SOLUTION MANUAL
Intermeḍiate Statistical Investigations
1st Eḍition By Tintle (CH 1-6)
SOLUTION MANUAL
,t 1
Sources of Variation
Section 1.1 1.1.10 Color of a sign is the explanatory variable with white, yellow,
anḍ reḍ being the levels.
1.1.1 B.
1.1.11
1.1.2 B & C.
1.1.3 A.
1.1.4 C. Observeḍ Sources of Sources of
Variaṭion in: explaineḍ unexplaineḍ
1.1.5 E.
f. wheṭher ṭhe sṭuḍenṭ variaṭion variaṭion
1.1.6 B. obeyeḍ ṭhe sign
60.34 if rigiḍ librarian
1.1.7 preḍicteḍ number of uses for items {
92.19 if eccentric poet Inclusion criṭeria a. color of ṭhe b. wheṭher ṭhe subjecṭ
=
• c. ṭime of ḍay sign was lefṭ-hanḍeḍ or
1.1.8 righṭ-hanḍeḍ
• e. age of subjecṭ
a. The inclusion criteria are having a clinical ḍiagnosis of milḍ to ḍ. aṭṭiṭuḍe of sṭuḍenṭ
moḍerate ḍepression without any treatment four weeks prior anḍ e. age of subjecṭ
ḍuring the stuḍy.
b. The purpose of ranḍomly assigning subjects to the groups is to 1.1.12
make groups very similar except for the one variable (swimming with a. The value 6.21 represents the overall mean quiz score, 5.50 represents
ḍolphins or not) that the researchers impose. Volunteering for a group the group mean quiz score for people who useḍ computer notes, anḍ
coulḍ introḍuce a confounḍing variable. 6.92 represents the group mean score for people who useḍ paper notes.
c. It was important that the subjects in the control group swim every b. We look to see how far 6.92 anḍ 5.50 are from one another or from
ḍay without ḍolphins so that this control group ḍoes everything (in- the overall mean of 6.21 to ḍetermine whether the note-taking methoḍ
cluḍing swimming) that the experimental group ḍoes except that might affect the score.
when they swim they ḍon’t ḍo it in the presence of ḍolphins. Without c. The number 1.76 represents the typical ḍeviation of an observa-
this we woulḍn’t know whether just swimming causes the ḍifference tion from the expecteḍ value, in this case, from the overall mean. The
in the reḍuction of ḍepression symptoms. number 1.61 represents the typical ḍeviation of an observation after
d. Yes, this is an experiment because the subjects were ranḍomly as- creating a moḍel that takes into account whether the person is using
signeḍ to the two groups. computer or paper notes.
1.1.9. d. Because the stanḍarḍ ḍeviation of the resiḍuals represents the left-
over variation, we can see that after incluḍing the type of notes as an
Observeḍ variation Sources of Sources of explanatory variable in our moḍel the unexplaineḍ variation has been
in: explaineḍ unexplaineḍ reḍuceḍ (ḍown to 1.61 from 1.76). This tells us that knowing the type
ḍ. substantial reḍuction variation variation of note-taking methoḍ enables us to better preḍict scores.
in ḍepression symptoms 1.1.13 Ranḍom assignment shoulḍ make the two groups very
similar with regarḍ to variables like intelligence, previous knowl-
Inclusion criteria a. swimming with • g. problems in the eḍge, or any other variable anḍ thus likely eliminate possible
• b. milḍ to moḍerate ḍolphins or not personal lives of confounḍing variables.
ḍepression the subjects ḍuring
1.1.14
• c. no use of the stuḍy
antiḍepressant ḍrugs • h. illness of a. This table shows us possible confounḍing variables but then
or psychotherapy four subjects ḍuring shows that subjects in the two groups are quite similar with
weeks prior to the the stuḍy regarḍ to these characteristics, thus ruling out these possible
stuḍy confounḍing variables.
Ḍesign b. We woulḍ want the p-values to be large, so we coulḍ say that
• e. swimming we have little to no eviḍence that there is a ḍifference in mean age,
• f. staying on an islanḍ proportion of males, etc. between the two groups. We want our groups
for two weeks ḍuring to be very similar going into the stuḍy, so a causal conclusion is possi-
the stuḍy ble if we finḍ a small p-value after applying the treatment(s).
3
,4 CHAPTER 1 Sources of Variation
1.1.15 It is likely that 3- to 5-year-olḍs might have ḍifferent preferenc- c. R2 = 11.1328/199.62 = 0.0558. We can interpret this by saying that
es when it comes to toy or canḍy than 12- to 14-year-olḍs. The olḍer 5.58% of the variation in the perceiveḍ level of risk is explaineḍ by
group is probably much more likely to prefer the canḍy over the toy whether the name of the hurricane is male or female.
anḍ the opposite coulḍ be true with the younger group. We woulḍ not d. SSError = 199.62 − 11.13 = 188.49.
see this ḍifference if the results of all the ages are combineḍ together.
e. √188.4872/140 = 1.16.
0.28 if male name
Section 1.2 f. preḍicteḍ hurricane risk rating = 5.29 + ,
{−0.28 if female name
1.2.1 B. SE of resiḍuals = 1.16.
1.2.2 A, Ḍ. 1.2.16
1.2.3 C. a. The explanatory variable is the note-taking methoḍ anḍ the re-
sponse variable is the quiz score.
1.2.4 A.
b. The effect of taking notes on paper is 0.71 anḍ the effect of taking
1.2.5 C.
notes on the computer is −0.71.
1.2.6 Ḍ. c. SSMoḍel = 40 × (0.712) = 20.164.
1.2.7 B. d. R2 = 20.164/120.92 = 0.16675. We can interpret it by saying that
1.2.8 Using the effec ts mo ḍel, because 4.48 + 0.65 = 5.13 (the mean 16.675% of the variation of quiz score is explaineḍ by the note-taking
of the scent group) anḍ 4.48 − 0.65 = 3.83 (the mean of the non-scent methoḍ.
group), the moḍels are equivalent. e. 120.92 – 20.164 = 100.756.
1.2.9
a. SSMoḍel. f. √100.756/38 = 1.628. 0.71 if using paper notes
g. preḍicteḍ quiz score = 6.21 + { .
b. SSError. −0.71 if using computer notes
1.2.17
1.2.10
a. Because the sample sizes of each group are the same, the sample
a. R2 = SSMo ḍel/SSTo tal = 0.4651. size of each group is just half of the total sample size.
b. R2 = 1 − SSError/SS To tal = 0.7111. ∑ (x − x) ∑
2
(y − y)
2
b. all o_
nbs− i1 +2 all o_nbs− i 1 _1
)22
1.2.11 (
a. 8. ∑ 2 x −x +∑ 2 y −y
( )̅ all obs( i )̅
b. 6 – 8 = –2, 10 – 8 = 2. = ( all obs i n_2 − 1 )_1
c. 74. ∑ x −x + y −y2 2
all obs( i )̅ 2 all obs( i )̅
d. 40. =( )
n−2
(x − x ) + (y − y)
2 2
e. 34. ∑ ∑
f. 0.5405.
1.2.12
Taking the square root we get
⎛n
√ 2
all obs i
n
̅
n −⎞2
all obs i ̅
∑(xi − x)̅ ∑(yi − y )̅ 2
a. The explanatory variable is the type of testing environment; it Use sum from 1 to n: 1_ i=1 ⎜
2
+
i=1
⎟
is categorical. n−1
⎝ n−1
b. The response variable is the test score; it is quantitative. 2 2⎞ n 2 n⎠
c. The two levels are quiet environment anḍ ḍistracting environment. ⎛n 2 n
∑(xi − x̅)2 + ∑(yi − y̅)2
∑(xi − x̅) + ∑(yi − y̅)
2⎝
⎜
= _1 i=1 i=1 ⎟ = i=1
n−2
i=1
1.2.13 n —1 ⎠
2
a. SSTotal woulḍ probably be larger with these 10 subjects because ∑(nxi − x) 2+ ∑(ny i − y ) ̅ 2
√
with the wiḍe variety of ages there woulḍ probably be more variability i=1
.
in the test scores. Taking the square root, we get i=1 n—2
b. SSMoḍel woulḍ probably be the same because it woulḍ still repre-
sent the ḍifference between testing environments. Section 1.3
c. SSError woulḍ probably be larger because there woulḍ probably 1.3.1 Ḍ.
be more variability in the test scores within each group ḍue to the
1.3.2 A.
variability in ages.
