,Solution manual for Intermediate Statistical
Investigations 1e Tintle
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, CHAPTER 1
Sources of Variation
Section 1.1 1.1.10 Color of asignistheexplanatoryvariablewithwhite,yellow,
andredbeingthelevels.
1.1.1 B.
1.1.11
1.1.2 B & C.
1.1.3 A. Observed Sources of Sources of
1.1.4 C. Variation in: explained unexplained
1.1.5 E. f.whetherthestudent variation variation
obeyedthesign
1.1.6 B.
{ 92.19if eccentric poet
60.34if rigid librarian Inclusion criteria a.colorof the b.whetherthesubject
1.1.7 predicted number of uses for items=
• c.timeof day sign wasleft-handedor
1.1.8 right-handed
• e.ageof subject
a. The inclusion criteria are having a clinical diagnosis of mild to d.attitudeof student
moderate depression without any treatment four weeks prior and e.ageof subject
duringthestudy.
1.1.12
b. The purpose of randomly assigning subjects to the groups is to
makegroupsverysimilarexceptfortheonevariable(swimmingwith a. Thevalue6.21representstheoverallmeanquizscore,5.50represents
dolphinsornot)thattheresearchersimpose.Volunteeringforagroup the group mean quiz score for people who used computer notes, and
couldintroduceaconfoundingvariable. 6.92representsthegroupmeanscoreforpeoplewhousedpapernotes.
c. Itwasimportantthatthesubjectsinthecontrolgroupswimevery b. Welooktoseehowfar6.92and5.50arefromoneanotherorfrom
day withoutdolphinssothatthiscontrolgroupdoeseverything(in- theoverallmeanof 6.21todeterminewhetherthenote-takingmethod
cluding swimming) that the experimental group does except that mightaffectthescore.
whentheyswimtheydon’tdoitinthepresenceof dolphins.Without c. The number 1.76 represents the typical deviation of an observa-
thiswewouldn’tknowwhetherjustswimmingcausesthedifference tionfromtheexpectedvalue,inthiscase,fromtheoverallmean.The
inthereductionof depressionsymptoms. number1.61representsthetypicaldeviationof anobservationafter
d. Yes,thisisanexperimentbecausethesubjectswererandomlyas- creatingamodelthattakesintoaccountwhetherthepersonisusing
signedtothetwogroups. computerorpapernotes.
1.1.9. d. Becausethestandarddeviationof theresidualsrepresentstheleft-
overvariation,wecanseethatafterincludingthetypeof notesasan
Observed variation Sources of Sources of explanatoryvariableinourmodeltheunexplainedvariationhasbeen
in: explained unexplained reduced(downto1.61from1.76).Thistellsusthatknowingthetype
d.substantialreduction variation variation of note-takingmethodenablesustobetterpredictscores.
indepressionsymptoms 1.1.13 Random assignment should make the two groups very
similar with regard to variables like intelligence, previous knowl-
Inclusion criteria a.swimmingwith • g.problemsinthe edge, or any other variable and thus likely eliminate possible
• b.mildtomoderate dolphinsornot personallivesof confoundingvariables.
depression thesubjectsduring
thestudy 1.1.14
• c.nouseof
antidepressantdrugs • h.illnessof a. This table shows us possible confounding variables but then
orpsychotherapyfour subjectsduring shows that subjects in the two groups are quite similar with
weekspriortothe thestudy regard to these characteristics, thus ruling out these possible
study confoundingvariables.
Design
b. We would want the p-values to be large, so we could say that
• e.swimming we havelittletonoevidencethatthereisadifferenceinmeanage,
• f.stayingonanisland
proportion of males,etc.betweenthetwogroups.Wewantourgroups
fortwoweeksduring tobeverysimilargoingintothestudy,soacausalconclusionispossi-
thestudy
bleif wefindasmallp-valueafterapplyingthetreatment(s).
3
c01InstructorSolutions.indd 3 16/10/20 6:49
, 4 C HA PTER 1 Sources of Variation
1.1.15 Itislikelythat3-to5-year-oldsmighthavedifferentpreferenc- c. R2=11.1328/199.62=0.0558.Wecaninterpretthisbysayingthat
es when it comes to toy or candy than 12- to 14-year-olds. The older 5.58% of the variation in the perceived level of risk is explained by
group is probably much more likely to prefer the candy over the toy whetherthenameof thehurricaneismaleorfemale.
andtheoppositecouldbetruewiththeyoungergroup.Wewouldnot d. SSError=199.62−11.13=188.49.
seethisdifferenceif theresultsof alltheagesarecombinedtogether. _____________
e. √
188.4872/140
=1.16.
{−0.28iffemalename
0.28ifmalename
Section 1.2 f. predicted
hurricaneriskrating=5.29+
,
1.2.1 B. SEofresiduals=1.16.
1.2.2 A, D. 1.2.16
1.2.3 C. a. The explanatory variable is the note-taking method and the re-
sponsevariableisthequizscore.
1.2.4 A.
b. Theeffectof takingnotesonpaperis0.71andtheeffectof taking
1.2.5 C.
notesonthecomputeris−0.71.
1.2.6 D.
c. SSModel=40×(0.712)=20.164.
1.2.7 B.
d. R2=20.164/120.92=0.16675.Wecaninterpretitbysayingthat
1.2.8 Using the effects model, because 4.48 + 0.65 = 5.13 (the mean 16.675%of thevariationof quizscoreisexplainedbythenote-taking
of thescentgroup)and4.48−0.65=3.83(themeanof thenon-scent method.
group),themodelsareequivalent.
e. 120.92–20.164=100.756.
1.2.9 ___________
100.756/38
f. √
=1.628.
{−0.71if using computer notes
a. SSModel. 0.71 if using paper notes
g. predicted
quiz score=6.21+ .
b. SSError.
1.2.17
1.2.10
a. Becausethesamplesizesof eachgrouparethesame,thesample
a. R2=SSModel/SSTotal=0.4651. sizeof eachgroupisjusthalf of thetotalsamplesize.
b. R2=1−SSError/SSTotal=0.7111.
∑ (x i−x̅)2 _____________
∑allobs(yi−y̅)2 _
( )2
b. _____________
allobs
n
+
1
1.2.11 _
−1 n −1
_
2 2
a. 8.
∑ (x −x) 2+∑allobs(y i−y̅ )2 _
( )2
allobs i ̅ _
= ____________________________ 1
b. 6–8=–2,10–8=2. n −1
2
c. 74.
=(____________________________ )
∑ (x −x) 2+∑allobs(y i−y̅)2
d. 40. allobs i ̅
n−2
e. 34. ________________________________________
∑allobs(x i−x̅) 2+∑allobs(yi−y̅ )2
f. 0.5405. √
Takingthesquarerootweget ______________________________________
n−2
1.2.12 ⎛ ⎞
⎜ ⎟
n n
∑(xi−x̅)2 ∑(yi−y̅)2
a. The explanatory variable is the type of testing environment; it 1 __________
Usesumfrom1ton:_ i=1
+__________
n −1
i=1 __
iscategorical. 2 __ ⎝ n −1 ⎠
2 2
b. Theresponsevariableisthetestscore;itisquantitative. ⎛ ⎞
⎜ ⎟
n n n n
∑(xi−x̅)2+∑(yi−y̅)2 ∑(xi−x̅)2+∑(yi−y̅)2
c. Thetwolevelsarequietenvironmentanddistractingenvironment.
1 _____________________
=_ i=1
i=1
=_____________________
i=1
i=1
1.2.13
2
⎝ n −1
__
⎠
n−2
2 _____________________
n n
√
a. SSTotal would probably be larger with these 10 subjects because ∑(xi−x̅) 2+∑(yi−y̅)2
withthewidevarietyof agestherewouldprobablybemorevariability _____________________
Takingthesquareroot,weget i=1 i=1
.
n−2
inthetestscores.
b. SSModelwouldprobablybethesamebecauseitwouldstillrepre-
sentthedifferencebetweentestingenvironments. Section 1.3
c. SSError would probably be larger because there would probably 1.3.1 D.
be more variability in the test scores within each group due to the 1.3.2 A.
variabilityinages.
1.3.3 D.
1.2.14 Thevarianceof thescoresinthedistractingenvironmentis2.5
1.3.4 A.
andthevarianceof thescoresinthedistractingenvironment
_
is6.The
squarerootof theaverageof thesetwovariancesis√4.25
= 2.06. The 1.3.5 A.
_
SSErroris34,sothestandarderrorof theresidualsis√34/8 =2.06. 1.3.6 The validity conditions are not met because the male sample
1.2.15 sizeissmallandthedistributionof thenumberof flip-flopsownedby
themalesisquiteskewedtotheright.
a. The explanatory variably is whether the name of the hurricane is
maleorfemaleandtheresponseistheperceivedrisklevel. 1.3.7
____________________
b. The effect of naming the hurricane Christina is 5.01 − 5.29 = (24.382+
a. √ 36.992)/2
=31.33.
−0.28 andtheeffectof namingthehurricaneChristopheris 5.57 − 92.16______________
− 60.34
b. t=__
___________________
=4.06.
5.29=0.28.TheSSModelis142(0.282)=11.1328. 31.33√
1/32+1/32
c01InstructorSolutions.indd 4 16/10/20 6:49
, Solutions to Exercises 5
c. Yes, there is strong evidence that average creativity is different be- 1.3.14
tween “rigid librarians” and “eccentric poets” because the t-statistic is a. The paper method mean is 6.92 points and the computer method
larger than 2. mean is 5.50 points, so the paper method tends to give a higher score.
