CHAPTER 1
Sources of Variation
Section 1.1 1.1.10 Color of a sign is the explanatory variable with white, yellow,
and red being the levels.
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. whether the student variation variation
obeyed the sign
1.1.6 B.
60.34 if rigid librarian Inclusion criteria a. color of the b. whether the subject
1.1.7 predicted number of uses for items =
{ 92.19 if eccentric poet sign was left-handed or
• c. time of day
1.1.8 right-handed
• e. age of subject
a. The inclusion criteria are having a clinical diagnosis of mild to d. attitude of student
moderate depression without any treatment four weeks prior and e. age of subject
during the study.
1.1.12
b. The purpose of randomly assigning subjects to the groups is to
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
dolphins or not) that the researchers impose. Volunteering for a group the group mean quiz score for people who used computer notes, and
could introduce a confounding variable. 6.92 represents the group mean score for people who used paper notes.
c. It was important that the subjects in the control group swim every b. We look to see how far 6.92 and 5.50 are from one another or from
day without dolphins so that this control group does everything (in- the overall mean of 6.21 to determine whether the note-taking method
cluding swimming) that the experimental group does except that might affect the score.
when they swim they don’t do it in the presence of dolphins. Without c. The number 1.76 represents the typical deviation of an observa-
this we wouldn’t know whether just swimming causes the difference tion from the expected value, in this case, from the overall mean. The
in the reduction of depression symptoms. number 1.61 represents the typical deviation of an observation after
d. Yes, this is an experiment because the subjects were randomly as- creating a model that takes into account whether the person is using
signed to the two groups. computer or paper notes.
1.1.9. d. Because the standard deviation of the residuals represents the left-
over variation, we can see that after including the type of notes as an
Observed variation Sources of Sources of explanatory variable in our model the unexplained variation has been
in: explained unexplained reduced (down to 1.61 from 1.76). This tells us that knowing the type
d. substantial reduction variation variation of note-taking method enables us to better predict scores.
in depression symptoms 1.1.13 Random assignment should make the two groups very
similar with regard to variables like intelligence, previous knowl-
Inclusion criteria a. swimming with • g. problems in the edge, or any other variable and thus likely eliminate possible
• b. mild to moderate dolphins or not personal lives of confounding variables.
depression the subjects during
the study 1.1.14
• c. no use of
antidepressant drugs • h. illness of a. This table shows us possible confounding variables but then
or psychotherapy four subjects during shows that subjects in the two groups are quite similar with
weeks prior to the the study regard to these characteristics, thus ruling out these possible
study confounding variables.
Design
b. We would want the p-values to be large, so we could say that
• e. swimming we have little to no evidence that there is a difference in mean age,
• f. staying on an island proportion of males, etc. between the two groups. We want our groups
for two weeks during to be very similar going into the study, so a causal conclusion is possi-
the study
ble if we find 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-olds might have different preferenc- c. R2 = 11.1328/199.62 = 0.0558. We can interpret this by saying that
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 whether the name of the hurricane is male or female.
and the opposite could be true with the younger group. We would not d. SSError = 199.62 − 11.13 = 188.49.
see this difference if the results of all the ages are combined together.
e. √ 188.4872/140 = 1.16.
0.28 if male name
Section 1.2 f. predicted hurricane risk rating = 5.29 +
{−0.28 if female name
,
1.2.1 B. SE of residuals = 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-
sponse variable is the quiz score.
1.2.4 A.
b. The effect of taking notes on paper is 0.71 and the effect of taking
1.2.5 C.
notes on the computer is −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. We can interpret it by saying that
1.2.8 Using the effects model, because 4.48 + 0.65 = 5.13 (the mean 16.675% of the variation of quiz score is explained by the note-taking
of the scent group) and 4.48 − 0.65 = 3.83 (the mean of the non-scent method.
group), the models are equivalent.