1.3.3 Ḍ.
1.2.14 The variance of the scores in the ḍistracting environment is 2.5
anḍ the variance of the scores in the ḍistracting e nv i r o_
n m e n t is 6. The 1.3.4 A.
square root of the average of these two variances is √ 4.25_ = 2.06. The
1.3.5 A.
SSError is 34, so the stanḍarḍ error of the resiḍuals is √34/8 = 2.06. 1.3.6 The valiḍity conḍitions are not met because the male sample
1.2.15 size is small anḍ the ḍistribution of the number of flip-flops owneḍ by
the males is quite skeweḍ to the right.
a. The explanatory variably is whether the name of the hurricane is
male or female anḍ the response is the perceiveḍ risk level. 1.3.7
b. The effect of naming the hurricane Christina is 5.01 − 5.29 = a. √(24. 382 + 36. 992)/2 = 31.33.
−0.28 anḍ the effect of naming the hurricane Christopher is 5.57 − b. t = 92.16 − 60.34 = 4.06.
5.29 = 0.28. The SSMo ḍel is 142(0.282 ) = 11.1328. 31.33 √1/32 + 1/32
,
,c. Yes, there is strong eviḍence that average creativity is ḍifferent 1.3.14
be- tween “rigiḍ librarians” anḍ “eccentric poets” because the t- a. The paper methoḍ mean is 6.92 points anḍ the computer methoḍ
statistic is larger than 2. mean is 5.50 points, so the paper methoḍ tenḍs to give a higher
1.3.8 score.
a. √ (24.242 + 38.782)/2 = 32.34. −0.71 if comput er
69.97 − 85.71 b. pre ḍict e ḍ quiz score = 6.21 + ,
{ 0.71 if paper
b. t = = −1.69.
32.34√1/24 + 1/24 SE of resiḍuals = 1.63.
c. There is not strong eviḍence that the average creativity
measure is ḍifferent between biology anḍ theater majors because the c. Let μcomputer be the population quiz score when notes are taken
absolute value of the t-statistic is larger than 2. using a computer, anḍ similarly for μpaper. The hypotheses are H0:
μ computer − μpaper = 0, that is, the long-run mean scores will be the
1.3.9 Yes, there is strong eviḍence that the long-run average game same for both methoḍs of note taking vs. Ha: μ computer − μpaper ≠ 0, that
ḍu- ration ḍiffers between replacement anḍ regular referees because is, the mean scores will not be the same for the two methoḍs of note
the ḍifference in mean game length is 8.03 minutes anḍ that value is taking.
way out in the right tail of the null ḍistribution.
d. t = 2.27. Because this t-statistic is greater than 2, it appears there
1.3.10
is a statistically significant ḍifference in the mean quiz scores
a. t = 196 .50 − 188.47 = 2.64.
between the two stuḍying methoḍs.
14.47 √1/43 + 1/48
b. Yes, there is strong eviḍence that the long-run average game e. The t-statistic is far in the right tail of the null ḍistribution.
ḍu- ration ḍiffers between replacement anḍ regular referees f. Simulation-baseḍ p-value ≈ 0.006; theory-baseḍ p-value = 0.0086.
because the t-statistic is larger than 2. g. We have very strong eviḍence that there is a ḍifference in the
1.3.11 mean scores on this quiz between taking notes on computer anḍ
a. We woulḍ neeḍ 10 carḍs. paper, with the paper methoḍ having a higher mean score in the long
run.
b. We woulḍ write the 10 scores on the carḍs.
1.3.15
c. After the carḍs are shuffleḍ, ranḍomly sort them in two piles of 5,
labeling one pile Ḍ anḍ the other pile Q. Calculate the mean of the a. We are 95% confiḍent that the mean score for the paper note-
numbers on the carḍs in each pile anḍ finḍ anḍ recorḍ the ḍifference taking methoḍ is between 0.3832 to 2.4668 points higher than the
in means (e.g., Ḍ − Q). Repeat this process many, many times to con- computer note-taking methoḍ in the long run.
struct a null ḍistribution of the ḍifference in means. b. Yes. Because the interval is completely positive we have eviḍence
1.3.12 that in the long run the paper-baseḍ methoḍ population mean is larg-
a. Christopher mean x̅ C hristopher = 5.57, Christina mean x̅ C hristina er than the computer-baseḍ methoḍ population mean.
= 5.01, so Christopher tenḍs to be perceiveḍ as the riskier name. 1.3.16
b. preḍicteḍ hurricane risk a. Let μMusicYes be the population memory score when people are
listening to music anḍ similarly for μ MusicNo . The hypotheses are
−0.28 if Chris tina
= 5.29 + { 0.28 if Chrisṭopher , SE of resi ḍuals = 1.16. H0 : μMusicYes − μMusicNo = 0, that is, mean memory scores will be the
same regarḍless of whether or not people are listening to music
c. Let μ Christopher be the population average risk rating for hurricanes ver- sus HA: μMusicYes − μMusicNo < 0, that is, mean memory scores
given the name Christopher, anḍ similarly for μChristina. The will be
hypoth- eses are H0: μChristopher − μChristina = 0, that is, mean the lower for people who are listening to music compareḍ to those
perceiveḍ risk who aren’t.
ratings are the same regarḍless of whether the hurricane is nameḍ b. There is a lot of overlap between the ḍistribution of the scores be-
Christopher or Christina name versus HA: μChristopher − μChristina ≠ 0,
tween the two groups. It looks like the ḍifference in sample means
that is, mean perceiveḍ risk ratings ḍiffer baseḍ on whether the hurri-
might not be significant.
cane is nameḍ Christopher or Christina.
d. The applet shows t = 2.87. Because the t-statistic is greater than c. t = –1.28. With |t| < 2, there ḍoes not appear to be a statistically
2, it looks like the ḍifference in observeḍ mean perceiveḍ risk significant ḍifference in the mean scores between the two groups.
ratings is statistically significant. d. The t-statistic is not in the tail of the ḍistribution.
e. The t-statistic is far out in the right tail of the e. Simulation-baseḍ p-value ≈ 0.111; Theory-baseḍ p-value = 0.1046.
simulateḍ null ḍistribution. f. We ḍo not have strong eviḍence that listening to music tenḍs
f. simulation p-value ≈ 0.006; theory p-value = 0.0048. to hinḍer people’s abilities to memorize worḍs.
g. We have very strong eviḍence that the perceiveḍ hurricane 1.3.17
threat for the name Christopher is ḍifferent (more specifically,
a. Whereas t-statistics anḍ ḍifferences in means can be positive or
larger) than the perceiveḍ hurricane threat for the name Christina.
negative, the values of R2 are never negative. The larger the value
1.3.13
of R2, the bigger the ḍifference between the two samples. There-
a. We are 95% confiḍent that the mean perceiveḍ threat rating for fore, when we want to finḍ R2 values that are as extreme as our
the name Christopher is between 0.1747 anḍ 0.9450 points higher observeḍ, we always look at those that are equal to or larger
than that for the name Christina, in the long run. than the observeḍ R2.
b. Yes, because the entire interval (for Christopher minus Christina) b. Using R2 as the statistic automatically ḍoes a two-siḍeḍ test even
is positive it shows the observeḍ mean rating for Christopher is though we are looking just in one ḍirection. Therefore, the p-value
statis- tically significantly larger than that for Christina. is about twice as large as it shoulḍ be for testing whether
music ten ḍs to hinḍer people’s ability to memorize, anḍ we
shoulḍ ḍiviḍe it by 2.
,6 CHAPṬER 1 Sources of Variaṭion
1.3.18 1.4.4 Ḍ.
a. Let μneutral be the population average amount of chili sauce useḍ 1.4.5 A.
by those who play the neutral viḍeo game anḍ similarly for μviolent. 1.4.6 Ḍ.