1.3.8
_____________________
−0.71 if computer
a. √(24.24 2 + 38.78 2)/2 = 32.34. b. predicted quiz score = 6.21 + ,
{ 0.71 if paper
− 85.71____ = −1.69.
69.97______________
b. t = _________________
32.34√1/24 + 1/24 SE of residuals = 1.63.
c. There is not strong evidence that the average creativity measure
is different between biology and theater majors because the absolute c. Let μcomputer be the population quiz score when notes are taken
value of the t-statistic is larger than 2. using a computer, and similarly for μpaper. The hypotheses are H0:
μcomputer − μpaper = 0, that is, the long-run mean scores will be the same
1.3.9 Yes, there is strong evidence that the long-run average game du-
for both methods of note taking vs. Ha: μcomputer − μpaper ≠ 0, that is, the
ration differs between replacement and regular referees because the
mean scores will not be the same for the two methods of note taking.
difference in mean game length is 8.03 minutes and that value is way
out in the right tail of the null distribution. d. t = 2.27. Because this t-statistic is greater than 2, it appears there is
a statistically significant difference in the mean quiz scores between
1.3.10
196 .50______________
− 188.47 = 2.64. the two studying methods.
a. t = _____________________
14.47 √1/43 + 1/48 e. The t-statistic is far in the right tail of the null distribution.
b. Yes, there is strong evidence that the long-run average game du- f. Simulation-based p-value ≈ 0.006; theory-based p-value = 0.0086.
ration differs between replacement and regular referees because the
g. We have very strong evidence that there is a difference in the mean
t-statistic is larger than 2.
scores on this quiz between taking notes on computer and paper, with
1.3.11 the paper method having a higher mean score in the long run.
a. We would need 10 cards. 1.3.15
b. We would write the 10 scores on the cards. a. We are 95% confident that the mean score for the paper note-taking
c. After the cards are shuffled, randomly sort them in two piles of 5, method is between 0.3832 to 2.4668 points higher than the computer
labeling one pile D and the other pile Q. Calculate the mean of the note-taking method in the long run.
numbers on the cards in each pile and find and record the difference
b. Yes. Because the interval is completely positive we have evidence
in means (e.g., D − Q). Repeat this process many, many times to con-
that in the long run the paper-based method population mean is larg-
struct a null distribution of the difference in means.
er than the computer-based method population mean.
1.3.12
1.3.16
a. Christopher mean xChristopher
̅ = 5.57, Christina mean x̅Christina =
a. Let μMusicYes be the population memory score when people are
5.01, so Christopher tends to be perceived as the riskier name.
listening to music and similarly for μMusicNo. The hypotheses are
b. predicted hurricane risk H0: μMusicYes − μMusicNo = 0, that is, mean memory scores will be the
−0.28 if Christina same regardless of whether or not people are listening to music ver-
= 5.29 + , SE of residuals = 1.16.
{ 0.28 if Christopher sus HA: μMusicYes − μMusicNo < 0, that is, mean memory scores will be
the lower for people who are listening to music compared to those
c. Let μChristopher be the population average risk rating for hurricanes who aren’t.
given the name Christopher, and similarly for μChristina. The hypoth-
eses are H0: μChristopher − μChristina = 0, that is, mean perceived risk b. There is a lot of overlap between the distribution of the scores be-
ratings are the same regardless of whether the hurricane is named tween the two groups. It looks like the difference in sample means
Christopher or Christina name versus HA: μChristopher − μChristina ≠ 0, might not be significant.
that is, mean perceived risk ratings differ based on whether the hurri- c. t = –1.28. With |t| < 2, there does not appear to be a statistically
cane is named Christopher or Christina. significant difference in the mean scores between the two groups.
d. The applet shows t = 2.87. Because the t-statistic is greater than 2, d. The t-statistic is not in the tail of the distribution.
it looks like the difference in observed mean perceived risk ratings is e. Simulation-based p-value ≈ 0.111; Theory-based p-value = 0.1046.
statistically significant.
f. We do not have strong evidence that listening to music tends to
e. The t-statistic is far out in the right tail of the simulated
hinder people’s abilities to memorize words.
null distribution.
1.3.17
f. simulation p-value ≈ 0.006; theory p-value = 0.0048.
a. Whereas t-statistics and differences in means can be positive or
g. We have very strong evidence that the perceived hurricane threat
negative, the values of R2 are never negative. The larger the value
for the name Christopher is different (more specifically, larger) than
of R2, the bigger the difference between the two samples. There-
the perceived hurricane threat for the name Christina.
fore, when we want to find R2 values that are as extreme as our
1.3.13 observed, we always look at those that are equal to or larger than
a. We are 95% confident that the mean perceived threat rating for the the observed R2.
name Christopher is between 0.1747 and 0.9450 points higher than b. Using R2 as the statistic automatically does a two-sided test even
that for the name Christina, in the long run. though we are looking just in one direction. Therefore, the p-value
b. Yes, because the entire interval (for Christopher minus Christina) is about twice as large as it should be for testing whether music
is positive it shows the observed mean rating for Christopher is statis- tends to hinder people’s ability to memorize, and we should divide
tically significantly larger than that for Christina. it by 2.
,6 C HA PT E R 1 Sources of Variation
1.3.18 1.4.4 D.
a. Let μneutral be the population average amount of chili sauce used 1.4.5 A.
by those who play the neutral video game and similarly for μviolent. 1.4.6 D.
The hypotheses are H0: μneutral − μviolent = 0, that is, in the long run
1.4.7
the average amount of chili sauce used will be the same regardless
of which video game is played vs. Ha: μneutral − μviolent < 0, those who a. C.
play the neutral video game will select less chili on average than those b. A.
who play the violent video game. d. B.
b. Yes, the violent condition has some very large chili sauce amounts e. A.
compared to the neutral condition and their mean is 16.12 vs. a mean
of 9.06 for the neutral group. 1.4.8 B.
c. t = −2.96. Because |t| > 2 there appears to be a significant differ- 1.4.9 A.
ence in the amount of chili sauce used by the two groups. 1.4.10 B.
d. The observed t-statistic is far out in left the tail. 1.4.11 B.
e. Simulation-based p-value ≈ 0.004; theory-based p-value = 0.0019. 1.4.12 C.
f. We have very strong evidence that people tend to put more chili 1.4.13
sauce into the recipe (and thus be more aggressive) after they play a a. The F-statistic will increase and the p-value will decrease.
violent video game than when they play a non-violent one.
b. The F-statistic will decrease and the p-value will increase.
1.3.19
1.4.14
a. The SD should be around 0.37 which is a bit larger than 0.32. a. 4.
_ _ _
b. i. xnoscent = 4.52; xscent – xnoscent = 0.04. b. 93.
_ _ _
ii. xnoscent = 3.96; xscent – xnoscent = 1.04. c. 0.018.
iii. If the mean of the scent group is unusually large, the mean of d. 0.536.
the no scent group should be unusually small and the differ-
1.4.15 The F-statistic is much larger than 4, so there is strong evi-
ence in means should be unusually large.
dence that the groups are significantly different.
c. If we are forcing some of the simulated differences in means to be
unusually large (either positive or negative), we are making the vari- Source of Sum of Mean
ability of the null distribution (or the SD of the null) a bit larger than Variation DF Squares Squares F
in should be compared to what we should get when we are sampling Model 2 35.05 17.53 10.01
from independent populations.
Error 54 94.53 1.75
d. The SD should be around 0.31 which is very close to 0.32. 56 129.58 19.28
Total
e. i. Through shuffling, you should get two groups that are typical-
ly quite similar and hence should have similar means, on av- 1.4.16
erage. The difference in these two similar means should then a. There were 3 groups.
be zero, on average. Therefore, this type of null distribution
should be centered on zero. b. The total sample size was 81.
ii. If we are sampling from two independent populations, we Sums of Mean
should get two means that are typically close to the two pop- Source DF squares squares F
ulation means. Because our sample means are being used as Model 2 227.63 113.81 7.08
the estimates for the population means, on average, we should
Error 78 1,253.26 16.07
get our two sample means back when we resample. The dif-
ference in these should be the difference in our two sample Total 80 1,480.89 129.88
means, on average, or 1.292.
1.4.17
1.3.20
a. The response variable is the amount of money spent on meals
a. Only one combination would produce a result as extreme as
and the explanatory variable is the type of music playing. The ex-
−83.77, placing the nine largest times in one group and the nine
perimental units are the customers eating at the restaurant during
smallest times in the other group.
the study.
b. C(18,9) = 18!/(9!)2 = 48,620.
b. To compute the effects, we compare the group means to the LS
c. 1/48,620 ≈ 0.0000206. mean: (21.69 + 21.91 + 24.13)/3 = 22.576. The effect for no music is
d. The simulation-based, theory-based, and exact p-values are all –£0.886, for pop music is –£0.666, and for classical music is £1.554.
quite similar as the p-values are all extremely small. These numbers tell us how much each group mean is above or below
the overall mean.
