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**Unformatted text preview: **:Iaora I MST P.on.oz.ooso A coffee manufacturer has three different kinds of coffee beans that it develops. It believes that the average amount of caffeine In one of their cups ofcoffee does not depend upon which type of bean is used. A
onerway ANOVA test Is conducted to test this hypothesis. For each of the three types of beans, a sample of 50 cups of beans are collected, and the mass of caffeine present in each cup is synthesized and
measured. The following table shows the relevant ﬁgures in the ANOVA test: SUM of squares Degrees of freedom Mean squares
Between groups 478.50 2 235.45 1.6600 0.1931
21,204.75 147 144.25
21,683.65 Within groups
Total One assumption in ANDVA testing is that the standard deviations of the different populations being tested are ali the same. 0. Based on the above results, the estimate for d in this ANOVA test is approximately: 144.25
21,204.75
145.62
12.01 - 21.88
239.45
15.47
21,683.65
147.25
475.90 3 Feedback [I] outol‘ 1] This Is not correct. [0 IznltolI 1] This Is not com.
The estimate for a in this AND“! tea ls approain'rately 12.01. Discussion When conducting an RNOVA test, It is assumed that the population aandaro deviations across all of the differmt populations are equal. The sample data oollecoed produces an estimate for this Population
standard deviation. This estimate is the ﬁuare rod ofthe mean square within the groups. In this qumbn, the mean muare within the groups (MSE) is given as 1.44.25. Thardbre the estimate for the population standard deviation, o, Is.- Estimate fora = v'MSE
= €144.25
= 12.01041215...
= 12.01 Rounded as iaststep This eairnate. and the mean square within the groom (which Is an estimate for the Dominion variance. 0:) are vet'if Important to ANOVA teetlntl. The aqnlﬂcanoe ofthese emanates can be seen In the fact
that in an AMEN-ﬂ hm the mean square between groups (MSG) empared to the mean square M‘tl'rin groups (M55). In partlwlar, the best ﬁaﬁiit is the ratio of MSG ID NEE. In WA testing, MSE is an
estimate for the population vananoe. And under the null hypothesis [which states that all ofthe population means are equivalent} the variance beth gmps should be the same as the variance within
QI'DUDS, 56 W36 ShOuId be sim||ar DD MSE. In other words, when conducting an ANNA oest. the crudel part of the out lnmlm Dating whether or not the eve ofvariatlon bemeen the dlﬂerent samplﬁ drawn (MSG) is aowunoed for by tile sample's
estimate for the population variance, MSE. So this estimate for the population variance, and the estimate for the pomlation standard deviation. are an essential part at nNCNA testing. : 3 or 3 to: HST.CP.AOV.02.0100 A medical research team is investigating the effect of diFferent brands ol' aspirin on body metabolism. Three brands of aspirin are tested. known to the team simplyr as
A1. A; and A3. For each brand. 10 test subjects are each g'u'en a dose ol' the aspirin. 111e body metabolism is then measured in each subject (the measurement results in a score out of 151: for each subject, the faster the metabolism, the higier the score). The data from these samples are included below: 55.65 55.28 52.82 52.13 5?.35 MI
54.25 52.45 52.79 55.57 5434 A 51.55 58.53 53.0? 55.15 5-1.1:l 54.35 53.55 5039 5?.5
54.55 55.51 54.45 52.72 54.04
54.53 55.93 55.77 51.?7 55.44 A one-way ANOVA test is conducted to deterrnlne whether there Is any dlﬁerenoe between the averages for bodyI metabolism ([11, p2, p3} for the three dﬁerent
aspirin brands. In particular. the null and alternative hymthesesfor this test are: i 1 H0: 111 = H2 = “3
Ha: Not all three means are equal
A level of signiﬁanoe of 0.05 is used. a) Calmlate the F test statistic for this ANOVA tea. Give your answer as a decimal to 2 decimal places.
F = 2.5?3 b) Calmlate the Malue for this tat statistic. Give your answer as a decimal to 3 decimal places.