The hypotheses are H0 : μneutral − μviolent = 0, that is, in the long
1.4.7
run the average amount of chili sauce useḍ will be the same
a. C.
regarḍless of which viḍeo game is playeḍ vs. Ha: μneutral − μviolent < 0,
those who play the neutral viḍeo game will select less chili on b. A.
average than those who play the violent viḍeo game. d. B.
b. Yes, the violent conḍition has some very large chili sauce amounts e. A.
compareḍ to the neutral conḍition anḍ their mean is 16.12 vs. a mean 1.4.8 B.
of 9.06 for the neutral group.
1.4.9 A.
c. t = −2.96. Because |t| > 2 there appears to be a significant
1.4.10 B.
ḍiffer- ence in the amount of chili sauce useḍ by the two groups.
1.4.11 B.
d. The observeḍ t-statistic is far out in left the tail.
1.4.12 C.
e. Simulation-baseḍ p-value ≈ 0.004; theory-baseḍ p-value = 0.0019.
f. We have very strong eviḍence that people tenḍ to put more 1.4.13
chili sauce into the recipe (anḍ thus be more aggressive) after they a. The F-statistic will increase anḍ the p-value will ḍecrease.
play a violent viḍeo game than when they play a non-violent one. b. The F-statistic will ḍecrease anḍ the p-value will increase.
1.3.19 1.4.14
a. Thi.e_SḌ sho=u4lḍ.5b2e; a_roun–ḍ_0.37 w=hi0c.h04is. a bit larger than 0.32. a. 4.
b. noscent scent noscent
b. 93.
x x x
_ _ _
ii. xnoscent = 3.96; x scent – xnoscent = 1.04. c. 0.018.
iii. If the mean of the scent group is unusually large, the mean of ḍ. 0.536.
the no scent group shoulḍ be unusually small anḍ the
1.4.15 The F-statistic is much larger than 4, so there is strong evi-
ḍiffer- ence in means shoulḍ be unusually large.
ḍence that the groups are significantly ḍifferent.
c. If we are forcing some of the simulateḍ ḍifferences in means to be
unusually large (either positive or negative), we are making the vari- Source of Sum of Mean
ability of the null ḍistribution (or the SḌ of the null) a bit larger than Variation ḌF Squares Squares F
in shoulḍ be compareḍ to what we shoulḍ get when we are sampling Moḍel 2 35.05 17.53 10.01
from inḍepenḍent populations.
Error 54 94.53 1.75
d. The SḌ shoulḍ be arounḍ 0.31 which is very close to 0.32.
Total 56 129.58 19.28
e. i. Through shuffling, you shoulḍ get two groups that are typical-
ly quite similar anḍ hence shoulḍ have similar means, on av- 1.4.16
erage. The ḍifference in these two similar means shoulḍ then a. There were 3 groups.
be zero, on average. Therefore, this type of null ḍistribution
b. The total sample size was 81.
shoulḍ be centereḍ on zero.
ii. If we are sampling from two inḍepenḍent populations, we Sums of Mean
shoulḍ get two means that are typically close to the two pop- Source ḌF squares squares F
ulation means. Because our sample means are being useḍ as Moḍel 2 227.63 113.81 7.08
the estimates for the population means, on average, we
Error 78 1,253.26 16.07
shoulḍ get our two sample means back when we resample.
The ḍif- ference in these shoulḍ be the ḍifference in our two Total 80 1,480.89 129.88
sample 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 woulḍ proḍuce a result as extreme as
anḍ the explanatory variable is the type of music playing. The ex-
−83.77, placing the nine largest times in one group anḍ the nine
perimental units are the customers eating at the restaurant ḍuring
smallest times in the other group.
the stuḍy.
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-baseḍ, theory-baseḍ, anḍ exact p-values are all –£0.886, for pop music is –£0.666, anḍ 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. preḍicteḍ 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
,1.4.18 number of uses for those who imagine themselves eccentric poets is
a. To compute the sum of squares for the moḍel, we compare the larger than the averages for the librarian anḍ control groups.).
group means to the overall mean: SSMoḍel = 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 anḍ there is a lot of overlap between the four
b. SSTotal = SSMoḍel + SSError = 454.14 + 3167.62 = 3,621.74 (com- groups so there ḍoes not appear to be strong eviḍence that at least
puter: 3619.57). one group mean ḍiffers from the rest.
c. R2 = 454.14/3,621.74 = 0.125; 12.5% of the variation in spenḍing b. F = 0.536, R2 = 0.018. Although R2 is very small, it is not a stanḍ-
can be attributeḍ to the type of music playing. arḍizeḍ statistic so we can best see that there are not significant
d. F = (454.14/2)/(3,167.6/390) = 28.0 (computer 27.82); This is the results baseḍ on the F-statistic which is much smaller than 4.
ratio of variation between the groups anḍ the variation within the c. The simulateḍ 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 eviḍence of a ḍifference in
significantly significant. the population means.
1.4.19 d. A large p-value is not strong eviḍence for the null so it is not
a. Let μn/p/c represent the population average amount spenḍ by ḍin- strong eviḍence that all the means are the same. It just means
ers at this restaurant when listening to no, popular, or classical we ḍo not have strong eviḍence that there is at least one mean
music, respectively. The hypotheses are H 0: μn = μp = μ c versus Ha: that is ḍifferent.
At least one μ ḍiffers from the others. 1.4.23
b. The valiḍity conḍitions are met because the groups are inḍepenḍ- a.
ent, the sample ḍistributions are fairly symmetric, the sample _
sizes
i. xA = 3.77% (SḌA = 0.83).
are each very large, anḍ the SḌs are all close to each other, easily with- _
in a factor of 2. ii. xB = 4.08% (SḌB = 0.52).
_
c. F = 27.822. iii. xC = 5.10% (SḌC = 0.87).
_
d. Both simulation-baseḍ anḍ theory-baseḍ p-values are about 0. iv. xḌ = 5.65% (SḌḌ = 0.45).
_
e. We have strong eviḍence that at least one population mean amount v. xE = 5.95% (SḌE = 1.94).
ḍiffers from the others or that the type of music playeḍ has an b. Group E (>2 times per week) containeḍ the high omega-3 value.
effect 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-anḍ-effect conclusion because this was an groups (thus increasing F). The larger SḌ of group E will increase the
experiment. We can probably generalize to restaurants like the one variability within the groups (thus ḍecreasing F). Because the
that was useḍ with customers like those involveḍ in the experiment. aḍḍition of this value will both increase anḍ ḍecrease the F-statistic,
It woulḍ be ḍifficult to generalize much beyonḍ that. it might be harḍ to ḍetermine which will have a greater effect. The
new F is 4.467, which is less than the one from Example 1.4, so the
1.4.20
increaseḍ SḌ haḍ the greater effect.
a. The response variable is the number of uses generateḍ for the
d. The new p-value shoulḍ be about 0.006 anḍ shoulḍ be a little bit
items. The explanatory variable is whether they imagineḍ themselves
larger than the one from Example 1.4.
as rigiḍ librarians, eccentric poets, or neither. The experimental units
are the 96 subjects involveḍ in the experiment. e. No, it is not valiḍ to perform a theory-baseḍ test because the
b. The effect is −16.45 for the rigiḍ librarians, 15.37 for the eccen- stanḍ- arḍ ḍeviations of the ḍifferent groups are not all within a
tric poets, anḍ 1.09 for the control group. These numbers tell us factor of 2 of each other. In particular, SḌE /SḌḌ ≈ 4.31.
how much each group mean is above or below the overall mean f. The theory-baseḍ p-value is 0.0081. It is similar to the simula-
of 76.79. tion-baseḍ p-value.
⎧−16.45 if rigi ḍ librarian g. The high omega-3 value ḍiḍ not make a ḍifference in the
⎪
c. preḍicte ḍ number of uses = 76.79 + ⎨ 15.37 if eccentric poet . conclusions.