Section 1.4 c. predicted amount of money spent
1.4.1 C.
−£0.886 if no music
⎪
1.4.2 E. = £22.58 + ⎨−£0.666 if pop music .
⎪
1.4.3 B. ⎩ £1.554 if classical music
, Solutions to Exercises 7
1.4.18 number of uses for those who imagine themselves eccentric poets is
a. To compute the sum of squares for the model, we compare the larger than the averages for the librarian and control groups.).
group means to the overall mean: SSModel = 131(21.69 − 22.52)2 1.4.22
+ 142(21.91 − 22.52)2 + 120(24.13 − 22.52)2 = 1454.14 (computer
a. From the graphs in the applets, the means of the groups appear
451.95); this is a measure of variability between the groups.
roughly the same and there is a lot of overlap between the four groups
b. SSTotal = SSModel + SSError = 454.14 + 3167.62 = 3,621.74 (com- so there does not appear to be strong evidence that at least one group
puter: 3619.57). mean differs from the rest.
c. R2 = 454.14/3,621.74 = 0.125; 12.5% of the variation in spending b. F = 0.536, R2 = 0.018. Although R2 is very small, it is not a stand-
can be attributed to the type of music playing. ardized statistic so we can best see that there are not significant results
d. F = (454.14/2)/(3,167.6/390) = 28.0 (computer 27.82); This is the based on the F-statistic which is much smaller than 4.
ratio of variation between the groups and the variation within the c. The simulated p-value using either the F-statistic or R2 is about
groups. Because the F-statistic is much larger than 4 these results are 0.66. This confirms that there is not much evidence of a difference in
significantly significant. the population means.
1.4.19 d. A large p-value is not strong evidence for the null so it is not
a. Let μn/p/c represent the population average amount spend by din- strong evidence that all the means are the same. It just means we
ers at this restaurant when listening to no, popular, or classical music, do not have strong evidence that there is at least one mean that
respectively. The hypotheses are H0: μn = μp = μc versus Ha: At least is different.
one μ differs from the others. 1.4.23
b. The validity conditions are met because the groups are independ- a.
ent, the sample distributions are fairly symmetric, the sample sizes _
are each very large, and the SDs are all close to each other, easily with- i. xA = 3.77% (SDA = 0.83).
_
in a factor of 2. ii. xB = 4.08% (SDB = 0.52).
_
c. F = 27.822. iii. xC = 5.10% (SDC = 0.87).
_
d. Both simulation-based and theory-based p-values are about 0. iv. xD = 5.65% (SDD = 0.45).
_
e. We have strong evidence that at least one population mean amount v. xE = 5.95% (SDE = 1.94).
differs from the others or that the type of music played has an effect b. Group E (>2 times per week) contained the high omega-3 value.
on the amount of money spent at the restaurant.
c. The larger mean for group E increases the variability between the
f. We can make a cause-and-effect conclusion because this was an groups (thus increasing F). The larger SD of group E will increase the
experiment. We can probably generalize to restaurants like the one variability within the groups (thus decreasing F). Because the addition
that was used with customers like those involved in the experiment. It of this value will both increase and decrease the F-statistic, it might be
would be difficult to generalize much beyond that. hard to determine which will have a greater effect. The new F is 4.467,
which is less than the one from Example 1.4, so the increased SD had
1.4.20
the greater effect.
a. The response variable is the number of uses generated for the
d. The new p-value should be about 0.006 and should be a little bit
items. The explanatory variable is whether they imagined themselves
larger than the one from Example 1.4.
as rigid librarians, eccentric poets, or neither. The experimental units
are the 96 subjects involved in the experiment. e. No, it is not valid to perform a theory-based test because the stand-
ard deviations of the different groups are not all within a factor of 2 of
b. The effect is −16.45 for the rigid librarians, 15.37 for the eccen-
each other. In particular, SDE/SDD ≈ 4.31.
tric poets, and 1.09 for the control group. These numbers tell us
how much each group mean is above or below the overall mean f. The theory-based p-value is 0.0081. It is similar to the simula-
of 76.79. tion-based p-value.
⎧−16.45 if rigid librarian g. The high omega-3 value did not make a difference in the
⎪
c. predicted number of uses = 76.79 + ⎨ 15.37 if eccentric poet . conclusions.
⎪
⎩ 1.09 if control 1.4.24
1.4.21 a. R2 = SSModel/SSTotal, 1 – R2 = 1 – (SSModel/SSTotal) = (SSTotal –
a. SSModel = 32(16.452) + 32(15.372) + 32(1.092) = 16256.88; this is a R2 n − k = (SSModel/SSTotal)/
SSModel)/SSTotal, so [_ × _
measure of the variability between the groups. 1 − R 2] [ k − 1 ]
b. SSTotal = SSModel + SSError = 109,240.9. n−k .
[1 – (SSTotal – SSModel)/SSTotal] × [_
k − 1]
c. R2 = SSModel/SSTotal = 0.149. This tells us that 15% of the varia-
tion in the number of uses for the items can be explained by what the SSModel/SSTotal n−k =
b. [____________________________] × [_
subject imagines themselves as. (SSTotal − SSModel)/SSTotal k − 1]
d. F = (16,256.88/2) / (92,984.01/93) = 8.13. This is the ratio of varia-
SSModel
_____________________ n−k
_ SSModel
_ n−k
_
tion between the groups and the variation within the groups. Because [ (SSTotal − SSModel) ] × [ k − 1 ] = [ SSError ] × [ k − 1 ]
this F-statistic is much larger than 4, we have very strong evidence
that at least one of the population mean number of uses for these SSModel × _ SSModel/(k − 1)
n − k = _________________
= [_ .
items is different from the others. (More specifically, the average k − 1 ] [ SSError ] [ SSError/(n − k ]
,8 C HA PT E R 1 Sources of Variation
1.4.25 For these data, MSModel is 40/1 = 40 and MSError = 34/8 = 1.5.8
10 − 6_
4.25, so the F-statistic = 40/4.25 = 9.41. The t-statistic = ______________________ a. A 95% confidence interval for the population average score is 6.3 ±
_ _
2 √ 4.25 × √ _ 1+_ 1
2 × (12.45/√27) ≈ 6.3 ± 4.79 = (1.51, 11.09); the validity conditions
= 3.068 and 3.068 = 9.41. 5 5
are met because the sample size is fairly large.
1.4.26 _ _ b. Yes, because the interval is completely positive, there is strong ev-
(x1 − x2) 2 idence that, on average, people tend to pick a face that is more attrac-
a. If we can show that MSModel = ________ then,
_1 _1 tive than their own when they are asked to identify their own face.
n1 + n2
_ _ 2 _ _ 1.5.9
(x1___________
− x2) (x1 − x2) 2 ___________
t2 = _____________ = ____________ = a. A 95% prediction interval is 6.3 ± 2 × (12.45) × √1 + 1/27 ≈ 6.3
1 1
( sp(√_
1 1
_ sp 2(_ _
n1 + n2 ) ) n1 + n2 ) ± 25.36 = (−19.06, 31.66); the validity conditions are met because we
_ _ were told the distribution of the results was fairly symmetric.
2
(x1 − x2)
_________________ MSModel = F .
=_ b. The prediction interval is trying to capture 95% of the individual
MSError _ 1 + _
1 MSError
( n1 n2 ) results in the long run while the confidence interval is trying to cap-
ture the average result in the long run.
_ _ _ _
b. MSModel = n1 (x1 − x) 2 + n2 (x2 − x) 2 1.5.10
_ _ 2 _ _ 2 a. The applet reports a p-value of 0.0000, so there is strong evidence
_ n1 x1 + n2 x2 _ n1 x1 + n2 x2
= n1 (x1 − __________ __________
n1 + n2 ) + n2 (x2 − n1 + n2 ) at least one type of background music results in a different long-
run mean amount of money spent; the validity conditions are met
_ _ 2 because the sample sizes are fairly large, and all four groups have
_ (n1 + n2) __________
n x +n x
= n1 (x1 ____________ − 1n1 + n2 2 ) + similar SD values.
n
( 1 + n 2) 1 2
_ _ 2 b. The 95% confidence intervals are Classical–Pop: (£1.52, £2.91),
_ (n1 + n2) __________
n x +n x
n2 (x2 ____________ − 1n1 + n2 2 ) Classical–None: (£1.73, £3.14), and Pop–None: (–£0.4590, £0.8985).
(n1 + n2) 1 2
c. We can be 95% confident that, on average, customers will spend
_ _ _ _ 2
n1 x1 + n2 x1 − n1 x1 − n2 x2
______________________
between £1.52 and £2.91 more per evening meal when classical music
= n1 ( n1 + n2 )+ is playing than when pop music is playing at the restaurant.
_ _ _ _ 2 d. The mean meal cost when classical music is playing is significantly
n1 x2 + n2 x2 − n1 x1 − n2 x2
n2 (______________________
n1 + n2 ) greater than when either pop or no music is playing.
_ _ 2 _ _ 2 e. Letters plot:
n2 x1 − n2 x2 n1 x2 − n1 x1
= n1 (___________
n +n ) + n 2 (
__________
n +n )
1 2 1 2
_ _ 2 _ _ 2 Music Group Mean Letters
x1 − x2 x2 − x1
= n1 n22 (_______
n1 + n2 ) + n 2 n 2 _______
1 ( n1 + n2 ) Classical £24.13 a
Pop £21.91 b
_ _ 2 _ _ 2
x1 − x2 x1 − x2
= (_______
n1 + n2
2 2
) (n1 n2 + n1 n2) = (
_______
) (n1 n2)(n1 + n2)
n1 + n2
None £21.69 b
_ _ _ _ _ _
(x1 − x2) 2 (x1 − x2) 2 (x1 − x2) 2 1.5.11
= _________ (n 1 n 2 )( n 1 + n 2 ) = ________
(n 1 n 2) = _________ .