P—ualue = 0.095 e) In the ﬁeld below, datrlbe the mulls of this ANOVA tat. In your dﬁcrlptlon, include a decision of whether or not the null hypothals Is rejeﬁ:edf and interpret
this decision In relation to mnduslpns that the meditzl team can draw. 1:} [n the ﬁeld balow, describe the FEle of this ANCWA test. In your description, include a decision of whether or not the null hypothesis is rejected. and interpret
this decision in relation to oondusions that the medical team can draw. IOl'I I'IO Feedback [3 out of 3] a) You are correct.
I!) You are correct.
1:) You have answered this essay.
Please review the model solution below. Discussion a) Using appropriate statistical soltware, the F test statistic can be calculated to be the ratio of the mean square between g'oups and the mean square within
groups. Therefore the F test statistic can be calculated uQng the following formula: W MSG = mean square error between groups = 4105285333...
MSE = mean square error within groups = 1829063111...
F = F test statistic = unknown MSG
MSE 4105285333...
1829063111... 2.57251 12...
= 2.57 Rounded as last step .ﬂ
ll MSG = mean square error between groups = 4105285333...
MSE = mean square error wimin groups = 1829063111...
F = Ftest statistic = unknown SG
MSE _ 4105235333. ..
_ 1829063111. .. 25725112...
2.57 Rounded as last step F: b} using appropriate statistical software, the P-yalue for this test statistic can be derived as 0.095 (rounded to 3 decimal places). all Model solution In this ANCWA test, the null hypothesis being tested Is that the brand of aspirin used does not affect body metabolism, or In other words that the averages
for body metabolism are the same for all three brands of aspirin. with a level of signiﬁcance of 0.05. a P-value of 0.095 is calculated. Since this P-value is
less than the level of signiﬁcancer the null hypothesis is rejected. Since the null hypothais ls rejected, it can be mnduded (with a level of signiﬁcance of 0.05) that the altematlve hypothesis is true. That Is, it can he
concluded that the brand of aspirin does have an effect on body metabolism. In other words, it can be ounciuded that the effects of the aspirin brands on
body metabolism are not all the same. 2| 1 nf : 1p: iris-hi: PAOUJDIJIOHI A racing car driver ls Investigating whether the type D‘ fuel he uses In his Gr aﬁects his Overall speed. He uses three different types OfﬁJel. ‘l'he drlver runs an experlment where he fllls his tar with ead'i type of
hit! 20 times. and each time records the time taken to drive one lap of a track. A one-way allow testis conducted to test the null hypothesis that type of fuel is not a signiﬁcant factor in determining speed. The
result of the test Is that the null hypothiﬁls is not rejeaed. The con cluslun that follows from this test Is that:
it cannot be ruled out that the type oi fuel is an insigniﬁcant factor in determining speed
It has heen proven mat the type of fuel can sometimes be a significant factor in determining speed it has been proven that the type of fuel is a signiﬁcant factor in determining speed
it has been prayen that the type of fuel is an insigniﬁcant factor in determining speed [1 out or 1] YOU are wrrect. Disc iis slcin
The drlver runs an ANCNA test to detainlne whether the type nfl’uel used Is a signlﬂcarit factor In detmninlng the drivers speed. The null and altemat'ive hypothm for sudi a tat ale:
H9: Tyne of hid Is not a signlllcarit factor In determining Speed
Ha: Type of fuel Is a slgnlliiant factor In determining speed
Slum the null hypothﬁls Is not w, thls means that the tﬁl‘ failed to reject the assertlon that type of fuel Is not a slgilﬂcant l'aqor In deterrnlnlng speed. in other m, Warm“ he nil” out “ﬁt the typo ﬁnial s an lnllmlﬂd‘lnt actor in ammuning lpled. Note that it hasn't been prwan that type of iuel is not a signiﬁcant factor. lil's factor simply llull't pun ruled out. That Is what it