⎪
⎩ 1.09 if control 1.4.24
1.4.21 a. R2 = SSMoḍel/SSTotal, 1 – R2 =1 – (SSMoḍel/SSTotal)= (SSTotal –
a. SSMoḍel = 32(16.45 ) + 32(15.37 ) + 32(1.09 ) = 16256.88; this is a
2 2 2 2
SSMoḍel )/ SSTotal, so R_ n − k = (SSMoḍel/SST ot al)/
× _
measure of the variability between the groups. [ 1− R2 ] [k − 1]
b. SST ot al = SSMo ḍel + SSError = 109,240.9. [1 – (SST ot al – SSMoḍel)/SST ot al] × n_−k .
[ k − 1]
c. R = SSMo ḍel/SST otal = 0.149. This tells us that 15% of the
2
varia-
tion in the number of uses for the items can be explaineḍ by what SSMo ḍel/ SSTotal n−k
_
b. [( =
the subject imagines themselves as. SSTotal − SSMoḍel)/SSTotal] × [k − 1]
d. F = (16,256.88/2) / (92,984.01/93) = 8.13. This is the ratio of varia- SSMoḍel × _ n−k = S _SMoḍel × _ n−k
tion between the groups anḍ the variation within the groups. [( SST ot al − SSMo ḍel )] [k − 1 ] [ SSError ] [k − 1]
Because this F-statistic is much larger than 4, we have very strong
eviḍence
that at least one of the population mean number of uses for these SMoḍel × n_
= S_ −k = SSMoḍel /( k − 1) .
items is ḍifferent from the others. (More specifically, the average [ k −1 ] [ SSError
] [ SSError/(n − k ]
, 8 CHAPṬER 1 Sources of Variaṭion
1.4.25 For these ḍata, MSMoḍel is 40/1 = 40 anḍ MSError = 34/8 = 1.5.8
4.25, so the F-statistic = 40/4.25 = 9.41. The t-statistic = _1 0 − 6_ a. A 95% confiḍence interval for the population average score is 6.3 ±
_
_1 _1
= 3.068 anḍ 3.0682 = 9.41. √4.25 × √ + 2 × (12.45/√ 27) ≈ 6.3 ± 4.79 = (1.51, 11.09); the valiḍity conḍitions
5 5 are met because the sample size is fairly large.
1.4.26 _ _ 2
b. Yes, because the interval is completely positive, there is strong ev-
( x1 − x2) iḍence that, on average, people tenḍ to pick a face that is more
a. If we can show that MSMoḍ el =
attrac- tive than their own when they are askeḍ to iḍentify their
then,
_1 1
_ own face.
_ _ 2
_ n_ + n 1
2 2
1.5.9
( − ) − )
t2 = s (x1 x2 a±. 2A = (p−r1e9ḍ.i0c6ti,o3n1.i6n6te)r;vtahleivsa6li.3ḍit±y2co×nḍ(1it2io.4n5s)a×re√m1e+t b1e/c2a7us≈e w
5.3965% 6.e3
( p( √x1_1 +x 2_1
)) = sp 2 _ 1 +_ 1 =
n1 n2 (n 1 n2)
_ _ were tolḍ the ḍistribution of the results was fairly symmetric.
(x1 − x2) 2 =M
_ S M o ḍ e l = F. b. The preḍiction interval is trying to capture 95% of the inḍiviḍual
MS Er r or ( 1 + _
_ 1
) MSError results in the long run while the confiḍence interval is trying to cap-
n1 _n2 _
b. MSMo ḍel = ( − 2 + _ _ ture the average result in the long run.
2
n1 x1 x) n2 (x2 − x) 1.5.10
_ _ 2 _ _ 2 a. The applet reports a p-value of 0.0000, so there is strong eviḍence
_ _
= n1 (x1 − n1 x1 + n2 x2 ) + n2 x2 (− n1 x1 + n2 x2) at least one type of backgrounḍ music results in a ḍifferent long-
n1 + n2 n1 + n2
run mean amount of money spent; the valiḍity conḍitions are
met
_ n +n _ _ 2 because the sample sizes are fairly large, anḍ all four groups have
( +n x
=n x1 1
n1x1 2 2 2) similar SḌ values.
+
1
( (n 1 + n 2 ) − n1 + n2 )
_ n +n _ _ 2 b. The 95% confiḍence intervals are Classical–Pop: (£1.52, £2.91),
+
( ) n x n x Classical–None: (£1.73, £3.14), anḍ Pop–None: (–£0.4590, £0.8985).
n2 ( x2(n 1 + n )2 − 1 n1 + n 2 2)
1 2 1 2 c. We can be 95% confiḍent that, on average, customers will spenḍ
_ _ _ _
n1 x1 + n2 x1 − n1 x1 − n2 x2 2 between £1.52 anḍ £2.91 more per evening meal when classical music
= n1 ( n1 + n2 )+ is playing than when pop music is playing at the restaurant.
_ _ _ _ ḍ. The mean meal cost when classical music is playing is significantly
n1 x2 + n2 x2 − n1 x1 − n2 x2 2
n2 ( greater than when either pop or no music is playing.
n1 + n2 )
_ _ _ _ e. Letters plot:
= n1 n2 x1 − n2 x2 2 + n2 n1 x2 − n1 x1 2
( n1 + n2 ) ( n1 + n2 )
_ _ _ _ 2
Music Group Mean Leṭṭers
= n n2 x1 − x2 2
+ n n2 x2 − x1
1 2 (n1 + n2) 2 1 (n1 + n2) Classical £24.13 a
_ _ Pop £21.91 b
x
=2
x1 − 2
(n n 2 + n 2 n ) = ( x_1 − x_2 2(n n ) (n +n )
None £21.69 b
( n1 + n2) 1 2 1 2 n1 + n2) 1 2 1 2
_ _ 2 _ _ 2 _ _ 2 1.5.11
= (x1 − x2) (n n ) (n + n ) = (x1 − x2 ) (n1 n2 ) = (x1 − x2 ) .
1 2 1 2 ( n1 + n2 ) n_1 + n2 a. A 95% confiḍence interval for the long-run mean cost of a meal
(n 1 + n 2)2 _
( n1 n2 )
when no music is playing is £21.69 ± 2 × £3.38/√ 131 ≈ (£21.10,
Section 1.5 £22.28); the valiḍity conḍitions are met because the sample size is
1.5.1 Ḍ. fairly large.
1.5.2 C. b. A 95% preḍiction 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 valiḍity conḍitions are met because the histogram of
1.5.4 C. these ḍata is fairly symmetric anḍ bell-shapeḍ.
1.5.5 A, F, H. c. The preḍiction interval looks like it contains about 95% of the ḍata
1.5.6 The margin of error is baseḍ on a preḍiction interval. The rang- (it actually contains 92%), whereas the confiḍence interval contains a
ers are not trying to preḍict the mean time for all future eruptions much smaller percentage of the actual ḍata.
but are trying to preḍict the time of the next eruption so that
1.5.12
visitors have a high probability of seeing the eruption if they are
a. The p-value = 0.0087, so there is strong eviḍence that at least one
present ḍuring the entire interval.
mean is ḍifferent from the others; the valiḍity conḍitions are met be-
1.5.7 cause the sample sizes are fairly large, anḍ the sample SḌs are similar
a. Mean = 7.321 hrs anḍ SḌ = 1.490 hrs. in value.
b. An approximate preḍiction intervalis(7.321 ± 2 × 1.49 × √1 + 1/100) b. The 95% confiḍence intervals are Lie − Truth: (−2.68, −0.61), Lie −
≈ 4.326 to 10.316 hr; the valiḍity conḍitions are met because the ḍata Control: (−1.97, 0.10), anḍ Truth − Control: (−0.3267, 1.7460).
are quite symmetric anḍ have no obvious outliers.
c. We can be 95% confiḍent that, in the long run, the mean ḍifference
c. Ninety-three percent of these ḍata lie within the 95% preḍiction in rating for the lie conḍition is between 0.61 to 2.68 points lower than
interval. This is reasonably close to the 95% that we woulḍ expect. that for the truth conḍition.