(n1 + n2)
2 (n1 + n2) _n1 + n2 a. A 95% confidence interval for the long-run mean _ cost of a meal
( n1 n2 )
when no music is playing is £21.69 ± 2 × £3.38/√131 ≈ (£21.10,
Section 1.5 £22.28); the validity conditions are met because the sample size is
1.5.1 D. fairly large.
1.5.2 C. b. A 95% prediction interval for the long run ____________
cost of a meal when
1.5.3 C. no music is playing is £21.69 ± 2 × £3.38 × √1 + 1/131 ≈ (£14.90,
£28.48); the validity conditions are met because the histogram of
1.5.4 C. these data is fairly symmetric and bell-shaped.
1.5.5 A, F, H.
c. The prediction interval looks like it contains about 95% of the data
1.5.6 The margin of error is based on a prediction interval. The rang- (it actually contains 92%), whereas the confidence interval contains a
ers are not trying to predict the mean time for all future eruptions but much smaller percentage of the actual data.
are trying to predict the time of the next eruption so that visitors have
1.5.12
a high probability of seeing the eruption if they are present during the
a. The p-value = 0.0087, so there is strong evidence that at least one
entire interval.
mean is different from the others; the validity conditions are met be-
1.5.7 cause the sample sizes are fairly large, and the sample SDs are similar
a. Mean = 7.321 hrs and SD = 1.490 hrs. in value.
____________
b. An approximate prediction interval is (7.321 ± 2 × 1.49 × √1 + 1/100 ) b. The 95% confidence intervals are Lie − Truth: (−2.68, −0.61), Lie −
≈ 4.326 to 10.316 hr; the validity conditions are met because the data Control: (−1.97, 0.10), and Truth − Control: (−0.3267, 1.7460).
are quite symmetric and have no obvious outliers. c. We can be 95% confident that, in the long run, the mean difference in
c. Ninety-three percent of these data lie within the 95% prediction rating for the lie condition is between 0.61 to 2.68 points lower than that
interval. This is reasonably close to the 95% that we would expect. for the truth condition.
, Solutions to Exercises 9
d. The lie condition has a mean that is significantly less than the truth earnings; the validity conditions are met because the sample size is
condition. Nothing else is significantly different. fairly large and the SDs are all within a factor of 2 of each other.
e. Letters plot: c. Letters plot:
Condition Group mean Letters Estimated
Lie −0.90 A Education level group mean Letters
Control 0.03 AB Doctorate $97.40K A
Truth 0.74 B Master’s $66.00K B
Bachelor’s $55.20K B
1.5.13
Associate $36.80K C
a. A 95% confidence interval for the long run difference in
_
mean rat-
Some College $32.51K C
ings between bottled and tap water is 0.03 ± 2 × 1.975/√31 ≈ 0.03 ±
0.71 = (−0.679, 0.739); the validity conditions are met because the
sample size is fairly large. 1.5.16
b. A 95% prediction interval for the___________
difference in bottled and tap wa- a. A 95% confidence interval for the mean _
amount earned by those
ter ratings is 0.03 ± 2 × 1.975 × √1 + 1/31 ≈ 0.03 ± 4.01 = (−3.98, with doctorates is $97.4K ± 2 × $40.5K/√50 ≈ $97.4K ± $11.455K =
4.04); the validity conditions are met because the dotolot of these data ($85.94K, $108.86K); the validity conditions are met because the sam-
is fairly symmetric and bell-shaped. ple size is fairly large.
c. The prediction interval looks like it contains about 95% of the data b. A 95% prediction interval for the amount
___________earned by those with
(it actually contains 30/31 = 96.8%), whereas the confidence interval doctorates is $97.4K ± 2 × $40.5K × √1 + 1/50 ≈ $97.4K ± $81.81K
contains a much smaller percentage. = ($15.59K, $179.21K); the validity conditions may not be met in this
case because the distribution appears to be skewed to the right with a
1.5.14
5.50 if computer few large outliers.
a. predicted quiz score = , SE of residuals = 1.63.
{6.92 if paper c. There are 46/50 or 92% of those with doctoral degrees in this sam-
ple contained in the prediction interval.
b. A 95% confidence interval
_
for the long-run mean score using paper
notes is 6.92 ± 2 × 1.07/√20 ≈ 6.92 ± 0.384 = (6.44 to 7.40). 1.5.17
−0.713 if computer a. A 95% prediction interval for the mean amount earned by those
___________
c. predicted quiz score = 6.213 + , with associate degrees is $36.8K ± 2 × $28.5K × √1 + 1/50 ≈ $36.8K
{ 0.713 if paper
SE of residuals = 1.63. ± $57.57K = (−$20.77K, $94.37K); the validity conditions are suspect
because the distribution looks skewed right.
d. A 95% confidence interval for the
_
long-run mean effect when using
paper notes is 0.71 ± 2 × 1.07/√20 = 0.71 ± 0.384 = (0.23, 1.19). We b. There are 49/50 or 98% of these data within the prediction interval.
can use the same standard deviation because the distribution of ef- c. This is such a bad fit because the distribution of salaries is highly
fects is the same as the distribution of scores but slid down 6.21 units. skewed to the right. This method is only valid when we have a bell-
1.5.15 shaped distribution.
a. d. The concerns aren’t as great for a confidence interval. Even though
the distribution is skewed, the associated sampling distribution
should be quite symmetric with a sample size as large as 50.
Some Asso Bach Mast Doct
1.5.18
_ _ _ _ _
a. Each margin of error is 2s/√n, so y2 − y1 = 2 × 2s/√n = 4s/√n.
____________ _ _ _
b. The margin of error is 2s √1/n + 1/n = 2s √2/n = 2 √2 s/√n.
Degree
_
c. With 4 > 2 √2, the answer to part (a) is larger than part (b). Both
_ _
of the answers represent y2 − y1 but only the confidence interval for
the difference in means uses the correctly pooled SE in the margin
of error expression. If the individual means were just a tiny bit
closer together the single mean intervals from part (a) would over-
0 100 200 lap, however the difference in means interval from part (b) would be
Earnings ($K) completely positive.
Section 1.6
Education level Group mean Group SD
1.6.1 A, C, D.
Doctorate $97.40K $40.50K
1.6.2 A.
Master’s $66.00K $38.40K
1.6.3 C.
Bachelor’s $55.20K $32.20K
1.6.4
Associate $36.80K $28.50K
a Increase alpha level.
Some College $32.51K $20.80K
b. Increase sample size.
b. The F-statistic is 31.534 and the p-value is < 0.0001, so there is c. Decrease number of groups comparing.
strong evidence of an association between the levels of education and d. Decrease variability within each group.
, 10 C HA PT E R 1 Sources of Variation
1.6.5 f. ≈ 0.870.
a. Just under 20. g. As the effect size (difference in mean heart rates) increases, power
b. Just under 40. of the test increases.
c. Just under 50.
Difference in mean
d. Just under 60.
heart rates (bpm) 5 10 15 20 25
e. Just over 75.
Power 0.395 0.870 0.995 1.000 1.000
f. As sample size per group increases, power of the test increases line-
arly at first, but then plateaus. (The relationship looks logarithmic, or a
1.6.12
power between 0 and 1.).
a. Now the rejection region should be a difference in means of about
1.6.6
8.2 or more. In this case, the power is roughly 0.189.
a. This will decrease the power of the test.
b. Using a significance level of 0.10 would increase the power.
b. Power = 0.73.
c. Now the rejection region is about ≈ 0.503.
c. The power would be very close to 1.
d. As level of significance increases, the power of the test increases.
d. The relationship between power and sample size looks linear for
most values of the sample size, with no plateauing visible even with Level of significance 0.001 0.01 0.05 0.10 0.15
each sample being even as large as 120.
Power 0.028 0.189 0.431 0.503 0.611
1.6.7
a. 1 ≤ SD ≤ 3. 1.6.13
b. About 5. a. About 0.238.
c. 4 ≤ SD ≤ 4.5. b. Increase power.
d. A little more than 5.5. c. About 0.510.
e. As SD increases, power decreases. For very small SDs the power d. As number per group increases, power of the test increases.
of the test is one, then it begins to decrease somewhat linearly as the
SDs increase. Sample size per
1.6.8 group 5 10 15 20 25
a. 81%. Power 0.238 0.395 0.510 0.644 0.672
b. 59.1%.
1.6.14
c. 0.10 < α < 0.15.
a. About 0.88.
d. As the level of significance increases, the power of the test increas-
b. Power will decrease.
es. Power doesn’t increase linearly, but in steps.
c. About 0.253.
1.6.9 A difference between 15 and 20 mL/d.
d. As SD within each group increases, power of the test decreases.
1.6.10
a. The differences in the group means and overall mean are the same Standard deviation
in Scenarios 1 and 2. per group 4 8 12 16 20
b. There is greater variability within the groups in Scenario 1. 0.880 0.427 0.253 0.201 0.153
Power
c. Scenario 2 will have the larger F-statistic.
d. Scenario 2 will be more likely to have a statistically
significant result. End of Chapter 1 Exercises
e. As the variability within the groups decreases, the F-statistic in- 1.CE.1.
creases, as does the power of the test. a. H0: μregular = μfilled and Ha: μregular ≠ μfilled; p-value = 0.003. Because
1.6.11 the p-value is less than 0.01, there is very strong evidence against the
null and for the alternative that there is an association between the
a. The rejection region is any difference in means of 5.9 or greater.
type of soup bowl and the amount of soup consumed with the secretly
b. A difference in mean heart rates of 7 bpm will be in the rejection refilled soup bowl resulting in a higher average consumption of
region so you would conclude that the two treatment means are sig- soup (oz).
nificantly different from each other.
b. The samples are independent of each other (randomly assigned
c. A difference of 4 pbm is not in the rejection region, so you would to bowl type), sample sizes are greater than 20 and there is no strong
conclude it is plausible that the two treatment means do not differ skewness in the data.
from each other.
c. Theory-based two-sided p-value is 0.0032, which is very close to
d. P(Type I error) = 0.05. the simulation-based p-value. This is expected as validity conditions
e. ≈ 0.395. were met to perform the theory-based test.
