,Solution manual for Introduction to Statistical
Investigations, 2e Tintle
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, CHAPTER 1
Significance: How
Strong Is the Evidence?
Section 1.1 e. ½ of 1,383 or about 691 or 692
1.1.1 C. 1.1.11
1.1.2 A. a. Parameter
1.1.3 C. b. Statistic
1.1.4 A. c. The correct matches are shown below:
1.1.5 B. Column A Column B
1.1.6 D.
Coin flip Kevin shoots a field goal
1.1.7
Heads Kevin makes his field goal
a. The true proportion of times the racket lands face up
Tails Kevin misses his field goal
b. Parameter
Chance of heads Long-run proportion of field goals Kevin makes
c. 0.50
One repetition One set of 100 field goal shots by Kevin
d. 48 out of 100 does not constitute strong evidence that the spin-
1.1.12
ning process is not fair, because if the spinning process was fair (50%
chance of racket landing face up), getting 48 out of 100 spins landing a. Parameter
face up is a typical result. b. Statistic
e. Plausible that the spinning process is fair c. The correct matches are shown below:
1.1.8
Column A Column B
a. 24 out of 100 does constitute strong evidence that the spinning pro-
Coin flip Author plays a game of Minesweeper
cess is not fair, because if the spinning process was fair (50% chance of
racket landing face up), getting 24 out of 100 spins landing face up is Heads Author wins a game
an atypical result. Tails Author loses a game
b. Statistically significant evidence that spinning is not fair Chance of heads Long-run proportion of games that the author wins
1.1.9 One repetition One set of 20 Minesweeper games played
a. LeBron’s long-run proportion of making a field goal by author
b. Statistic 1.1.13
c. 0.50 a. 100 dots
d. Flip a coin 1,095 times and record the number of heads. Repeat b. Each dot represents the number of times out of 20 attempts the
this 1,000 times keeping track of the number of heads in each set author wins a game of Minesweeper when the probability that the
of 1,095. author wins is 0.50.
e. Approximately ½ of 1,095 (547 or 548) will be one of the most c. 10, because that is what will happen on average if the author plays
likely values. 20 games and wins 50% of her games.
1.1.10 d. No, we are not convinced that the author’s long-run proportion of
a. Parameter winning at Minesweeper is above 0.50 because 12 is a fairly typical
b. Statistic outcome for the number of wins out of 20 games when the long-run
proportion of winning is 0.50. Stated another way, 0.50 is a plausible
c. 1,383
value for the long-run proportion of games that the author wins Mine-
d. The number of heads (or proportion of heads) out of 1,383 flips sweeper based on the author getting 12 wins in 20 games.
5
c01Solutions.indd 5 10/16/20 9:15
, 6 C HA PTER 1 Significance: How Strong Is the Evidence
e. No, 0.50 is just a plausible (reasonable) explanation for the data. c. Yes, the simulation analysis gives strong evidence that the woman
Other explanations are possible (e.g., the author’s long-run propor- is not simply guessing. If she were guessing she’d rarely get 8 out of
tion of wins could be 0.55). 8 correct.
f. Yes, it means that there were special circumstances when the author d. Statistically significance evidence she is not guessing
played these 20 games and so these 20 games may not be a good rep- 1.1.18
resentation of the author’s long-run proportion of wins in Minesweeper.
a. The conclusion you’ve drawn is incorrect, because 5 out of 8 is a
1.1.14 likely result if someone is just guessing. In particular, when you do
a. 100 a simulation with probability of success = 0.5, sample size (n) = 8,
b. Each dot represents the number of times out of 10 attempts the getting 5 heads happens quite frequently.
toast lands buttered side down when the probability that the toast b. No, this does not prove that you cannot tell the difference. It’s plausi-
lands buttered side down is 0.50. ble (believable) you are not guessing, but we haven’t proven it.