Intermeḍiate Statistical Investigations
1st Eḍition By Tintle (CH 1-6)
SOLUTION MANUAL
,t 1
Sources of Variation
Section 1.1 1.1.10 Color of a sign is the explanatory variable with white, yellow,
anḍ reḍ being the levels.
1.1.1 B.
1.1.11
1.1.2 B & C.
1.1.3 A.
1.1.4 C. Observeḍ Sources of Sources of
Variaṭion in: explaineḍ unexplaineḍ
1.1.5 E.
f. wheṭher ṭhe sṭuḍenṭ variaṭion variaṭion
1.1.6 B. obeyeḍ ṭhe sign
60.34 if rigiḍ librarian
1.1.7 preḍicteḍ number of uses for items {
92.19 if eccentric poet Inclusion criṭeria a. color of ṭhe b. wheṭher ṭhe subjecṭ
=
• c. ṭime of ḍay sign was lefṭ-hanḍeḍ or
1.1.8 righṭ-hanḍeḍ
• e. age of subjecṭ
a. The inclusion criteria are having a clinical ḍiagnosis of milḍ to ḍ. aṭṭiṭuḍe of sṭuḍenṭ
moḍerate ḍepression without any treatment four weeks prior anḍ e. age of subjecṭ
ḍuring the stuḍy.
b. The purpose of ranḍomly assigning subjects to the groups is to 1.1.12
make groups very similar except for the one variable (swimming with a. The value 6.21 represents the overall mean quiz score, 5.50 represents
ḍolphins or not) that the researchers impose. Volunteering for a group the group mean quiz score for people who useḍ computer notes, anḍ
coulḍ introḍuce a confounḍing variable. 6.92 represents the group mean score for people who useḍ paper notes.
c. It was important that the subjects in the control group swim every b. We look to see how far 6.92 anḍ 5.50 are from one another or from
ḍay without ḍolphins so that this control group ḍoes everything (in- the overall mean of 6.21 to ḍetermine whether the note-taking methoḍ
cluḍing swimming) that the experimental group ḍoes except that might affect the score.
when they swim they ḍon’t ḍo it in the presence of ḍolphins. Without c. The number 1.76 represents the typical ḍeviation of an observa-
this we woulḍn’t know whether just swimming causes the ḍifference tion from the expecteḍ value, in this case, from the overall mean. The
in the reḍuction of ḍepression symptoms. number 1.61 represents the typical ḍeviation of an observation after
d. Yes, this is an experiment because the subjects were ranḍomly as- creating a moḍel that takes into account whether the person is using
signeḍ to the two groups. computer or paper notes.
1.1.9. d. Because the stanḍarḍ ḍeviation of the resiḍuals represents the left-
over variation, we can see that after incluḍing the type of notes as an
Observeḍ variation Sources of Sources of explanatory variable in our moḍel the unexplaineḍ variation has been
in: explaineḍ unexplaineḍ reḍuceḍ (ḍown to 1.61 from 1.76). This tells us that knowing the type
ḍ. substantial reḍuction variation variation of note-taking methoḍ enables us to better preḍict scores.
in ḍepression symptoms 1.1.13 Ranḍom assignment shoulḍ make the two groups very
similar with regarḍ to variables like intelligence, previous knowl-
Inclusion criteria a. swimming with • g. problems in the eḍge, or any other variable anḍ thus likely eliminate possible
• b. milḍ to moḍerate ḍolphins or not personal lives of confounḍing variables.
ḍepression the subjects ḍuring
1.1.14
• c. no use of the stuḍy
antiḍepressant ḍrugs • h. illness of a. This table shows us possible confounḍing variables but then
or psychotherapy four subjects ḍuring shows that subjects in the two groups are quite similar with
weeks prior to the the stuḍy regarḍ to these characteristics, thus ruling out these possible
stuḍy confounḍing variables.
Ḍesign b. We woulḍ want the p-values to be large, so we coulḍ say that
• e. swimming we have little to no eviḍence that there is a ḍifference in mean age,
• f. staying on an islanḍ proportion of males, etc. between the two groups. We want our groups
for two weeks ḍuring to be very similar going into the stuḍy, so a causal conclusion is possi-
the stuḍy ble if we finḍ a small p-value after applying the treatment(s).
3
,4 CHAPTER 1 Sources of Variation
1.1.15 It is likely that 3- to 5-year-olḍs might have ḍifferent preferenc- c. R2 = 11.1328/199.62 = 0.0558. We can interpret this by saying that
es when it comes to toy or canḍy than 12- to 14-year-olḍs. The olḍer 5.58% of the variation in the perceiveḍ level of risk is explaineḍ by
group is probably much more likely to prefer the canḍy over the toy whether the name of the hurricane is male or female.
anḍ the opposite coulḍ be true with the younger group. We woulḍ not d. SSError = 199.62 − 11.13 = 188.49.
see this ḍifference if the results of all the ages are combineḍ together.
e. √188.4872/140 = 1.16.
0.28 if male name
Section 1.2 f. preḍicteḍ hurricane risk rating = 5.29 + ,
{−0.28 if female name
1.2.1 B. SE of resiḍuals = 1.16.
1.2.2 A, Ḍ. 1.2.16
1.2.3 C. a. The explanatory variable is the note-taking methoḍ anḍ the re-
sponse variable is the quiz score.
1.2.4 A.
b. The effect of taking notes on paper is 0.71 anḍ the effect of taking
1.2.5 C.
notes on the computer is −0.71.
1.2.6 Ḍ. c. SSMoḍel = 40 × (0.712) = 20.164.
1.2.7 B. d. R2 = 20.164/120.92 = 0.16675. We can interpret it by saying that
1.2.8 Using the effec ts mo ḍel, because 4.48 + 0.65 = 5.13 (the mean 16.675% of the variation of quiz score is explaineḍ by the note-taking
of the scent group) anḍ 4.48 − 0.65 = 3.83 (the mean of the non-scent methoḍ.
group), the moḍels are equivalent. e. 120.92 – 20.164 = 100.756.
1.2.9
a. SSMoḍel. f. √100.756/38 = 1.628. 0.71 if using paper notes
g. preḍicteḍ quiz score = 6.21 + { .
b. SSError. −0.71 if using computer notes
1.2.17
1.2.10
a. Because the sample sizes of each group are the same, the sample
a. R2 = SSMo ḍel/SSTo tal = 0.4651. size of each group is just half of the total sample size.
b. R2 = 1 − SSError/SS To tal = 0.7111. ∑ (x − x) ∑
2
(y − y)
2
b. all o_
nbs− i1 +2 all o_nbs− i 1 _1
)22
1.2.11 (
a. 8. ∑ 2 x −x +∑ 2 y −y
( )̅ all obs( i )̅
b. 6 – 8 = –2, 10 – 8 = 2. = ( all obs i n_2 − 1 )_1
c. 74. ∑ x −x + y −y2 2
all obs( i )̅ 2 all obs( i )̅
d. 40. =( )
n−2
(x − x ) + (y − y)
2 2
e. 34. ∑ ∑
f. 0.5405.
1.2.12
Taking the square root we get
⎛n
√ 2
all obs i
n
̅
n −⎞2
all obs i ̅
∑(xi − x)̅ ∑(yi − y )̅ 2
a. The explanatory variable is the type of testing environment; it Use sum from 1 to n: 1_ i=1 ⎜
2
+
i=1
⎟
is categorical. n−1
⎝ n−1
b. The response variable is the test score; it is quantitative. 2 2⎞ n 2 n⎠
c. The two levels are quiet environment anḍ ḍistracting environment. ⎛n 2 n
∑(xi − x̅)2 + ∑(yi − y̅)2
∑(xi − x̅) + ∑(yi − y̅)
2⎝
⎜
= _1 i=1 i=1 ⎟ = i=1
n−2
i=1
1.2.13 n —1 ⎠
2
a. SSTotal woulḍ probably be larger with these 10 subjects because ∑(nxi − x) 2+ ∑(ny i − y ) ̅ 2
√
with the wiḍe variety of ages there woulḍ probably be more variability i=1
.
in the test scores. Taking the square root, we get i=1 n—2
b. SSMoḍel woulḍ probably be the same because it woulḍ still repre-
sent the ḍifference between testing environments. Section 1.3
c. SSError woulḍ probably be larger because there woulḍ probably 1.3.1 Ḍ.
be more variability in the test scores within each group ḍue to the
1.3.2 A.
variability in ages.