Investigations 1e Tintle
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, CHAPTER 1
Sources of Variation
Section 1.1 1.1.10 Color of asignistheexplanatoryvariablewithwhite,yellow,
andredbeingthelevels.
1.1.1 B.
1.1.11
1.1.2 B & C.
1.1.3 A. Observed Sources of Sources of
1.1.4 C. Variation in: explained unexplained
1.1.5 E. f.whetherthestudent variation variation
obeyedthesign
1.1.6 B.
{ 92.19if eccentric poet
60.34if rigid librarian Inclusion criteria a.colorof the b.whetherthesubject
1.1.7 predicted number of uses for items=
• c.timeof day sign wasleft-handedor
1.1.8 right-handed
• e.ageof subject
a. The inclusion criteria are having a clinical diagnosis of mild to d.attitudeof student
moderate depression without any treatment four weeks prior and e.ageof subject
duringthestudy.
1.1.12
b. The purpose of randomly assigning subjects to the groups is to
makegroupsverysimilarexceptfortheonevariable(swimmingwith a. Thevalue6.21representstheoverallmeanquizscore,5.50represents
dolphinsornot)thattheresearchersimpose.Volunteeringforagroup the group mean quiz score for people who used computer notes, and
couldintroduceaconfoundingvariable. 6.92representsthegroupmeanscoreforpeoplewhousedpapernotes.
c. Itwasimportantthatthesubjectsinthecontrolgroupswimevery b. Welooktoseehowfar6.92and5.50arefromoneanotherorfrom
day withoutdolphinssothatthiscontrolgroupdoeseverything(in- theoverallmeanof 6.21todeterminewhetherthenote-takingmethod
cluding swimming) that the experimental group does except that mightaffectthescore.
whentheyswimtheydon’tdoitinthepresenceof dolphins.Without c. The number 1.76 represents the typical deviation of an observa-
thiswewouldn’tknowwhetherjustswimmingcausesthedifference tionfromtheexpectedvalue,inthiscase,fromtheoverallmean.The
inthereductionof depressionsymptoms. number1.61representsthetypicaldeviationof anobservationafter
d. Yes,thisisanexperimentbecausethesubjectswererandomlyas- creatingamodelthattakesintoaccountwhetherthepersonisusing
signedtothetwogroups. computerorpapernotes.
1.1.9. d. Becausethestandarddeviationof theresidualsrepresentstheleft-
overvariation,wecanseethatafterincludingthetypeof notesasan
Observed variation Sources of Sources of explanatoryvariableinourmodeltheunexplainedvariationhasbeen
in: explained unexplained reduced(downto1.61from1.76).Thistellsusthatknowingthetype
d.substantialreduction variation variation of note-takingmethodenablesustobetterpredictscores.
indepressionsymptoms 1.1.13 Random assignment should make the two groups very
similar with regard to variables like intelligence, previous knowl-
Inclusion criteria a.swimmingwith • g.problemsinthe edge, or any other variable and thus likely eliminate possible
• b.mildtomoderate dolphinsornot personallivesof confoundingvariables.
depression thesubjectsduring
thestudy 1.1.14
• c.nouseof
antidepressantdrugs • h.illnessof a. This table shows us possible confounding variables but then
orpsychotherapyfour subjectsduring shows that subjects in the two groups are quite similar with
weekspriortothe thestudy regard to these characteristics, thus ruling out these possible
study confoundingvariables.
Design
b. We would want the p-values to be large, so we could say that
• e.swimming we havelittletonoevidencethatthereisadifferenceinmeanage,
• f.stayingonanisland
proportion of males,etc.betweenthetwogroups.Wewantourgroups
fortwoweeksduring tobeverysimilargoingintothestudy,soacausalconclusionispossi-
thestudy
bleif wefindasmallp-valueafterapplyingthetreatment(s).
3
c01InstructorSolutions.indd 3 16/10/20 6:49
, 4 C HA PTER 1 Sources of Variation
1.1.15 Itislikelythat3-to5-year-oldsmighthavedifferentpreferenc- c. R2=11.1328/199.62=0.0558.Wecaninterpretthisbysayingthat
es when it comes to toy or candy than 12- to 14-year-olds. The older 5.58% of the variation in the perceived level of risk is explained by
group is probably much more likely to prefer the candy over the toy whetherthenameof thehurricaneismaleorfemale.
andtheoppositecouldbetruewiththeyoungergroup.Wewouldnot d. SSError=199.62−11.13=188.49.
seethisdifferenceif theresultsof alltheagesarecombinedtogether. _____________
e. √
188.4872/140
=1.16.
{−0.28iffemalename
0.28ifmalename
Section 1.2 f. predicted
hurricaneriskrating=5.29+
,
1.2.1 B. SEofresiduals=1.16.
1.2.2 A, D. 1.2.16
1.2.3 C. a. The explanatory variable is the note-taking method and the re-
sponsevariableisthequizscore.
1.2.4 A.
b. Theeffectof takingnotesonpaperis0.71andtheeffectof taking
1.2.5 C.
notesonthecomputeris−0.71.
1.2.6 D.
c. SSModel=40×(0.712)=20.164.
1.2.7 B.
d. R2=20.164/120.92=0.16675.Wecaninterpretitbysayingthat
1.2.8 Using the effects model, because 4.48 + 0.65 = 5.13 (the mean 16.675%of thevariationof quizscoreisexplainedbythenote-taking
of thescentgroup)and4.48−0.65=3.83(themeanof thenon-scent method.
group),themodelsareequivalent.
e. 120.92–20.164=100.756.
1.2.9 ___________
100.756/38
f. √
=1.628.
{−0.71if using computer notes
a. SSModel. 0.71 if using paper notes
g. predicted
quiz score=6.21+ .
b. SSError.
1.2.17
1.2.10
a. Becausethesamplesizesof eachgrouparethesame,thesample
a. R2=SSModel/SSTotal=0.4651. sizeof eachgroupisjusthalf of thetotalsamplesize.
b. R2=1−SSError/SSTotal=0.7111.
∑ (x i−x̅)2 _____________
∑allobs(yi−y̅)2 _
( )2
b. _____________
allobs
n
+
1
1.2.11 _
−1 n −1
_
2 2
a. 8.
∑ (x −x) 2+∑allobs(y i−y̅ )2 _
( )2
allobs i ̅ _
= ____________________________ 1
b. 6–8=–2,10–8=2. n −1
2
c. 74.
=(____________________________ )
∑ (x −x) 2+∑allobs(y i−y̅)2
d. 40. allobs i ̅
n−2
e. 34. ________________________________________
∑allobs(x i−x̅) 2+∑allobs(yi−y̅ )2
f. 0.5405. √
Takingthesquarerootweget ______________________________________
n−2
1.2.12 ⎛ ⎞
⎜ ⎟
n n
∑(xi−x̅)2 ∑(yi−y̅)2
a. The explanatory variable is the type of testing environment; it 1 __________
Usesumfrom1ton:_ i=1
+__________
n −1
i=1 __
iscategorical. 2 __ ⎝ n −1 ⎠
2 2
b. Theresponsevariableisthetestscore;itisquantitative. ⎛ ⎞
⎜ ⎟
n n n n
∑(xi−x̅)2+∑(yi−y̅)2 ∑(xi−x̅)2+∑(yi−y̅)2
c. Thetwolevelsarequietenvironmentanddistractingenvironment.
1 _____________________
=_ i=1
i=1
=_____________________
i=1
i=1
1.2.13
2
⎝ n −1
__
⎠
n−2
2 _____________________
n n
√
a. SSTotal would probably be larger with these 10 subjects because ∑(xi−x̅) 2+∑(yi−y̅)2
withthewidevarietyof agestherewouldprobablybemorevariability _____________________
Takingthesquareroot,weget i=1 i=1
.
n−2
inthetestscores.
b. SSModelwouldprobablybethesamebecauseitwouldstillrepre-
sentthedifferencebetweentestingenvironments. Section 1.3
c. SSError would probably be larger because there would probably 1.3.1 D.
be more variability in the test scores within each group due to the 1.3.2 A.
variabilityinages.
1.3.3 D.
1.2.14 Thevarianceof thescoresinthedistractingenvironmentis2.5
1.3.4 A.
andthevarianceof thescoresinthedistractingenvironment
_
is6.The
squarerootof theaverageof thesetwovariancesis√4.25
= 2.06. The 1.3.5 A.
_
SSErroris34,sothestandarderrorof theresidualsis√34/8 =2.06. 1.3.6 The validity conditions are not met because the male sample
1.2.15 sizeissmallandthedistributionof thenumberof flip-flopsownedby
themalesisquiteskewedtotheright.
a. The explanatory variably is whether the name of the hurricane is
maleorfemaleandtheresponseistheperceivedrisklevel. 1.3.7
____________________
b. The effect of naming the hurricane Christina is 5.01 − 5.29 = (24.382+
a. √ 36.992)/2
=31.33.
−0.28 andtheeffectof namingthehurricaneChristopheris 5.57 − 92.16______________
− 60.34
b. t=__
___________________
=4.06.
5.29=0.28.TheSSModelis142(0.282)=11.1328. 31.33√
1/32+1/32
c01InstructorSolutions.indd 4 16/10/20 6:49
, Solutions to Exercises 5
c. Yes, there is strong evidence that average creativity is different be- 1.3.14
tween “rigid librarians” and “eccentric poets” because the t-statistic is a. The paper method mean is 6.92 points and the computer method
larger than 2. mean is 5.50 points, so the paper method tends to give a higher score.