1.2.9 e. 120.92 – 20.164 = 100.756.
a. SSModel. f. √100.756/38 = 1.628. 0.71 if using paper notes
g. predicted 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 = SSModel/SSTotal = 0.4651. size of each group is just half of the total sample size.
b. R2 = 1 − SSError/SSTotal = 0.7111. ∑ (x − x )2 ∑ (y − y )2
1.2.11 b. all obs i
n + all _obs i
n _1
( 2
∑ 2 − 1x − x 2 + ∑ 2 − 1y − y)2
_
a. 8.
all obs( i
b. 6 – 8 = –2, 10 – 8 = 2. = all obs( i )̅ ̅) 1
_
n−1
( _
2 )2
c. 74. ∑all obs(xi − x̅) + ∑all obs(yi − y)̅
2 2
d. 40. =( )
n−2
e. 34.
f. 0.5405. Taking the square root we get
⎛∑n (x i − x̅) 2 ∑
√
∑ all obs (x i − x̅) 2 + ∑ all obs (y i − y̅) 2
n−2
(y i − y̅) 2⎞
1.2.12 n
a. The explanatory variable is the type of testing environment; it Use sum from 1 to n: 1 ⎜ + ⎟
_
2 i=1 i=1
is categorical. ⎝ n−1 n−1
b. The response variable is the test score; it is quantitative. 2 2⎞ n 2 2 n⎠
⎛ n 2 n 2
∑ (x i − x̅) + ∑ (y i − y̅) ∑ (x i − x̅) + ∑ (y i − y̅)
c. The two levels are quiet environment and distracting environment.
1.2.13 2
⎜
= _1 i=1
n —1
i=1 ⎟ = i=1
n−2
i=1
⎝ ⎠
2
a. SSTotal would probably be larger with these 10 subjects because n 2 n 2
√
∑( x i − x ) ̅ + ∑( y i − y ) ̅
with the wide variety of ages there would probably be more variability Taking the square root, we get i=1 i=1
.
n —2
in the test scores.
b. SSModel would probably be the same because it would still repre-
sent the difference between testing environments. 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.
variability in ages.
1.3.3 D.
1.2.14 The variance of the scores in the distracting environment is 2.5
1.3.4 A.
and the variance of the scores in the distracting enviro_
nment is 6. The
square root of the average of these two variances is √4.2_5 = 2.06. The 1.3.5 A.
SSError is 34, so the standard error of the residuals is √34/8 = 2.06. 1.3.6 The validity conditions are not met because the male sample
1.2.15 size is small and the distribution of the number of flip-flops owned by
the males is quite skewed to the right.
a. The explanatory variably is whether the name of the hurricane is
male or female and the response is the perceived 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 and the effect of naming the hurricane Christopher is 5.57 − 92.16 − 60.34
b. t = = 4.06.
5.29 = 0.28. The SSModel is 142(0.282) = 11.1328. 31.33 √1/32 + 1/32
, 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. √(24.242 + 38.782)/2 = 32.34. b. predicted quiz score = 6.21 + ,
{ 0.71 if paper
b. t = 69.97 − 85.71 = −1.69.
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
1.3.10 a statistically significant difference in the mean quiz scores between
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 x̅Christopher = 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 + { 0.28 if Christopher , SE of residuals = 1.16. 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 μ Christopher be the population average risk rating for hurricanes who aren’t.
given the name Christopher, and similarly for μChristina. The hypoth-
b. There is a lot of overlap between the distribution of the scores be-
eses are H0: μChristopher − μChristina = 0, that is, mean perceived risk
tween the two groups. It looks like the difference in sample means
ratings are the same regardless of whether the hurricane is named
Christopher or Christina name versus H A: 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
the perceived hurricane threat for the name Christina. of R2, the bigger the difference between the two samples. There-
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 CHAPTER 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 μ violent. 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.
b. i. noscent = 4.52; scent – noscent = 0.04. a. 4.
b. 93.
x x x
_ _ _
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.
Total 56 129.58 19.28
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
get our two sample means back when we resample. The dif- Error 78 1,253.26 16.07
ference in these should be the difference in our two sample Total 80 1,480.89 129.88
means, on average, or 1.292.
1.3.20 1.4.17
a. Only one combination would produce a result as extreme as a. The response variable is the amount of money spent on meals
−83.77, placing the nine largest times in one group and the nine and the explanatory variable is the type of music playing. The ex-
smallest times in the other group. perimental units are the customers eating at the restaurant during
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
Sources of Variation
Section 1.1 1.1.10 Color of a sign is the explanatory variable with white, yellow,
and red being the levels.