c. 5, because that is what will happen on average if the toast is c. Applet inputs are: probability of success (π) = 0.5, sample size
dropped 10 times and 50% of the drops it lands buttered side down. (n) = 16, number of samples = 1000. Applet output suggests that 14
d. No, we are not convinced that the long-run proportion of times the out of 16 is a fairly unlikely result (~2 out of 1000 times). Thus, this
toast lands buttered side down is above 0.50 because 7 is a fairly typical result also provides strong evidence that the person actually has abil-
outcome for the number of times landing buttered side down out of ity better than random guessing. The applet value for π stays the same
10 drops of toast when the long-run proportion of times it lands but- because 0.50 still represents guessing, and n = 16 now because there
tered side down is 0.50. Stated another way, 0.50 is a plausible value for are 16 cups of tea.
the long-run proportion of times that the toast lands buttered side down 1.1.19
based on getting 7 times landing buttered side down in 10 drops. a. The long-run proportion of times that Zwerg chooses the correct
e. No, 0.50 is just a plausible (reasonable) explanation for the data. object
Other explanations are possible (e.g., the long-run proportion of times b. Zwerg is just guessing or Zwerg is choosing the correct object be-
the toast lands buttered side down could be 0.60). cause she understands the cue.
1.1.15 c. 37 out of 48 attempts seems fairly unlikely to happen by chance, be-
a. Statistic cause 24 out of 48 is what we would expect to happen in the long run.
b. Parameter d. 50%
c. Yes, it is possible to get 17 out of 20 first serves in if Mark was just 1.1.20
as likely to make his first serve as to miss it. a. 37 times out of 48 attempts
d. Getting 17 out of 20 first serves in if Mark was just as likely to make b. Applet input: probability of success is 0.5, sample size is 48, num-
the serve as to miss it is like flipping a coin 20 times and getting heads ber of samples is 1,000
17 times. This is fairly unlikely, so 17 out of 20 first serves in is not a 160
very plausible outcome if Mark is just as likely to make his first serve
as to miss it.
1.1.16 120
a. Observational unit is each cup, variable is whether the tea or
milk was poured first.
80
b. The long-run proportion of times the woman correctly identifies a
cup
c. 8, p̂ = 1.0 40
d. Yes, it’s possible she could get 8 out of 8 correct if she was just
randomly guessing with each cup.
0
e. Getting 8 out of 8 correct if she was randomly guessing is like flip- 10 12 14 16 18 20 22 24 26 28 30 32 34 36
ping a coin 8 times and getting heads every time—a fairly unlikely
Number of successes
result. Thus, 8 out of 8 seems unlikely.
1.1.17 c. Yes, it appears as if the chance model is wrong, as it is highly un-
likely to obtain a value as large as 37 when there is a 50% chance of
a. Toss a coin 8 times to represent the 8 cups of tea. Heads represents
picking the correct object.
a correct identification of what was poured first, tea or milk, and tails
represents an incorrect identification of what was poured first. Count the d. We have strong evidence that Zwerg can correctly follow this type
number of heads in the 8 tosses, this represents the number of correct of direction more than 50% of the time.
identifications of what was poured first out of the 8 cups. Repeat this e. The results are statistically significant because we have strong evi-
process many times (1,000). You will end up with a distribution of the dence that the chance model is incorrect.
number of correct identifications out of 8 cups when the chance of a cor- 1.1.21
rect identification is 50%. If 8 correct out of 8 cups rarely occurs, then it is
a. Zwerg is just guessing or Zwerg is picking up on the experimenter
unlikely that the woman was just guessing as to what was poured first.
cue to make a choice.
b. Using the applet shows that 8 out of 8 occurs rarely by chance (~4
b. 26 out of 48 seems like the kind of thing that could happen just by
times out of 1,000), confirming the fact that 8 out of 8 is quite unlikely
chance because 24 out of 48 is what we would expect on average in the
to occur just by chance.
long run.
c01Solutions.indd 6 10/16/20 9:15
, Solutions to Problems 7
c. 50% b. 17 out of 30 seems like the kind of thing that could happen just by
1.1.22 chance because 15 out of 30 is what we would expect on average in the
long run.
a. 26 times out of 48 attempts
c. 50%
b. Applet input: probability of success is 0.5, sample size is 48, num-
ber of samples is 1,000. This distribution is centered at 24. 1.1.26
a. 17 times out of 30 attempts
180 0.3300 b. Applet input: probability of success is 0.5, sample size is 30, num-
ber of samples is 1,000. The distribution is centered at 15.
c. We cannot conclude the chance model is wrong because a value as
120
large or larger than 17 is fairly likely.
d. We do not have strong evidence that Janine can land the majority
of her serves in-bounds when serving right-handed.