1.3.8
_____________________
−0.71 if computer
a. √(24.24 2 + 38.78 2)/2 = 32.34. b. predicted quiz score = 6.21 + ,
{ 0.71 if paper
− 85.71____ = −1.69.
69.97______________
b. t = _________________
32.34√1/24 + 1/24 SE of residuals = 1.63.
c. There is not strong evidence that the average creativity measure
is different between biology and theater majors because the absolute c. Let μcomputer be the population quiz score when notes are taken
value of the t-statistic is larger than 2. using a computer, and similarly for μpaper. The hypotheses are H0:
μcomputer − μpaper = 0, that is, the long-run mean scores will be the same
1.3.9 Yes, there is strong evidence that the long-run average game du-
for both methods of note taking vs. Ha: μcomputer − μpaper ≠ 0, that is, the
ration differs between replacement and regular referees because the
mean scores will not be the same for the two methods of note taking.
difference in mean game length is 8.03 minutes and that value is way
out in the right tail of the null distribution. d. t = 2.27. Because this t-statistic is greater than 2, it appears there is
a statistically significant difference in the mean quiz scores between
1.3.10
196 .50______________
− 188.47 = 2.64. the two studying methods.
a. t = _____________________
14.47 √1/43 + 1/48 e. The t-statistic is far in the right tail of the null distribution.
b. Yes, there is strong evidence that the long-run average game du- f. Simulation-based p-value ≈ 0.006; theory-based p-value = 0.0086.
ration differs between replacement and regular referees because the
g. We have very strong evidence that there is a difference in the mean
t-statistic is larger than 2.
scores on this quiz between taking notes on computer and paper, with
1.3.11 the paper method having a higher mean score in the long run.
a. We would need 10 cards. 1.3.15
b. We would write the 10 scores on the cards. a. We are 95% confident that the mean score for the paper note-taking
c. After the cards are shuffled, randomly sort them in two piles of 5, method is between 0.3832 to 2.4668 points higher than the computer
labeling one pile D and the other pile Q. Calculate the mean of the note-taking method in the long run.
numbers on the cards in each pile and find and record the difference
b. Yes. Because the interval is completely positive we have evidence
in means (e.g., D − Q). Repeat this process many, many times to con-
that in the long run the paper-based method population mean is larg-
struct a null distribution of the difference in means.
er than the computer-based method population mean.
1.3.12
1.3.16
a. Christopher mean xChristopher
̅ = 5.57, Christina mean x̅Christina =
a. Let μMusicYes be the population memory score when people are
5.01, so Christopher tends to be perceived as the riskier name.
listening to music and similarly for μMusicNo. The hypotheses are
b. predicted hurricane risk H0: μMusicYes − μMusicNo = 0, that is, mean memory scores will be the
−0.28 if Christina same regardless of whether or not people are listening to music ver-
= 5.29 + , SE of residuals = 1.16.
{ 0.28 if Christopher sus HA: μMusicYes − μMusicNo < 0, that is, mean memory scores will be
the lower for people who are listening to music compared to those
c. Let μChristopher be the population average risk rating for hurricanes who aren’t.
given the name Christopher, and similarly for μChristina. The hypoth-
eses are H0: μChristopher − μChristina = 0, that is, mean perceived risk b. There is a lot of overlap between the distribution of the scores be-
ratings are the same regardless of whether the hurricane is named tween the two groups. It looks like the difference in sample means
Christopher or Christina name versus HA: μChristopher − μChristina ≠ 0, might not be significant.
that is, mean perceived risk ratings differ based on whether the hurri- c. t = –1.28. With |t| < 2, there does not appear to be a statistically
cane is named Christopher or Christina. significant difference in the mean scores between the two groups.
d. The applet shows t = 2.87. Because the t-statistic is greater than 2, d. The t-statistic is not in the tail of the distribution.
it looks like the difference in observed mean perceived risk ratings is e. Simulation-based p-value ≈ 0.111; Theory-based p-value = 0.1046.
statistically significant.
f. We do not have strong evidence that listening to music tends to
e. The t-statistic is far out in the right tail of the simulated
hinder people’s abilities to memorize words.
null distribution.
1.3.17
f. simulation p-value ≈ 0.006; theory p-value = 0.0048.
a. Whereas t-statistics and differences in means can be positive or
g. We have very strong evidence that the perceived hurricane threat
negative, the values of R2 are never negative. The larger the value
for the name Christopher is different (more specifically, larger) than
of R2, the bigger the difference between the two samples. There-
the perceived hurricane threat for the name Christina.
fore, when we want to find R2 values that are as extreme as our
1.3.13 observed, we always look at those that are equal to or larger than
a. We are 95% confident that the mean perceived threat rating for the the observed R2.
name Christopher is between 0.1747 and 0.9450 points higher than b. Using R2 as the statistic automatically does a two-sided test even
that for the name Christina, in the long run. though we are looking just in one direction. Therefore, the p-value
b. Yes, because the entire interval (for Christopher minus Christina) is about twice as large as it should be for testing whether music
is positive it shows the observed mean rating for Christopher is statis- tends to hinder people’s ability to memorize, and we should divide
tically significantly larger than that for Christina. it by 2.
,6 C HA PT E R 1 Sources of Variation
1.3.18 1.4.4 D.
a. Let μneutral be the population average amount of chili sauce used 1.4.5 A.
by those who play the neutral video game and similarly for μviolent. 1.4.6 D.
The hypotheses are H0: μneutral − μviolent = 0, that is, in the long run
1.4.7
the average amount of chili sauce used will be the same regardless
of which video game is played vs. Ha: μneutral − μviolent < 0, those who a. C.
play the neutral video game will select less chili on average than those b. A.
who play the violent video game. d. B.
b. Yes, the violent condition has some very large chili sauce amounts e. A.
compared to the neutral condition and their mean is 16.12 vs. a mean
of 9.06 for the neutral group. 1.4.8 B.
c. t = −2.96. Because |t| > 2 there appears to be a significant differ- 1.4.9 A.
ence in the amount of chili sauce used by the two groups. 1.4.10 B.
d. The observed t-statistic is far out in left the tail. 1.4.11 B.
e. Simulation-based p-value ≈ 0.004; theory-based p-value = 0.0019. 1.4.12 C.
f. We have very strong evidence that people tend to put more chili 1.4.13
sauce into the recipe (and thus be more aggressive) after they play a a. The F-statistic will increase and the p-value will decrease.
violent video game than when they play a non-violent one.
b. The F-statistic will decrease and the p-value will increase.
1.3.19
1.4.14
a. The SD should be around 0.37 which is a bit larger than 0.32. a. 4.
_ _ _
b. i. xnoscent = 4.52; xscent – xnoscent = 0.04. b. 93.
_ _ _
ii. xnoscent = 3.96; xscent – xnoscent = 1.04. c. 0.018.
iii. If the mean of the scent group is unusually large, the mean of d. 0.536.
the no scent group should be unusually small and the differ-
1.4.15 The F-statistic is much larger than 4, so there is strong evi-
ence in means should be unusually large.
dence that the groups are significantly different.
c. If we are forcing some of the simulated differences in means to be
unusually large (either positive or negative), we are making the vari- Source of Sum of Mean
ability of the null distribution (or the SD of the null) a bit larger than Variation DF Squares Squares F
in should be compared to what we should get when we are sampling Model 2 35.05 17.53 10.01
from independent populations.
Error 54 94.53 1.75
d. The SD should be around 0.31 which is very close to 0.32. 56 129.58 19.28
Total
e. i. Through shuffling, you should get two groups that are typical-
ly quite similar and hence should have similar means, on av- 1.4.16
erage. The difference in these two similar means should then a. There were 3 groups.
be zero, on average. Therefore, this type of null distribution
should be centered on zero. b. The total sample size was 81.
ii. If we are sampling from two independent populations, we Sums of Mean
should get two means that are typically close to the two pop- Source DF squares squares F
ulation means. Because our sample means are being used as Model 2 227.63 113.81 7.08
the estimates for the population means, on average, we should
Error 78 1,253.26 16.07
get our two sample means back when we resample. The dif-
ference in these should be the difference in our two sample Total 80 1,480.89 129.88
means, on average, or 1.292.
1.4.17
1.3.20
a. The response variable is the amount of money spent on meals
a. Only one combination would produce a result as extreme as
and the explanatory variable is the type of music playing. The ex-
−83.77, placing the nine largest times in one group and the nine
perimental units are the customers eating at the restaurant during
smallest times in the other group.
the study.
b. C(18,9) = 18!/(9!)2 = 48,620.
b. To compute the effects, we compare the group means to the LS
c. 1/48,620 ≈ 0.0000206. mean: (21.69 + 21.91 + 24.13)/3 = 22.576. The effect for no music is
d. The simulation-based, theory-based, and exact p-values are all –£0.886, for pop music is –£0.666, and for classical music is £1.554.
quite similar as the p-values are all extremely small. These numbers tell us how much each group mean is above or below
the overall mean.
