thc null hypothesis H0: p: po, a raldom sanple ofn observations ofxis taken, and the sample mean is t . Find, ir te.ms ofFo , o and r, the set ofvalues of i which will result in I l0 not being rejected in favour of Hl : p < Fo at the 20% level of significance. [Solnl: Ho:
P:
lrl
Hr:p
If
ofsignificancc. Ho is not rejected, p-value >
d.:
Use lnvNorm to find the critical valuc z"nd"ulIhen ,-value = 0.02
n& d DRHtd pdf (
invHcrm(4. 82)
1: nonna l
I
2: iroFr'lalcdf ( ( Si nuHorr'i 4: tFdf ( 5: tcdft 5: XzFdf(
7+XZcdf(
Thus thc criticai valuc: So if2-r,aluc >
, i-e. area to the
lell ol'this value will be {
, then zr".r >
In othcr words, so as l1ot to rejcct Ho,
ztc.r
> ?.nri."j
'^ "tr- x r
l1o
xpa
Therefore, the ser or vatucs or
;
which will result
Do it yoursclf What would the set ofvalues of
;
be
lr ']re R r 7,, 2.o-il7+!. I J')
in H6 not being rcjected in favour of I-Il is
ifthe outcome ofthc
c29
1t
test is to reject [Io instead?
Example 5 The table below shows the number ofcell phoncs owncd by 100 households. Numbcr ofcell phones per household. x Numbcr of households with a certain rrurnber of cell phones,
0
t6
t
29
2
3
34
14
4 6
Birsed on the above data, test the claim that the population mean is not equal
level
ofsig
5 1
to l-60 at the 5%
ficance.
lSolnl: Step 1: Store the values of)r (number ofcell phones) in and the values of/(number of households) in L2.
IC
?9
Ll
lt 6
tU
Stcp 2 (find mcan alrd thc standard dcviation):
P,"r.
FrAT-l
to select
l:
. highlisht
(
AL.
and
0,"*
[TG--l
l-Var Stats.
{7} =
EDIT EffIlE TE5T5 lEl-Uan sLeLs 2:z-UaF SteLE 3: lled-l'led 4: L inReg( ax+b) 5: SuadReg 6: Cut icRes
TI0uertRe.l
1-Uar 5ltste Lr,L 2
Step 3: Selcct the list
@l r."*
i''r
Ll
followcd by L2 by prcssing:
[,]
[-ND
1-Uan st'a'lr5 x=1, 68
lE4
5x=1 . 144861 15s sx=1 . 139122469
In=1AB
lGllrecl.
I
Note (important): (I S, shown in tho GC is the unbiased estinated population standard deviation s.
) (2) o, shown in thc CC is the sanple standard deviation. (3) If the lists L I and L2 are selected in the wrong order, you will get r :
c29
12
l5 instead of
100.
Let p To
denote thc avqage numbcr ofcell phones owned bya family.
test H0: ll=
against
InPL: D€La tslGlEl
I{t: lr*
s0i1-5 o:1.145
n! 1.58 nr 186
.(r:
'Iwo-tailed test at
iiEiii
Since samplc size of 100 is large,
ir:"
Z-Test
!r+1.€' z=. 6986899563 E=. 4847456641 n= 1gE
by
(
undcr
Ha. )f - Ni
\
Sincep : ard conclude that there is claim that the population mean is not equal to 1.60 at ofsignil'icancc.
Alternati!elv { usc critical re!'ion r (Rcfcr to Example l) 'l-hus Cdtical Region: Roject tto if t,.,r < Sincc the lest statistic
zr.sr
wc cqual to 1.60
to
or
:,!sr >
= :rnd cooclude {hat there is
at
to clairr that thc population Dlean is not
ofsignjficancc.
z,e:0.698' Do it yourself Wrat ifanother sct ofvalues wcrc addcd and that thc nufiber ofhouseholds with 5 handphones wcre changcd as fbllows? ]
Nu-t .. ofcett pl.
i
\s!.-s!ellL'
.
I Numbcr ofhuuscholds wrth a I nurnber ot ecll phones,
(crtrr
/
What would thc outconc ofyour tcsl be?
c29
r3
(4.2) Use of t-Test in Signiflcance testins for nopulation mean. u. usine a small sample(n < 30) of unknown variance. THE I- DISTRIBUTION
DtrFINITION The l-distribution is s],rnmctical, bell-shaped, and similar to the standard normal curve. [t differs from the standard nonnal curvc, howeveq in that it has an additional parameter, called degrees of freedom, which changes its shape. DEGREES OF' FREEDO]\,I Degrees of freedom, usually syrnbolized by y , is a parameter of the l-distibution which can be any real number greatcr thao zcro (0.0). Setting the value of y defines a particular member of the family of I distributions. A member of the family of I distributioos with a smaller y has more aroa in the tails ofthe distribution than the one with a larger v. The ellect
of v on thc I distribution is illustrated in the drree I distributions below.