means to not rider: the null hrpothals. The null hypothﬁls has not been yawn, it simpiy hasri’t been relaxed. 3 ID: MST.CF.AOV.02.0DAO The manufacturers at Vopstra are interested in testing whetherthe typical monthly talk time used by their customers depends on the type of phone that they own. A toial of 3D typlcal monthly talktlmes were recorded for customers using 3 different types of phones: 10 for Rollamoto (R); 1!] for Simsang (S),- and in for Hokia (H). The null and alternathe hypotheses for a oneiway analysis of variance (ANOVA) have been deﬁned as Hg: ii“ = u; = pH and Ha: not all the population means are equal. At a level of signiﬁcance ofDDS, the critical value corresponding to this onesway ANCNA ls 3.35. a) The data have been summarlzed in the table shown. Calculate the test statlstic (F) corresponding to the one-way data. Give your answer to 2 decimal places. Degrees of sum of Mean sauna freedom squares squares
Treatment 2 1,601,259 800,635 2.72
Random error 27 7,951,239 294,490
Total 29 9,552,508 h) Using the test statistl: for Vopstra's one-way ANOVA and a 95% conﬁdence level, Vnpstra should the null hypothesls. 3 Feedback [2 outor 2] a) You are mrrect. b) You are curred. _ Hamel: [2 outn‘l’lj a) You are mnect. b} You are comet. Calculation
a) The value oi the test statistic can be calou|ated using the lcllc-ving loanula'. mm
MSG MSE = 800,635
294,490 2.7187171...
- 2.72 Rounded as last step F- b) The one-wait MOW! carried out by Vcostra is a anentalled test because you are OnltrI concerned With whether the value at F is too Barge. For this one-tailed test, the null hvoothesls Is relaxed lfthe value
or r is greater than the critical value mrraﬁxmring tn the level of signiﬁcance of the test. This is heause the level or signiﬁcance dating the staan by which minions decide that there is enough
evlduloe tlc wogﬁt that the null hypothesis Is raise. The crltlal value that mesmds Do the let-'9 oi slgnlﬁanaa Mthe tat was given as 3.35.1'hls value was taken [tom a table ofcrltlal values of F at a level o! slgnllianue of 0.05 with 2 numerator
degrees of lteerlom and 2? denominator degrees off'reedom. The numerator degrees ofn'eedom are calculated as the number ofgroups minus 1 (that Is. 3 - 1 = 2). The denominator degrees of ﬁ'eedom
are calculated as the number oi data points minus the numba' «groups (that is. 30 - 3 - 2?). Aoconino to the value 01' F [of Houston's one-war WA, the null hypothesis should not he reject-ed at a
95% conﬁdence level since 1.?2 s 3.35. :I 3 Di 3 ID: MST.GP.AOV.01.0DZO Ajob recruitment ﬁrm runs assessment days for thelr clients. where potential employees are given a series of tests. Part of this assessment day ls a multiplerchoice exam on general knowledge. The ﬁrm ls trying
to determine if the time of day that the test is glven is a factor in how well candidates perform. Astudy is done, in which three random samples of people are given the same test, with one sample given the test
at 9:00 am, one sample given the test at midday, and the ﬁnal sample given the test at 3 :00 cm. Each sample has 60 candidates. A one-way ANOVA test is performed. The following table shows the relevant ﬁgures in the ANOVA test: Sum of squares Degrees offreedom Mean squares Between groups 1,455.42 2 727.71 3.4837 0.0328
Wlthln groups 36,973.53 177 208.89
Total 38,428.95 179 Calculate the coefﬁcient of determination, R2. GIVE your answer as a percentage to Z declrnal places. R2: 25.4 % 3 Feedback [II out of 1] Thlsls not correct. R2 : 3.19% Calculation [0 auto! 1] This is not Correct. R2 - 3.79% ﬁkulaﬁnn The ocelﬁcieut afdebemlination is the ratio of the sum of squarﬁ between the was and the total sum “squares across the 3 samples. Therefore the deﬁcient cl deten'ninaticn can be calculated by using
the l'oliowing formula: mute: 556
R2 E
= 1,455.12
33,428.95
0.0387301...