Section 1.4 c. predicted amount of money spent
1.4.1 C.
−£0.886 if no music
⎪
1.4.2 E. = £22.58 + ⎨−£0.666 if pop music .
⎪
1.4.3 B. ⎩ £1.554 if classical music
, Solutions to Exercises 7
1.4.18 number of uses for those who imagine themselves eccentric poets is
a. To compute the sum of squares for the model, we compare the larger than the averages for the librarian and control groups.).
group means to the overall mean: SSModel = 131(21.69 − 22.52)2 1.4.22
+ 142(21.91 − 22.52)2 + 120(24.13 − 22.52)2 = 1454.14 (computer
a. From the graphs in the applets, the means of the groups appear
451.95); this is a measure of variability between the groups.
roughly the same and there is a lot of overlap between the four groups
b. SSTotal = SSModel + SSError = 454.14 + 3167.62 = 3,621.74 (com- so there does not appear to be strong evidence that at least one group
puter: 3619.57). mean differs from the rest.
c. R2 = 454.14/3,621.74 = 0.125; 12.5% of the variation in spending b. F = 0.536, R2 = 0.018. Although R2 is very small, it is not a stand-
can be attributed to the type of music playing. ardized statistic so we can best see that there are not significant results
d. F = (454.14/2)/(3,167.6/390) = 28.0 (computer 27.82); This is the based on the F-statistic which is much smaller than 4.
ratio of variation between the groups and the variation within the c. The simulated p-value using either the F-statistic or R2 is about
groups. Because the F-statistic is much larger than 4 these results are 0.66. This confirms that there is not much evidence of a difference in
significantly significant. the population means.
1.4.19 d. A large p-value is not strong evidence for the null so it is not
a. Let μn/p/c represent the population average amount spend by din- strong evidence that all the means are the same. It just means we
ers at this restaurant when listening to no, popular, or classical music, do not have strong evidence that there is at least one mean that
respectively. The hypotheses are H0: μn = μp = μc versus Ha: At least is different.
one μ differs from the others. 1.4.23
b. The validity conditions are met because the groups are independ- a.
ent, the sample distributions are fairly symmetric, the sample sizes _
are each very large, and the SDs are all close to each other, easily with- i. xA = 3.77% (SDA = 0.83).
_
in a factor of 2. ii. xB = 4.08% (SDB = 0.52).
_
c. F = 27.822. iii. xC = 5.10% (SDC = 0.87).
_
d. Both simulation-based and theory-based p-values are about 0. iv. xD = 5.65% (SDD = 0.45).
_
e. We have strong evidence that at least one population mean amount v. xE = 5.95% (SDE = 1.94).
differs from the others or that the type of music played has an effect b. Group E (>2 times per week) contained the high omega-3 value.
on the amount of money spent at the restaurant.
c. The larger mean for group E increases the variability between the
f. We can make a cause-and-effect conclusion because this was an groups (thus increasing F). The larger SD of group E will increase the
experiment. We can probably generalize to restaurants like the one variability within the groups (thus decreasing F). Because the addition
that was used with customers like those involved in the experiment. It of this value will both increase and decrease the F-statistic, it might be
would be difficult to generalize much beyond that. hard to determine which will have a greater effect. The new F is 4.467,
which is less than the one from Example 1.4, so the increased SD had
1.4.20
the greater effect.
a. The response variable is the number of uses generated for the
d. The new p-value should be about 0.006 and should be a little bit
items. The explanatory variable is whether they imagined themselves
larger than the one from Example 1.4.
as rigid librarians, eccentric poets, or neither. The experimental units
are the 96 subjects involved in the experiment. e. No, it is not valid to perform a theory-based test because the stand-
ard deviations of the different groups are not all within a factor of 2 of
b. The effect is −16.45 for the rigid librarians, 15.37 for the eccen-
each other. In particular, SDE/SDD ≈ 4.31.
tric poets, and 1.09 for the control group. These numbers tell us
how much each group mean is above or below the overall mean f. The theory-based p-value is 0.0081. It is similar to the simula-
of 76.79. tion-based p-value.
⎧−16.45 if rigid librarian g. The high omega-3 value did not make a difference in the
⎪
c. predicted number of uses = 76.79 + ⎨ 15.37 if eccentric poet . conclusions.
⎪
⎩ 1.09 if control 1.4.24
1.4.21 a. R2 = SSModel/SSTotal, 1 – R2 = 1 – (SSModel/SSTotal) = (SSTotal –
a. SSModel = 32(16.452) + 32(15.372) + 32(1.092) = 16256.88; this is a R2 n − k = (SSModel/SSTotal)/
SSModel)/SSTotal, so [_ × _
measure of the variability between the groups. 1 − R 2] [ k − 1 ]
b. SSTotal = SSModel + SSError = 109,240.9. n−k .
[1 – (SSTotal – SSModel)/SSTotal] × [_
k − 1]
c. R2 = SSModel/SSTotal = 0.149. This tells us that 15% of the varia-
tion in the number of uses for the items can be explained by what the SSModel/SSTotal n−k =
b. [____________________________] × [_
subject imagines themselves as. (SSTotal − SSModel)/SSTotal k − 1]
d. F = (16,256.88/2) / (92,984.01/93) = 8.13. This is the ratio of varia-
SSModel
_____________________ n−k
_ SSModel
_ n−k
_
tion between the groups and the variation within the groups. Because [ (SSTotal − SSModel) ] × [ k − 1 ] = [ SSError ] × [ k − 1 ]
this F-statistic is much larger than 4, we have very strong evidence
that at least one of the population mean number of uses for these SSModel × _ SSModel/(k − 1)
n − k = _________________
= [_ .
items is different from the others. (More specifically, the average k − 1 ] [ SSError ] [ SSError/(n − k ]
,8 C HA PT E R 1 Sources of Variation
1.4.25 For these data, MSModel is 40/1 = 40 and MSError = 34/8 = 1.5.8
10 − 6_
4.25, so the F-statistic = 40/4.25 = 9.41. The t-statistic = ______________________ a. A 95% confidence interval for the population average score is 6.3 ±
_ _
2 √ 4.25 × √ _ 1+_ 1
2 × (12.45/√27) ≈ 6.3 ± 4.79 = (1.51, 11.09); the validity conditions
= 3.068 and 3.068 = 9.41. 5 5
are met because the sample size is fairly large.
1.4.26 _ _ b. Yes, because the interval is completely positive, there is strong ev-
(x1 − x2) 2 idence that, on average, people tend to pick a face that is more attrac-
a. If we can show that MSModel = ________ then,
_1 _1 tive than their own when they are asked to identify their own face.
n1 + n2
_ _ 2 _ _ 1.5.9
(x1___________
− x2) (x1 − x2) 2 ___________
t2 = _____________ = ____________ = a. A 95% prediction interval is 6.3 ± 2 × (12.45) × √1 + 1/27 ≈ 6.3
1 1
( sp(√_
1 1
_ sp 2(_ _
n1 + n2 ) ) n1 + n2 ) ± 25.36 = (−19.06, 31.66); the validity conditions are met because we
_ _ were told the distribution of the results was fairly symmetric.
2
(x1 − x2)
_________________ MSModel = F .
=_ b. The prediction interval is trying to capture 95% of the individual
MSError _ 1 + _
1 MSError
( n1 n2 ) results in the long run while the confidence interval is trying to cap-
ture the average result in the long run.
_ _ _ _
b. MSModel = n1 (x1 − x) 2 + n2 (x2 − x) 2 1.5.10
_ _ 2 _ _ 2 a. The applet reports a p-value of 0.0000, so there is strong evidence
_ n1 x1 + n2 x2 _ n1 x1 + n2 x2
= n1 (x1 − __________ __________
n1 + n2 ) + n2 (x2 − n1 + n2 ) at least one type of background music results in a different long-
run mean amount of money spent; the validity conditions are met
_ _ 2 because the sample sizes are fairly large, and all four groups have
_ (n1 + n2) __________
n x +n x
= n1 (x1 ____________ − 1n1 + n2 2 ) + similar SD values.
n
( 1 + n 2) 1 2
_ _ 2 b. The 95% confidence intervals are Classical–Pop: (£1.52, £2.91),
_ (n1 + n2) __________
n x +n x
n2 (x2 ____________ − 1n1 + n2 2 ) Classical–None: (£1.73, £3.14), and Pop–None: (–£0.4590, £0.8985).
(n1 + n2) 1 2
c. We can be 95% confident that, on average, customers will spend
_ _ _ _ 2
n1 x1 + n2 x1 − n1 x1 − n2 x2
______________________
between £1.52 and £2.91 more per evening meal when classical music
= n1 ( n1 + n2 )+ is playing than when pop music is playing at the restaurant.
_ _ _ _ 2 d. The mean meal cost when classical music is playing is significantly
n1 x2 + n2 x2 − n1 x1 − n2 x2
n2 (______________________
n1 + n2 ) greater than when either pop or no music is playing.
_ _ 2 _ _ 2 e. Letters plot:
n2 x1 − n2 x2 n1 x2 − n1 x1
= n1 (___________
n +n ) + n 2 (
__________
n +n )
1 2 1 2
_ _ 2 _ _ 2 Music Group Mean Letters
x1 − x2 x2 − x1
= n1 n22 (_______
n1 + n2 ) + n 2 n 2 _______
1 ( n1 + n2 ) Classical £24.13 a
Pop £21.91 b
_ _ 2 _ _ 2
x1 − x2 x1 − x2
= (_______
n1 + n2
2 2
) (n1 n2 + n1 n2) = (
_______
) (n1 n2)(n1 + n2)
n1 + n2
None £21.69 b
_ _ _ _ _ _
(x1 − x2) 2 (x1 − x2) 2 (x1 − x2) 2 1.5.11
= _________ (n 1 n 2 )( n 1 + n 2 ) = ________
(n 1 n 2) = _________ .