-3.18
l.gd
-1.96
3.lB
7t\
-1/5
Nole that the smaller the v , the flattcr thc shape of the distribution, resulting in greater arca tn the tails ofthe distribution
c29
-
14
RELATIONSHIP TO T HE NORMAL CURVE The standard nonnal curve is a special case of the /-distribution whcn y ) .o. For practical purposes, the /-distribution approaches the standard nomal distribution relatively quickly, such that wherl v = 30 the two are alnost idcntical.
Following are thc important properties ofthe l distribution.
l.
The I distribution is diffcre[t firr dift'erent sample sizes.
2.
The
I distribution is generally bell-shaped. but with smallcr sample sizes showing increased variability (flatter). In other words, thc dist.ibution is liss peaked than romal distribution and with thicker tails_ As the sarnple size incrcases, the distribution approaches a nonnal distribution. For,, > 30, the differences are negligiblc.
i
3.
The mcan is zcro (much likc the standar
4. The distribution is symmetrical about the mean. 5. 'fhc variance is greater than I, but approaches I as th(- sanple size increases . 6. The populati(D standard deviation is unknown_ 7. The population is essentially normal A /-distribution is used when (i) the distribution is nonnal
(ii)
(iii)
the sample size is sntall ( lcss than 30) and the pofulation \/ariaDce is unknown
Notation:
-{- (v)
means that the variable
Xhas
a
l-distribution with v degrccs offreetlom-
In general, the degrees of frccdom is the nuntber ofvalues that can vary afler c€rtain restrictions have been imposed on all values.
c29
15
Performine thc ,-tcst When a small sample is drawn from a normal population of unkno$n variance, to test Ho: p=po against FIr: p+F0 p
F>tlo at (r.
x
10070 level
the test statistic
is ?'=
ofsignificance,
r lo J
lbllows the I distribution with degrees offrccdom
equaltor-
1.
J; Example 6 (FM N1999/P2l7(i)) The manufacturers ofan clect.ic watcr heatcr claim that thci heatcrs will heat 500 litres ofwater from a temperaturc of l0"Cto a temperature of 35"C in, on average, no longer than 12 minute. In ordcr to tcst this clain, 14 randornly chosen heaters are bought and the times (r. minutes) to heat 500 litres of water from I0'Cto 35"C is measured- Correct to I dccimal place, the results are as follows-
1.3.2 12.2 11.4 14.5 11.6 t2.9
l2.l
t2_4 l0_3 12.3 11.8 11.0
StatiDg any assumptiorl necessary significancc lcvr:l using r r-te.l.
lSolnl: Lct F denotes the populatior
for validity, test the manufacturer's claim at the
mean timc taken to heat 500 litres of water
f.om lo'C to -l-5'L
,'\.ssumption: The time taken is rormally distributed. To test H0:
l= against Ht: l>
One-tailed t{est at 1070 significance level.
Under H0, test statistics is
I=
((x:0-l)
-
t(
IE 1Z-: 1t-q ).
1q-5 11.6 12"9 12_q
Lr( =13. ? Step 1: Store thrl valucs ofrc in
St"p Z,
r3.0
D.6
I'.".,
FTAi]
Ll.
and use right arrow key to select
cz9
16
l0(%
.
Step 3: Select <2> to select T-Test.
Step 4: At the T-Test screen, highlight
T-Test InFL:llElF 5taL5 lrI r 12 LietrLr Freq: 1
List: Ll Freq:1
Hiqhheljt
Hf[fl,ItA
p> p.
step 5: Highlight and press
H"
@l
T-TeEt u)12 t=- 855S2116s2 r=.20481864S5 *=12. ?357r4?9 5x=1.031509883
I n= 14 Sincep
value:
,
at falsc.
we
and conclude that therc is
signiticance lcvel tlut thc nanuacturer's claim is
AIlernali\ely (usc crilical retio ) Ref-er to the table ofcritical values llr tlie l-distribution in Mlll5, and nrw y
=
, we
obtaifi
as the
Thcrefirre Critical Region: rcject H. iftr.,r
critical value_
under column 2
:
0.90
jl
Since 4o1: , we and cotclude that there is signilicance level that flre mlrufucluter'i claim ii [alsc.
at
Do it yourclf llepeat or your own Example 6 with thc following sot ofvalues instead.
13.2 12.2 ll_4 l4.s ll.6 l0-l 9.6 9-9
12_9 12.4 t0.l 12.3 lt.8 r0-3 t2-0
c29
t7
Example 7 machine is programmcd to fill containers with 20009 each of a c€rtain t)pe of powder contents. A sample of eight containers is selected at random from a largc batch. The containcrs have powdor contents with masses r grams which are sunmarized by | (r - 2000) = 2'7 .4 and
A
:('
2ooo)' = 231.52.