3.7530093... Mummy by 100 to convert to a percentage
3.79% Rounded as last step The ooelllcient ofdeoermination. R2. is a quantitative measure oftrle proportion of variation In last soores that can be amounocd for by the dlﬂerenoe In time for the tut. That is. the fact that R2 is equal on
339% Isan as: 3.29% o! the total variatlon In test scores ls due to the difference betweEn the times 0! dav that the test ls cornoleoe. The remalhlnc 96.21% 0! the variation In test scores ls due to
variation between one antidote and another, nut-“tin the diﬁerent timﬁ oi day that the best is intimidate. Notice that the melﬁcient Dfdeoerrnlnaticn is 'nelatimahlI small' as a percentage. This is particularly intenﬁting given that the Fuaiue for the cat is equal to 0.0328. This P-valiue is quite Small, and could
suggest that oiHerence In time Is a factor In best performance. and yet only 3.79% ofthe variation In one: soorﬁ is nor-minted for by this inner! This highlights an Important Feature of the onemlay mam
tea: simohr because there Is compelling evidence that the average test socres will dllfer from one time at day to another. this does nct mean that the dlﬁerence is 'lal'oe', or that time oi dav accounts for the
majorityr of the variation in the test soores. nf3 I ST.CP.TWADV.O 1.00 l l} Salem the onrrect statement regarolng twan ANNA: [n two-an ANOVA. there are alwavs two pawlatlons being consldered. but there can be rnarl'.r factors.
In two-way ANDVA, there are always two factors being consldered. and there are always two populatlons being consloered.
ln two-way ANOVR, there are always two factors being consldered. but there can be many populations. > None of the above. _ Feecbaclt [ﬂoutnl' 1] This Is not arrest.
The correct statement regal-oan two-war ANOVA Is In may move. there are alwaw two factors l:an conﬁrmed. but here can be mlnv mutations. Dllcusalon The diamerlstic that distingulshes two-way ANOVA from one-war AND“! is the numberol ﬁner: being onnsldued. So In two-way Mal-FA tlnno are all-lava m helm! bolus oonlldenod. Thue mar
be many populations belng oonsloeretl. bemuse populations form the levels of the oMereht factors. and there may be many ofthese. as an mmole ofa two-way mow best with two factors and several populations. oohsloer the following scenan a reseald‘l glouo would Illte to test the effect of a newI drug on alertness. The group bellms that the effect may be clIlI'erent for males and females. So 50 men and 50 women are randomly selected.
E with 20 at each gender men a large dose nl’lhe true. 20 «each gentler given a Small dose «the true. and :0 cream gender given no dose ofthe drug. A wvslcal alertness tea: ls admlnlsoered to
E measule the alertness ol' the wtaeus. and the results are reooroed. In this best. the two famors are the drug heth heated and gender. The llrst tauor (the drug) has thlee levels In this test: ho dose. small dose and large dose. The seoond lauor (gender) has two levels: male
and female. Tm hwothm bang med In this scenario are: U.) The avenge Mel o! alertness is the same whether no dose. a small £1062. Or a laroe dose afthe drug is ad‘ninisterw: (2) The average leve| of
alertness is the same For men and women. Three diHeIenI populatlons are hean moored tor the ﬁrst hypothesis. and two dlll'erent populations are being oomoared for the seoond hypoth esls. So in this om. there are two radars belng Dated. though several populations are belng oompared. H5T.CP.TWAOV.02.0IJZO Given below are three different means plots for three different two—way ANOVA tests. Te|t1 Test: Select the tests in which there does not appear to be any:r interaction between the two factors: v’ «I Test 2
Test 1
Test 3 :| Feedback [2 out of 2] Legend J You are correct. .D'§¢P$5.‘PF . . . . . . . . . . Discussion Ileana plots and Intemﬂnns When a two-way AND“! test Is oonduueo, it is often helpltll to create a means plot from the samples oolleoted. to give a visual Idea of how signlllcant the two facogrs are, and whether or not there Is any
interactlon between the two fadors. T'ypullv, one ol the two factors will be represented along the horizontal Maris. That '5. olflerant polnts along this axis erI represent the oMarent levels cl one ol‘ the
honors. The vertical y-axls represents values for the eﬁeu being measured. To represent the second factor on the plot, different line: are drawn for the dlﬂerent levels or thls lauor. To detalnlrle whether or not a means plot suooeﬂs that there Is an Interaction between the two faaors. use the following rule: Mfactors mrlmteraallfnotall‘ofthe mﬂ‘erentrmesforsegmentsofrhose llnest the meansplotare parallel. Conversely. Matador: will‘norlnteract lfaﬂ ofthe milieth (andsegrnentsofthose
lines) in the means We areparall‘el. Note that thls Is a guide only. and the means plot Is enlllI used as a vlsual aid In movie testlng. A lonrlal ANOVA test should be conducted In order on depermlne whether there Is Inter-salon (or Irelther or the
fol-mots are slg'llflcant}. In this titrationf the two llnas on the means plot to: forTest 2 are parallel In the sense that the afferent segments are parallel. The means plot for this test is glven again below, wIth some labels attach ed: Motor!