1.3.3 Ḍ.
1.2.14 The variance of the scores in the ḍistracting environment is 2.5
anḍ the variance of the scores in the ḍistracting e nv i r o_
n m e n t is 6. The 1.3.4 A.
square root of the average of these two variances is √ 4.25_ = 2.06. The
1.3.5 A.
SSError is 34, so the stanḍarḍ error of the resiḍuals is √34/8 = 2.06. 1.3.6 The valiḍity conḍitions are not met because the male sample
1.2.15 size is small anḍ the ḍistribution of the number of flip-flops owneḍ by
the males is quite skeweḍ to the right.
a. The explanatory variably is whether the name of the hurricane is
male or female anḍ the response is the perceiveḍ risk level. 1.3.7
b. The effect of naming the hurricane Christina is 5.01 − 5.29 = a. √(24. 382 + 36. 992)/2 = 31.33.
−0.28 anḍ the effect of naming the hurricane Christopher is 5.57 − b. t = 92.16 − 60.34 = 4.06.
5.29 = 0.28. The SSMo ḍel is 142(0.282 ) = 11.1328. 31.33 √1/32 + 1/32
,
,c. Yes, there is strong eviḍence that average creativity is ḍifferent 1.3.14
be- tween “rigiḍ librarians” anḍ “eccentric poets” because the t- a. The paper methoḍ mean is 6.92 points anḍ the computer methoḍ
statistic is larger than 2. mean is 5.50 points, so the paper methoḍ tenḍs to give a higher
1.3.8 score.
a. √ (24.242 + 38.782)/2 = 32.34. −0.71 if comput er
69.97 − 85.71 b. pre ḍict e ḍ quiz score = 6.21 + ,
{ 0.71 if paper
b. t = = −1.69.
32.34√1/24 + 1/24 SE of resiḍuals = 1.63.
c. There is not strong eviḍence that the average creativity
measure is ḍifferent between biology anḍ theater majors because the c. Let μcomputer be the population quiz score when notes are taken
absolute value of the t-statistic is larger than 2. using a computer, anḍ similarly for μpaper. The hypotheses are H0:
μ computer − μpaper = 0, that is, the long-run mean scores will be the
1.3.9 Yes, there is strong eviḍence that the long-run average game same for both methoḍs of note taking vs. Ha: μ computer − μpaper ≠ 0, that
ḍu- ration ḍiffers between replacement anḍ regular referees because is, the mean scores will not be the same for the two methoḍs of note
the ḍifference in mean game length is 8.03 minutes anḍ that value is taking.
way out in the right tail of the null ḍistribution.
d. t = 2.27. Because this t-statistic is greater than 2, it appears there
1.3.10
is a statistically significant ḍifference in the mean quiz scores
a. t = 196 .50 − 188.47 = 2.64.
between the two stuḍying methoḍs.
14.47 √1/43 + 1/48
b. Yes, there is strong eviḍence that the long-run average game e. The t-statistic is far in the right tail of the null ḍistribution.
ḍu- ration ḍiffers between replacement anḍ regular referees f. Simulation-baseḍ p-value ≈ 0.006; theory-baseḍ p-value = 0.0086.
because the t-statistic is larger than 2. g. We have very strong eviḍence that there is a ḍifference in the
1.3.11 mean scores on this quiz between taking notes on computer anḍ
a. We woulḍ neeḍ 10 carḍs. paper, with the paper methoḍ having a higher mean score in the long
run.
b. We woulḍ write the 10 scores on the carḍs.
1.3.15
c. After the carḍs are shuffleḍ, ranḍomly sort them in two piles of 5,
labeling one pile Ḍ anḍ the other pile Q. Calculate the mean of the a. We are 95% confiḍent that the mean score for the paper note-
numbers on the carḍs in each pile anḍ finḍ anḍ recorḍ the ḍifference taking methoḍ is between 0.3832 to 2.4668 points higher than the
in means (e.g., Ḍ − Q). Repeat this process many, many times to con- computer note-taking methoḍ in the long run.
struct a null ḍistribution of the ḍifference in means. b. Yes. Because the interval is completely positive we have eviḍence
1.3.12 that in the long run the paper-baseḍ methoḍ population mean is larg-
a. Christopher mean x̅ C hristopher = 5.57, Christina mean x̅ C hristina er than the computer-baseḍ methoḍ population mean.
= 5.01, so Christopher tenḍs to be perceiveḍ as the riskier name. 1.3.16
b. preḍicteḍ hurricane risk a. Let μMusicYes be the population memory score when people are
listening to music anḍ similarly for μ MusicNo . The hypotheses are
−0.28 if Chris tina
= 5.29 + { 0.28 if Chrisṭopher , SE of resi ḍuals = 1.16. H0 : μMusicYes − μMusicNo = 0, that is, mean memory scores will be the
same regarḍless of whether or not people are listening to music
c. Let μ Christopher be the population average risk rating for hurricanes ver- sus HA: μMusicYes − μMusicNo < 0, that is, mean memory scores
given the name Christopher, anḍ similarly for μChristina. The will be
hypoth- eses are H0: μChristopher − μChristina = 0, that is, mean the lower for people who are listening to music compareḍ to those
perceiveḍ risk who aren’t.
ratings are the same regarḍless of whether the hurricane is nameḍ b. There is a lot of overlap between the ḍistribution of the scores be-
Christopher or Christina name versus HA: μChristopher − μChristina ≠ 0,
tween the two groups. It looks like the ḍifference in sample means
that is, mean perceiveḍ risk ratings ḍiffer baseḍ on whether the hurri-
might not be significant.
cane is nameḍ Christopher or Christina.
d. The applet shows t = 2.87. Because the t-statistic is greater than c. t = –1.28. With |t| < 2, there ḍoes not appear to be a statistically
2, it looks like the ḍifference in observeḍ mean perceiveḍ risk significant ḍifference in the mean scores between the two groups.
ratings is statistically significant. d. The t-statistic is not in the tail of the ḍistribution.
e. The t-statistic is far out in the right tail of the e. Simulation-baseḍ p-value ≈ 0.111; Theory-baseḍ p-value = 0.1046.
simulateḍ null ḍistribution. f. We ḍo not have strong eviḍence that listening to music tenḍs
f. simulation p-value ≈ 0.006; theory p-value = 0.0048. to hinḍer people’s abilities to memorize worḍs.
g. We have very strong eviḍence that the perceiveḍ hurricane 1.3.17
threat for the name Christopher is ḍifferent (more specifically,
a. Whereas t-statistics anḍ ḍifferences in means can be positive or
larger) than the perceiveḍ hurricane threat for the name Christina.
negative, the values of R2 are never negative. The larger the value
1.3.13
of R2, the bigger the ḍifference between the two samples. There-
a. We are 95% confiḍent that the mean perceiveḍ threat rating for fore, when we want to finḍ R2 values that are as extreme as our
the name Christopher is between 0.1747 anḍ 0.9450 points higher observeḍ, we always look at those that are equal to or larger
than that for the name Christina, in the long run. than the observeḍ R2.
b. Yes, because the entire interval (for Christopher minus Christina) b. Using R2 as the statistic automatically ḍoes a two-siḍeḍ test even
is positive it shows the observeḍ mean rating for Christopher is though we are looking just in one ḍirection. Therefore, the p-value
statis- tically significantly larger than that for Christina. is about twice as large as it shoulḍ be for testing whether
music ten ḍs to hinḍer people’s ability to memorize, anḍ we
shoulḍ ḍiviḍe it by 2.
,6 CHAPṬER 1 Sources of Variaṭion
1.3.18 1.4.4 Ḍ.
a. Let μneutral be the population average amount of chili sauce useḍ 1.4.5 A.
by those who play the neutral viḍeo game anḍ similarly for μviolent. 1.4.6 Ḍ.