(n1 + n2)
2 (n1 + n2) _n1 + n2 a. A 95% confidence interval for the long-run mean _ cost of a meal
( n1 n2 )
when no music is playing is £21.69 ± 2 × £3.38/√131 ≈ (£21.10,
Section 1.5 £22.28); the validity conditions are met because the sample size is
1.5.1 D. fairly large.
1.5.2 C. b. A 95% prediction interval for the long run ____________
cost of a meal when
1.5.3 C. no music is playing is £21.69 ± 2 × £3.38 × √1 + 1/131 ≈ (£14.90,
£28.48); the validity conditions are met because the histogram of
1.5.4 C. these data is fairly symmetric and bell-shaped.
1.5.5 A, F, H.
c. The prediction interval looks like it contains about 95% of the data
1.5.6 The margin of error is based on a prediction interval. The rang- (it actually contains 92%), whereas the confidence interval contains a
ers are not trying to predict the mean time for all future eruptions but much smaller percentage of the actual data.
are trying to predict the time of the next eruption so that visitors have
1.5.12
a high probability of seeing the eruption if they are present during the
a. The p-value = 0.0087, so there is strong evidence that at least one
entire interval.
mean is different from the others; the validity conditions are met be-
1.5.7 cause the sample sizes are fairly large, and the sample SDs are similar
a. Mean = 7.321 hrs and SD = 1.490 hrs. in value.
____________
b. An approximate prediction interval is (7.321 ± 2 × 1.49 × √1 + 1/100 ) b. The 95% confidence intervals are Lie − Truth: (−2.68, −0.61), Lie −
≈ 4.326 to 10.316 hr; the validity conditions are met because the data Control: (−1.97, 0.10), and Truth − Control: (−0.3267, 1.7460).
are quite symmetric and have no obvious outliers. c. We can be 95% confident that, in the long run, the mean difference in
c. Ninety-three percent of these data lie within the 95% prediction rating for the lie condition is between 0.61 to 2.68 points lower than that
interval. This is reasonably close to the 95% that we would expect. for the truth condition.
, Solutions to Exercises 9
d. The lie condition has a mean that is significantly less than the truth earnings; the validity conditions are met because the sample size is
condition. Nothing else is significantly different. fairly large and the SDs are all within a factor of 2 of each other.
e. Letters plot: c. Letters plot:
Condition Group mean Letters Estimated
Lie −0.90 A Education level group mean Letters
Control 0.03 AB Doctorate $97.40K A
Truth 0.74 B Master’s $66.00K B
Bachelor’s $55.20K B
1.5.13
Associate $36.80K C
a. A 95% confidence interval for the long run difference in
_
mean rat-
Some College $32.51K C
ings between bottled and tap water is 0.03 ± 2 × 1.975/√31 ≈ 0.03 ±
0.71 = (−0.679, 0.739); the validity conditions are met because the
sample size is fairly large. 1.5.16
b. A 95% prediction interval for the___________
difference in bottled and tap wa- a. A 95% confidence interval for the mean _
amount earned by those
ter ratings is 0.03 ± 2 × 1.975 × √1 + 1/31 ≈ 0.03 ± 4.01 = (−3.98, with doctorates is $97.4K ± 2 × $40.5K/√50 ≈ $97.4K ± $11.455K =
4.04); the validity conditions are met because the dotolot of these data ($85.94K, $108.86K); the validity conditions are met because the sam-
is fairly symmetric and bell-shaped. ple size is fairly large.
c. The prediction interval looks like it contains about 95% of the data b. A 95% prediction interval for the amount
___________earned by those with
(it actually contains 30/31 = 96.8%), whereas the confidence interval doctorates is $97.4K ± 2 × $40.5K × √1 + 1/50 ≈ $97.4K ± $81.81K
contains a much smaller percentage. = ($15.59K, $179.21K); the validity conditions may not be met in this
case because the distribution appears to be skewed to the right with a
1.5.14
5.50 if computer few large outliers.
a. predicted quiz score = , SE of residuals = 1.63.
{6.92 if paper c. There are 46/50 or 92% of those with doctoral degrees in this sam-
ple contained in the prediction interval.
b. A 95% confidence interval
_
for the long-run mean score using paper
notes is 6.92 ± 2 × 1.07/√20 ≈ 6.92 ± 0.384 = (6.44 to 7.40). 1.5.17
−0.713 if computer a. A 95% prediction interval for the mean amount earned by those
___________
c. predicted quiz score = 6.213 + , with associate degrees is $36.8K ± 2 × $28.5K × √1 + 1/50 ≈ $36.8K
{ 0.713 if paper
SE of residuals = 1.63. ± $57.57K = (−$20.77K, $94.37K); the validity conditions are suspect
because the distribution looks skewed right.
d. A 95% confidence interval for the
_
long-run mean effect when using
paper notes is 0.71 ± 2 × 1.07/√20 = 0.71 ± 0.384 = (0.23, 1.19). We b. There are 49/50 or 98% of these data within the prediction interval.
can use the same standard deviation because the distribution of ef- c. This is such a bad fit because the distribution of salaries is highly
fects is the same as the distribution of scores but slid down 6.21 units. skewed to the right. This method is only valid when we have a bell-
1.5.15 shaped distribution.
a. d. The concerns aren’t as great for a confidence interval. Even though
the distribution is skewed, the associated sampling distribution
should be quite symmetric with a sample size as large as 50.
Some Asso Bach Mast Doct
1.5.18
_ _ _ _ _
a. Each margin of error is 2s/√n, so y2 − y1 = 2 × 2s/√n = 4s/√n.
____________ _ _ _
b. The margin of error is 2s √1/n + 1/n = 2s √2/n = 2 √2 s/√n.
Degree
_
c. With 4 > 2 √2, the answer to part (a) is larger than part (b). Both
_ _
of the answers represent y2 − y1 but only the confidence interval for
the difference in means uses the correctly pooled SE in the margin
of error expression. If the individual means were just a tiny bit
closer together the single mean intervals from part (a) would over-
0 100 200 lap, however the difference in means interval from part (b) would be
Earnings ($K) completely positive.
Section 1.6
Education level Group mean Group SD
1.6.1 A, C, D.
Doctorate $97.40K $40.50K
1.6.2 A.
Master’s $66.00K $38.40K
1.6.3 C.
Bachelor’s $55.20K $32.20K
1.6.4
Associate $36.80K $28.50K
a Increase alpha level.
Some College $32.51K $20.80K
b. Increase sample size.
b. The F-statistic is 31.534 and the p-value is < 0.0001, so there is c. Decrease number of groups comparing.
strong evidence of an association between the levels of education and d. Decrease variability within each group.
, 10 C HA PT E R 1 Sources of Variation
1.6.5 f. ≈ 0.870.
a. Just under 20. g. As the effect size (difference in mean heart rates) increases, power
b. Just under 40. of the test increases.
c. Just under 50.
Difference in mean
d. Just under 60.
heart rates (bpm) 5 10 15 20 25
e. Just over 75.
Power 0.395 0.870 0.995 1.000 1.000
f. As sample size per group increases, power of the test increases line-
arly at first, but then plateaus. (The relationship looks logarithmic, or a
1.6.12
power between 0 and 1.).
a. Now the rejection region should be a difference in means of about
1.6.6
8.2 or more. In this case, the power is roughly 0.189.
a. This will decrease the power of the test.
b. Using a significance level of 0.10 would increase the power.
b. Power = 0.73.
c. Now the rejection region is about ≈ 0.503.
c. The power would be very close to 1.
d. As level of significance increases, the power of the test increases.
d. The relationship between power and sample size looks linear for
most values of the sample size, with no plateauing visible even with Level of significance 0.001 0.01 0.05 0.10 0.15
each sample being even as large as 120.
Power 0.028 0.189 0.431 0.503 0.611
1.6.7
a. 1 ≤ SD ≤ 3. 1.6.13
b. About 5. a. About 0.238.
c. 4 ≤ SD ≤ 4.5. b. Increase power.
d. A little more than 5.5. c. About 0.510.
e. As SD increases, power decreases. For very small SDs the power d. As number per group increases, power of the test increases.
of the test is one, then it begins to decrease somewhat linearly as the
SDs increase. Sample size per
1.6.8 group 5 10 15 20 25
a. 81%. Power 0.238 0.395 0.510 0.644 0.672
b. 59.1%.
1.6.14
c. 0.10 < α < 0.15.
a. About 0.88.
d. As the level of significance increases, the power of the test increas-
b. Power will decrease.
es. Power doesn’t increase linearly, but in steps.
c. About 0.253.
1.6.9 A difference between 15 and 20 mL/d.
d. As SD within each group increases, power of the test decreases.
1.6.10
a. The differences in the group means and overall mean are the same Standard deviation
in Scenarios 1 and 2. per group 4 8 12 16 20
b. There is greater variability within the groups in Scenario 1. 0.880 0.427 0.253 0.201 0.153
Power
c. Scenario 2 will have the larger F-statistic.
d. Scenario 2 will be more likely to have a statistically
significant result. End of Chapter 1 Exercises
e. As the variability within the groups decreases, the F-statistic in- 1.CE.1.
creases, as does the power of the test. a. H0: μregular = μfilled and Ha: μregular ≠ μfilled; p-value = 0.003. Because
1.6.11 the p-value is less than 0.01, there is very strong evidence against the
null and for the alternative that there is an association between the
a. The rejection region is any difference in means of 5.9 or greater.
type of soup bowl and the amount of soup consumed with the secretly
b. A difference in mean heart rates of 7 bpm will be in the rejection refilled soup bowl resulting in a higher average consumption of
region so you would conclude that the two treatment means are sig- soup (oz).
nificantly different from each other.
b. The samples are independent of each other (randomly assigned
c. A difference of 4 pbm is not in the rejection region, so you would to bowl type), sample sizes are greater than 20 and there is no strong
conclude it is plausible that the two treatment means do not differ skewness in the data.
from each other.
c. Theory-based two-sided p-value is 0.0032, which is very close to
d. P(Type I error) = 0.05. the simulation-based p-value. This is expected as validity conditions
e. ≈ 0.395. were met to perform the theory-based test.