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. whether the student variation variation
obeyed the sign
1.1.6 B.
60.34 if rigid librarian Inclusion criteria a. color of the b. whether the subject
1.1.7 predicted number of uses for items =
{ 92.19 if eccentric poet sign was left-handed or
• c. time of day
1.1.8 right-handed
• e. age of subject
a. The inclusion criteria are having a clinical diagnosis of mild to d. attitude of student
moderate depression without any treatment four weeks prior and e. age of subject
during the study.
1.1.12
b. The purpose of randomly assigning subjects to the groups is to
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
dolphins or not) that the researchers impose. Volunteering for a group the group mean quiz score for people who used computer notes, and
could introduce a confounding variable. 6.92 represents the group mean score for people who used paper notes.
c. It was important that the subjects in the control group swim every b. We look to see how far 6.92 and 5.50 are from one another or from
day without dolphins so that this control group does everything (in- the overall mean of 6.21 to determine whether the note-taking method
cluding swimming) that the experimental group does except that might affect the score.
when they swim they don’t do it in the presence of dolphins. Without c. The number 1.76 represents the typical deviation of an observa-
this we wouldn’t know whether just swimming causes the difference tion from the expected value, in this case, from the overall mean. The
in the reduction of depression symptoms. number 1.61 represents the typical deviation of an observation after
d. Yes, this is an experiment because the subjects were randomly as- creating a model that takes into account whether the person is using
signed to the two groups. computer or paper notes.
1.1.9. d. Because the standard deviation of the residuals represents the left-
over variation, we can see that after including the type of notes as an
Observed variation Sources of Sources of explanatory variable in our model the unexplained variation has been
in: explained unexplained reduced (down to 1.61 from 1.76). This tells us that knowing the type
d. substantial reduction variation variation of note-taking method enables us to better predict scores.
in depression symptoms 1.1.13 Random assignment should make the two groups very
similar with regard to variables like intelligence, previous knowl-
Inclusion criteria a. swimming with • g. problems in the edge, or any other variable and thus likely eliminate possible
• b. mild to moderate dolphins or not personal lives of confounding variables.
depression the subjects during
the study 1.1.14
• c. no use of
antidepressant drugs • h. illness of a. This table shows us possible confounding variables but then
or psychotherapy four subjects during shows that subjects in the two groups are quite similar with
weeks prior to the the study regard to these characteristics, thus ruling out these possible
study confounding variables.
Design
b. We would want the p-values to be large, so we could say that
• e. swimming we have little to no evidence that there is a difference in mean age,
• f. staying on an island proportion of males, etc. between the two groups. We want our groups
for two weeks during to be very similar going into the study, so a causal conclusion is possi-
the study
ble if we find 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-olds might have different preferenc- c. R2 = 11.1328/199.62 = 0.0558. We can interpret this by saying that
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 whether the name of the hurricane is male or female.
and the opposite could be true with the younger group. We would not d. SSError = 199.62 − 11.13 = 188.49.
see this difference if the results of all the ages are combined together.
e. √ 188.4872/140 = 1.16.
0.28 if male name
Section 1.2 f. predicted hurricane risk rating = 5.29 +
{−0.28 if female name
,
1.2.1 B. SE of residuals = 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-
sponse variable is the quiz score.
1.2.4 A.
b. The effect of taking notes on paper is 0.71 and the effect of taking
1.2.5 C.
notes on the computer is −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. We can interpret it by saying that
1.2.8 Using the effects model, because 4.48 + 0.65 = 5.13 (the mean 16.675% of the variation of quiz score is explained by the note-taking
of the scent group) and 4.48 − 0.65 = 3.83 (the mean of the non-scent method.
group), the models are equivalent.