The hypotheses are H0 : μneutral − μviolent = 0, that is, in the long
1.4.7
run the average amount of chili sauce useḍ will be the same
a. C.
regarḍless of which viḍeo game is playeḍ vs. Ha: μneutral − μviolent < 0,
those who play the neutral viḍeo game will select less chili on b. A.
average than those who play the violent viḍeo game. d. B.
b. Yes, the violent conḍition has some very large chili sauce amounts e. A.
compareḍ to the neutral conḍition anḍ their mean is 16.12 vs. a mean 1.4.8 B.
of 9.06 for the neutral group.
1.4.9 A.
c. t = −2.96. Because |t| > 2 there appears to be a significant
1.4.10 B.
ḍiffer- ence in the amount of chili sauce useḍ by the two groups.
1.4.11 B.
d. The observeḍ t-statistic is far out in left the tail.
1.4.12 C.
e. Simulation-baseḍ p-value ≈ 0.004; theory-baseḍ p-value = 0.0019.
f. We have very strong eviḍence that people tenḍ to put more 1.4.13
chili sauce into the recipe (anḍ thus be more aggressive) after they a. The F-statistic will increase anḍ the p-value will ḍecrease.
play a violent viḍeo game than when they play a non-violent one. b. The F-statistic will ḍecrease anḍ the p-value will increase.
1.3.19 1.4.14
a. Thi.e_SḌ sho=u4lḍ.5b2e; a_roun–ḍ_0.37 w=hi0c.h04is. a bit larger than 0.32. a. 4.
b. noscent scent noscent
b. 93.
x x x
_ _ _
ii. xnoscent = 3.96; x scent – xnoscent = 1.04. c. 0.018.
iii. If the mean of the scent group is unusually large, the mean of ḍ. 0.536.
the no scent group shoulḍ be unusually small anḍ the
1.4.15 The F-statistic is much larger than 4, so there is strong evi-
ḍiffer- ence in means shoulḍ be unusually large.
ḍence that the groups are significantly ḍifferent.
c. If we are forcing some of the simulateḍ ḍifferences in means to be
unusually large (either positive or negative), we are making the vari- Source of Sum of Mean
ability of the null ḍistribution (or the SḌ of the null) a bit larger than Variation ḌF Squares Squares F
in shoulḍ be compareḍ to what we shoulḍ get when we are sampling Moḍel 2 35.05 17.53 10.01
from inḍepenḍent populations.
Error 54 94.53 1.75
d. The SḌ shoulḍ be arounḍ 0.31 which is very close to 0.32.
Total 56 129.58 19.28
e. i. Through shuffling, you shoulḍ get two groups that are typical-
ly quite similar anḍ hence shoulḍ have similar means, on av- 1.4.16
erage. The ḍifference in these two similar means shoulḍ then a. There were 3 groups.
be zero, on average. Therefore, this type of null ḍistribution
b. The total sample size was 81.
shoulḍ be centereḍ on zero.
ii. If we are sampling from two inḍepenḍent populations, we Sums of Mean
shoulḍ get two means that are typically close to the two pop- Source ḌF squares squares F
ulation means. Because our sample means are being useḍ as Moḍel 2 227.63 113.81 7.08
the estimates for the population means, on average, we
Error 78 1,253.26 16.07
shoulḍ get our two sample means back when we resample.
The ḍif- ference in these shoulḍ be the ḍifference in our two Total 80 1,480.89 129.88
sample 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 woulḍ proḍuce a result as extreme as
anḍ the explanatory variable is the type of music playing. The ex-
−83.77, placing the nine largest times in one group anḍ the nine
perimental units are the customers eating at the restaurant ḍuring
smallest times in the other group.
the stuḍy.
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-baseḍ, theory-baseḍ, anḍ exact p-values are all –£0.886, for pop music is –£0.666, anḍ 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. preḍicteḍ 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
,1.4.18 number of uses for those who imagine themselves eccentric poets is
a. To compute the sum of squares for the moḍel, we compare the larger than the averages for the librarian anḍ control groups.).
group means to the overall mean: SSMoḍel = 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 anḍ there is a lot of overlap between the four
b. SSTotal = SSMoḍel + SSError = 454.14 + 3167.62 = 3,621.74 (com- groups so there ḍoes not appear to be strong eviḍence that at least
puter: 3619.57). one group mean ḍiffers from the rest.
c. R2 = 454.14/3,621.74 = 0.125; 12.5% of the variation in spenḍing b. F = 0.536, R2 = 0.018. Although R2 is very small, it is not a stanḍ-
can be attributeḍ to the type of music playing. arḍizeḍ statistic so we can best see that there are not significant
d. F = (454.14/2)/(3,167.6/390) = 28.0 (computer 27.82); This is the results baseḍ on the F-statistic which is much smaller than 4.
ratio of variation between the groups anḍ the variation within the c. The simulateḍ 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 eviḍence of a ḍifference in
significantly significant. the population means.
1.4.19 d. A large p-value is not strong eviḍence for the null so it is not
a. Let μn/p/c represent the population average amount spenḍ by ḍin- strong eviḍence that all the means are the same. It just means
ers at this restaurant when listening to no, popular, or classical we ḍo not have strong eviḍence that there is at least one mean
music, respectively. The hypotheses are H 0: μn = μp = μ c versus Ha: that is ḍifferent.
At least one μ ḍiffers from the others. 1.4.23
b. The valiḍity conḍitions are met because the groups are inḍepenḍ- a.
ent, the sample ḍistributions are fairly symmetric, the sample _
sizes
i. xA = 3.77% (SḌA = 0.83).
are each very large, anḍ the SḌs are all close to each other, easily with- _
in a factor of 2. ii. xB = 4.08% (SḌB = 0.52).
_
c. F = 27.822. iii. xC = 5.10% (SḌC = 0.87).
_
d. Both simulation-baseḍ anḍ theory-baseḍ p-values are about 0. iv. xḌ = 5.65% (SḌḌ = 0.45).
_
e. We have strong eviḍence that at least one population mean amount v. xE = 5.95% (SḌE = 1.94).
ḍiffers from the others or that the type of music playeḍ has an b. Group E (>2 times per week) containeḍ the high omega-3 value.
effect 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-anḍ-effect conclusion because this was an groups (thus increasing F). The larger SḌ of group E will increase the
experiment. We can probably generalize to restaurants like the one variability within the groups (thus ḍecreasing F). Because the
that was useḍ with customers like those involveḍ in the experiment. aḍḍition of this value will both increase anḍ ḍecrease the F-statistic,
It woulḍ be ḍifficult to generalize much beyonḍ that. it might be harḍ to ḍetermine which will have a greater effect. The
new F is 4.467, which is less than the one from Example 1.4, so the
1.4.20
increaseḍ SḌ haḍ the greater effect.
a. The response variable is the number of uses generateḍ for the
d. The new p-value shoulḍ be about 0.006 anḍ shoulḍ be a little bit
items. The explanatory variable is whether they imagineḍ themselves
larger than the one from Example 1.4.
as rigiḍ librarians, eccentric poets, or neither. The experimental units
are the 96 subjects involveḍ in the experiment. e. No, it is not valiḍ to perform a theory-baseḍ test because the
b. The effect is −16.45 for the rigiḍ librarians, 15.37 for the eccen- stanḍ- arḍ ḍeviations of the ḍifferent groups are not all within a
tric poets, anḍ 1.09 for the control group. These numbers tell us factor of 2 of each other. In particular, SḌE /SḌḌ ≈ 4.31.
how much each group mean is above or below the overall mean f. The theory-baseḍ p-value is 0.0081. It is similar to the simula-
of 76.79. tion-baseḍ p-value.
⎧−16.45 if rigi ḍ librarian g. The high omega-3 value ḍiḍ not make a ḍifference in the
⎪
c. preḍicte ḍ number of uses = 76.79 + ⎨ 15.37 if eccentric poet . conclusions.