Assuming a nomal distribution for the masses of the contents, show that there is significance ovidence, at the 57o levol, thal the machine is ove#illing the containers.
lSolnl:
Let p To
denote the average mass ofthe contetts ofa contain€r.
test Il0 : / :
against
Hl: ll>
Onc-tailed t-test
sarnfllll4!!
at
sarnpl(
((!:
lcvcl.
-.Lr'ooo'
n,."nr
U'hirscd vanance srrim.,rc
)
r' -
|
lf u
r )000-2003.425
zooor
'L.-81
[lloooll +
Urder H0,
test statislics is
T-Test InPLiEEtE EiGrEl LTg: 2UUU
x: ?88J.425
5x:4.4669 n:8
iiliiiii
r
-4
=
l= ,n.nuay
s = 4.4669
).
T-Test !1)2000 t=2. 1686993
F=. 8333725451
x=2883.425 5x=4.4669
ffi,
Sinccp-value:
aod concludc that there is at
significance level that tho machinc is overfilling th()
containcrs
c29
t8
Alternativelv (use critical resiont R€fer to the table of c.itical values for the t-distuibution in MF15, under column alld row y = 7 1-895 as the critical value. , we obtain Therefore Critical Region: reiect Ho Since
46t:
is
,
iflt*,
:
we
and
conclude that there significance level that the machine is overfi lling the containers.
c29
-
t9
at
z:
Example 8
'Family' packs of bacon slices are sold in 1.5 kg packs. A sample of 12 packs was selected at random and the masses, measured in kg noted. The following results were obtained: f r=17.S1, t"' = 26-4351 . Assuming that the masses of the packs follow a Normal distribution with variance o'/ , test (a) if o? is urknown (b) if o'z:0.0003
at
lyo level whether thc packs are significantly underweight
[Soln]:
Let p donote the average mass ofthc To test H0: /= against
packs ofbacon slices.
HI: l<
One-tailed test
at
level.
(c:
z-' - JJJI r I'. n12 r
Samplqdata: sample mean
unbrasert variancc esri,nur..'
l-ft, - ,ll" -
1
aga2
U'tl=--ft n )t2t\
rn.+rsz
f:
illl:8"'" *
T-Test
x: 1 . 4842 5x: . 015643 n: 12 lr: +rr 0 EE )rrr CalcuI ate Orar,r Since thep-value:
is si
I
0-OOO244697 ..> s = 0.015643
(a) Sincc a small sample is drawn from a Normal distrit ution :rnd t- tcst.
Under H0, test statistics is
tlll t2 l
-t(
o? is unklorvrr, we use a
).
T-Tes'l: B<1-5 t= -3.498868856 P= -Pg249E6979 *=1.4842 5x=.615643 n=12
I
t:
,we
-3-1989
and conclude that there
significancc lcvol that the packs arc gnihcantly undcrwci ght.
c29
-
20
Alternativelv (usc critical reqion) Refer to the table ofcdtical values for the t-distribution in MFl5, under columnp and row v ,\aeobtain2?18 asthccritical value Thereforc Critical Rcgron: rcjcct Ho Sincc
/,,.",
:
if
tr"i <
, we rejcct Ho and
conclude that there is significance levcl that the packs are signilicantly undcrweight.
(b) Since o7 =0.0003 is known, we usc a z-test-
Under H0, test statistics is 26,
=
-
Note: o =.,6.0003 =0.01732..... (
Z-Test
1'or the
N(
result below r,6.0003 has bcen used;
Z-Test p.t1.5 z= -J. 16 F=l.4891166E -4 x=t .4842
F,nnE . g1?3?858867
l[!t:8"t" 't: I . 4842 I:
u: 1? Lr:+ur-EGI ]Br LE l Cul EtEr UF.ir.l
n=12
!:-3.16
Sinccp-valuc:
and conclude that thore is
sigrificance level that the packs are signilicantly undcrweight
Alternativcly (usc critical resion)
i nvHorr'r {6. El }
Thus Critical Region: Rcject H"
Since:,..,: and concludc that there
is
if
2,.., <
,we at
signilicance level that the packs arc srgnifi cantly undefweight_
c29
- 2I
Summarw
(D
population
is
normally
distributed, population variance .' 2 i" knar*n-
(ii) A large
sample from a normal population, population variance d2 is unknown (compare this to (iii))
.Y
- 7v(lo.:)
F- lr@","
(iii) A small sample from a normal population, variance o
)
T = -------:--!-
is unknown
t(n
ll
G
(iv) A large sample (n > 50 as a By CLT, guide) fiom anv poDulation (not (1) X-lr@0,1) wrren Normal) n o2 koo*r4
(r)z= X-po,X
lto o
t; X-p"
(2)X - ir/(pn,-) when o'
n unknown, where s is the unbiased variance estimate.
End ofChapter
c29
-22