1.2.9 e. 120.92 – 20.164 = 100.756.
a. SSModel. f. √100.756/38 = 1.628. 0.71 if using paper notes
g. predicted 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 = SSModel/SSTotal = 0.4651. size of each group is just half of the total sample size.
b. R2 = 1 − SSError/SSTotal = 0.7111. ∑ (x − x )2 ∑ (y − y )2
1.2.11 b. all obs i
n + all _obs i
n _1
( 2
∑ 2 − 1x − x 2 + ∑ 2 − 1y − y)2
_
a. 8.
all obs( i
b. 6 – 8 = –2, 10 – 8 = 2. = all obs( i )̅ ̅) 1
_
n−1
( _
2 )2
c. 74. ∑all obs(xi − x̅) + ∑all obs(yi − y)̅
2 2
d. 40. =( )
n−2
e. 34.
f. 0.5405. Taking the square root we get
⎛∑n (x i − x̅) 2 ∑
√
∑ all obs (x i − x̅) 2 + ∑ all obs (y i − y̅) 2
n−2
(y i − y̅) 2⎞
1.2.12 n
a. The explanatory variable is the type of testing environment; it Use sum from 1 to n: 1 ⎜ + ⎟
_
2 i=1 i=1
is categorical. ⎝ n−1 n−1
b. The response variable is the test score; it is quantitative. 2 2⎞ n 2 2 n⎠
⎛ n 2 n 2
∑ (x i − x̅) + ∑ (y i − y̅) ∑ (x i − x̅) + ∑ (y i − y̅)
c. The two levels are quiet environment and distracting environment.
1.2.13 2
⎜
= _1 i=1
n —1
i=1 ⎟ = i=1
n−2
i=1
⎝ ⎠
2
a. SSTotal would probably be larger with these 10 subjects because n 2 n 2
√
∑( x i − x ) ̅ + ∑( y i − y ) ̅
with the wide variety of ages there would probably be more variability Taking the square root, we get i=1 i=1
.
n —2
in the test scores.
b. SSModel would probably be the same because it would still repre-
sent the difference between testing environments. 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.
variability in ages.
1.3.3 D.
1.2.14 The variance of the scores in the distracting environment is 2.5
1.3.4 A.
and the variance of the scores in the distracting enviro_
nment is 6. The
square root of the average of these two variances is √4.2_5 = 2.06. The 1.3.5 A.
SSError is 34, so the standard error of the residuals is √34/8 = 2.06. 1.3.6 The validity conditions are not met because the male sample
1.2.15 size is small and the distribution of the number of flip-flops owned by
the males is quite skewed to the right.
a. The explanatory variably is whether the name of the hurricane is
male or female and the response is the perceived 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 and the effect of naming the hurricane Christopher is 5.57 − 92.16 − 60.34
b. t = = 4.06.
5.29 = 0.28. The SSModel is 142(0.282) = 11.1328. 31.33 √1/32 + 1/32
, 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. √(24.242 + 38.782)/2 = 32.34. b. predicted quiz score = 6.21 + ,
{ 0.71 if paper
b. t = 69.97 − 85.71 = −1.69.
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
1.3.10 a statistically significant difference in the mean quiz scores between
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 x̅Christopher = 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 + { 0.28 if Christopher , SE of residuals = 1.16. 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 μ Christopher be the population average risk rating for hurricanes who aren’t.
given the name Christopher, and similarly for μChristina. The hypoth-
b. There is a lot of overlap between the distribution of the scores be-
eses are H0: μChristopher − μChristina = 0, that is, mean perceived risk
tween the two groups. It looks like the difference in sample means
ratings are the same regardless of whether the hurricane is named
Christopher or Christina name versus H A: 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
the perceived hurricane threat for the name Christina. of R2, the bigger the difference between the two samples. There-
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 CHAPTER 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 μ violent. 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.
b. i. noscent = 4.52; scent – noscent = 0.04. a. 4.
b. 93.
x x x
_ _ _
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.
Total 56 129.58 19.28
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
get our two sample means back when we resample. The dif- Error 78 1,253.26 16.07
ference in these should be the difference in our two sample Total 80 1,480.89 129.88
means, on average, or 1.292.
1.3.20 1.4.17
a. Only one combination would produce a result as extreme as a. The response variable is the amount of money spent on meals
−83.77, placing the nine largest times in one group and the nine and the explanatory variable is the type of music playing. The ex-
smallest times in the other group. perimental units are the customers eating at the restaurant during
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