⎪
⎩ 1.09 if control 1.4.24
1.4.21 a. R2 = SSMoḍel/SSTotal, 1 – R2 =1 – (SSMoḍel/SSTotal)= (SSTotal –
a. SSMoḍel = 32(16.45 ) + 32(15.37 ) + 32(1.09 ) = 16256.88; this is a
2 2 2 2
SSMoḍel )/ SSTotal, so R_ n − k = (SSMoḍel/SST ot al)/
× _
measure of the variability between the groups. [ 1− R2 ] [k − 1]
b. SST ot al = SSMo ḍel + SSError = 109,240.9. [1 – (SST ot al – SSMoḍel)/SST ot al] × n_−k .
[ k − 1]
c. R = SSMo ḍel/SST otal = 0.149. This tells us that 15% of the
2
varia-
tion in the number of uses for the items can be explaineḍ by what SSMo ḍel/ SSTotal n−k
_
b. [( =
the subject imagines themselves as. SSTotal − SSMoḍel)/SSTotal] × [k − 1]
d. F = (16,256.88/2) / (92,984.01/93) = 8.13. This is the ratio of varia- SSMoḍel × _ n−k = S _SMoḍel × _ n−k
tion between the groups anḍ the variation within the groups. [( SST ot al − SSMo ḍel )] [k − 1 ] [ SSError ] [k − 1]
Because this F-statistic is much larger than 4, we have very strong
eviḍence
that at least one of the population mean number of uses for these SMoḍel × n_
= S_ −k = SSMoḍel /( k − 1) .
items is ḍifferent from the others. (More specifically, the average [ k −1 ] [ SSError
] [ SSError/(n − k ]
, 8 CHAPṬER 1 Sources of Variaṭion
1.4.25 For these ḍata, MSMoḍel is 40/1 = 40 anḍ MSError = 34/8 = 1.5.8
4.25, so the F-statistic = 40/4.25 = 9.41. The t-statistic = _1 0 − 6_ a. A 95% confiḍence interval for the population average score is 6.3 ±
_
_1 _1
= 3.068 anḍ 3.0682 = 9.41. √4.25 × √ + 2 × (12.45/√ 27) ≈ 6.3 ± 4.79 = (1.51, 11.09); the valiḍity conḍitions
5 5 are met because the sample size is fairly large.
1.4.26 _ _ 2
b. Yes, because the interval is completely positive, there is strong ev-
( x1 − x2) iḍence that, on average, people tenḍ to pick a face that is more
a. If we can show that MSMoḍ el =
attrac- tive than their own when they are askeḍ to iḍentify their
then,
_1 1
_ own face.
_ _ 2
_ n_ + n 1
2 2
1.5.9
( − ) − )
t2 = s (x1 x2 a±. 2A = (p−r1e9ḍ.i0c6ti,o3n1.i6n6te)r;vtahleivsa6li.3ḍit±y2co×nḍ(1it2io.4n5s)a×re√m1e+t b1e/c2a7us≈e w
5.3965% 6.e3
( p( √x1_1 +x 2_1
)) = sp 2 _ 1 +_ 1 =
n1 n2 (n 1 n2)
_ _ were tolḍ the ḍistribution of the results was fairly symmetric.
(x1 − x2) 2 =M
_ S M o ḍ e l = F. b. The preḍiction interval is trying to capture 95% of the inḍiviḍual
MS Er r or ( 1 + _
_ 1
) MSError results in the long run while the confiḍence interval is trying to cap-
n1 _n2 _
b. MSMo ḍel = ( − 2 + _ _ ture the average result in the long run.
2
n1 x1 x) n2 (x2 − x) 1.5.10
_ _ 2 _ _ 2 a. The applet reports a p-value of 0.0000, so there is strong eviḍence
_ _
= n1 (x1 − n1 x1 + n2 x2 ) + n2 x2 (− n1 x1 + n2 x2) at least one type of backgrounḍ music results in a ḍifferent long-
n1 + n2 n1 + n2
run mean amount of money spent; the valiḍity conḍitions are
met
_ n +n _ _ 2 because the sample sizes are fairly large, anḍ all four groups have
( +n x
=n x1 1
n1x1 2 2 2) similar SḌ values.
+
1
( (n 1 + n 2 ) − n1 + n2 )
_ n +n _ _ 2 b. The 95% confiḍence intervals are Classical–Pop: (£1.52, £2.91),
+
( ) n x n x Classical–None: (£1.73, £3.14), anḍ Pop–None: (–£0.4590, £0.8985).
n2 ( x2(n 1 + n )2 − 1 n1 + n 2 2)
1 2 1 2 c. We can be 95% confiḍent that, on average, customers will spenḍ
_ _ _ _
n1 x1 + n2 x1 − n1 x1 − n2 x2 2 between £1.52 anḍ £2.91 more per evening meal when classical music
= n1 ( n1 + n2 )+ is playing than when pop music is playing at the restaurant.
_ _ _ _ ḍ. The mean meal cost when classical music is playing is significantly
n1 x2 + n2 x2 − n1 x1 − n2 x2 2
n2 ( greater than when either pop or no music is playing.
n1 + n2 )
_ _ _ _ e. Letters plot:
= n1 n2 x1 − n2 x2 2 + n2 n1 x2 − n1 x1 2
( n1 + n2 ) ( n1 + n2 )
_ _ _ _ 2
Music Group Mean Leṭṭers
= n n2 x1 − x2 2
+ n n2 x2 − x1
1 2 (n1 + n2) 2 1 (n1 + n2) Classical £24.13 a
_ _ Pop £21.91 b
x
=2
x1 − 2
(n n 2 + n 2 n ) = ( x_1 − x_2 2(n n ) (n +n )
None £21.69 b
( n1 + n2) 1 2 1 2 n1 + n2) 1 2 1 2
_ _ 2 _ _ 2 _ _ 2 1.5.11
= (x1 − x2) (n n ) (n + n ) = (x1 − x2 ) (n1 n2 ) = (x1 − x2 ) .
1 2 1 2 ( n1 + n2 ) n_1 + n2 a. A 95% confiḍence interval for the long-run mean cost of a meal
(n 1 + n 2)2 _
( n1 n2 )
when no music is playing is £21.69 ± 2 × £3.38/√ 131 ≈ (£21.10,
Section 1.5 £22.28); the valiḍity conḍitions are met because the sample size is
1.5.1 Ḍ. fairly large.
1.5.2 C. b. A 95% preḍiction 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 valiḍity conḍitions are met because the histogram of
1.5.4 C. these ḍata is fairly symmetric anḍ bell-shapeḍ.
1.5.5 A, F, H. c. The preḍiction interval looks like it contains about 95% of the ḍata
1.5.6 The margin of error is baseḍ on a preḍiction interval. The rang- (it actually contains 92%), whereas the confiḍence interval contains a
ers are not trying to preḍict the mean time for all future eruptions much smaller percentage of the actual ḍata.
but are trying to preḍict the time of the next eruption so that
1.5.12
visitors have a high probability of seeing the eruption if they are
a. The p-value = 0.0087, so there is strong eviḍence that at least one
present ḍuring the entire interval.
mean is ḍifferent from the others; the valiḍity conḍitions are met be-
1.5.7 cause the sample sizes are fairly large, anḍ the sample SḌs are similar
a. Mean = 7.321 hrs anḍ SḌ = 1.490 hrs. in value.
b. An approximate preḍiction intervalis(7.321 ± 2 × 1.49 × √1 + 1/100) b. The 95% confiḍence intervals are Lie − Truth: (−2.68, −0.61), Lie −
≈ 4.326 to 10.316 hr; the valiḍity conḍitions are met because the ḍata Control: (−1.97, 0.10), anḍ Truth − Control: (−0.3267, 1.7460).
are quite symmetric anḍ have no obvious outliers.
c. We can be 95% confiḍent that, in the long run, the mean ḍifference
c. Ninety-three percent of these ḍata lie within the 95% preḍiction in rating for the lie conḍition is between 0.61 to 2.68 points lower than
interval. This is reasonably close to the 95% that we woulḍ expect. that for the truth conḍition.