. luau
. Later?
..-
o .
.f’d.
H,
a r Lﬂlﬂl W 2 LMQIJ
Fem-I'll This means that there is no interaction between the two tenors involved. There are two wars of looking at why the fact that the IlnE are parallel suggleas that there is no interaction. These two ways
correspond to the two foams themselves, and why each [actor does not depend on the other l'actor. Itch! A one not Elma all Flmr I The enact or Famor A ran be oonsidenod by going norioontally along the means plotI because the oiflerent levels or Factor a appear along the horizontal x-axis. [n going from one level or Factor a do another
level of Factor A. the amount by which the eﬁ’ect change-s does not depend on whlch of the two different llne-B Is being considered. These two dlﬂerent hrlee remnt the two different levels or Factor B. So. In
other words. the ellett of Factor A doa not depend on what letrel of Factor Bis being considered. ‘I'hls Is because the Iinﬁ are parallel. Fleur B does not depend on Flcllor A Thls means that there is no Interaction between the two factors Involved. There are two ways or looking at why the fact that the lines are parallel suggests that there is no Intel-action. These two ways
oonaoend to the two factors themselves, and why each lactor does not depend on the other faaor. Factorndooenotﬂopandonracoora The effect or Fador a ran be oonsloareo by going horizontally along the means plot, bemuse the different levels of Factor 3 appear along the horizontal )r-axis. In goan from one level of FactorA to another
level of Fauor A. the amount by whlch the eﬁeu changes does not depend on whlch of the two drlferent lines is being consdered. These two dufferent lines represent the two dlﬂ'eralt levels of Fauor B. So, In
other words. the elfeo: ol‘ Factor A does not depend on what level of Factor B Is bean oonsidered. This is because the llnﬁ are Parallel. Factor Bdoee notoapalld on FacoorA An alternative way of seeing this is to focus on Fm B, and how it does not depend on Factor A. The dlllercnt levels of Factor B are the two dill'erent lines. So the elect of Factor B Is suggested by what
ch angﬁ oocur ln going from one llne In the means plot to the other line in the means plot. llotloe that this mange Is the same at all levels along the horizontal X-axle. ‘I'hae horizontal levels are the levels or
Famr A. In other words, the jump [ﬁlm one line on another line In the means plot is the same Jump, Indepeth or the level or Factor A being considered. again, this is because the lines are parallel. No Intencalon The above deeorlptlons explaln why the means plot for Test 2 suggests that there Is no In teramon bemoen the lamors In that men. The same arguments cannot be applied on Tan 1 and 3 In this quest-m.
and so the means plot for those tms suggest that there is interaction between the faaorsl'n dlose tests. :I: or: I not.ce.'rwaov.o2.oo:o One ofthe major lilTererlaﬁ between one‘way NOVA and two‘way move is the notion of interaction. The [allowing table prEents three Qatements in relation to interaction. For each statement, select whetl'
It is true or false: a] All factors eshl t interaction w' h one another. I - II] I! either of two factors is a significant factor, ...

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