60
e. The chance model (Janine lands 50% of her serves in-bounds when
serving right-handed) is a plausible explanation for the observed data
(17 out of 30).
0
12 15 18 21 24 27 30 33 36
26 f. This does not prove that Janine lands 50% of her right-handed
Number of heads short-serves in bounds. This is just one plausible explanation for Ja-
nine’s performance. We cannot rule out a 0.50 long-run proportion
c. We cannot conclude the chance model is wrong because a value as of serves in bounds as an explanation for Janine landing 17 out of 30
large or larger than 26 is fairly likely. serves in bounds.
d. We do not have strong evidence that Zwerg can correctly follow 1.1.27
this type of direction more than 50% of the time.
a. 0.50
e. The chance model (Zwerg guessing) is a plausible explanation for
the observed data (26 out of 48), because the observed outcome was b. 20
likely to occur under the chance model. c. 1,000 (or some large number)
f. Less convincing evidence that Zwerg can correctly follow this type d. 12 out 20 is a fairly likely value because it occurred frequently in
of direction more than 50% of the time. We could have anticipated this the simulated data.
because 26 out of 48 is closer to 24 out of 48 than is 37 out of 48.
1.1.28
g. This does not prove that Zwerg is just guessing. Guessing is just
one plausible explanation for Zwerg’s performance in this experiment. a. 0.50
We cannot rule out guessing as an explanation for Zwerg getting 26 b. 100
out of 48 correct. c. 1,000 (or some large number)
1.1.23 d. 60 out of 100 is somewhat unlikely because it occurred somewhat
a. The long-run proportion of times that Janine’s short serve lands in infrequently in the simulated data.
bounds when serving left-handed
e. The sample size was different (20 serves vs. 100 serves).
b. Janine has a 50-50 chance of landing in- bounds and so 23 out of 30
1.1.29 B.
happened by chance; Janine’s chance of landing her serve in bounds
is greater than 50%. 1.1.30
c. 23 out of 30 seems somewhat unlikely to occur if she has a 50-50 a. 12 coin flips
chance of landing the serve in-bounds b. It is unlikely, because at least 11 correct (heads) happened very
d. 50% rarely on 12 coin flips.
1.1.24 c. Yes, we have very strong evidence that Milne does better than
a. 23 out of 30 attempts just guess because 11 correct out of 12 is very unlikely to happen by
b. Applet input: probability of success is 0.5, sample size is 30, num- just guessing.
ber of samples is 1,000. Centered ~15 d. Even stronger evidence, because 12 correct is even more extreme
c. Yes, it appears as if the chance model is wrong, as it is highly un- than 11 correct.
likely to obtain a value as large as 23 when there is a 50% chance of e. Compared to 11, 12 is farther away in the tail and farther away
getting the serve in-bounds. from 6 (which is ½ of 12).
d. We have strong evidence that Janine can land the majority of her 1.1.31
serves in bounds.
a. 20 times
e. The results are statistically significant because we have strong evi-
dence that the chance model is incorrect. b. Yes, because 16 heads out 20 coin flips is very rare.
1.1.25 c. Yes, because 16 heads in 20 tosses is very rare, so we have strong
evidence that Mercury’s choices are not random.
a. Janine has a 50-50 chance of landing in- bounds and so 17 out of 30
happened by chance; Janine’s chance of landing her serve in-bounds 1.1.32
is greater than 50%. a. 20 times
c01Solutions.indd 7 10/16/20 9:15
, 8 C HA PTER 1 Significance: How Strong Is the Evidence
b. No, because 11 heads in 20 coin flips is not unusual. d. Because 4 of the 100 simulated outcomes gave a result of 7 or more,
c. No, we do not have evidence that Panzee’s choices are that dif- the p-value is 0.04.
ferent from what could have happened if Panzee was just randomly e. We have strong evidence that Sarah is not simply guessing, be-
choosing containers. cause 7 out 8 rarely occurs by chance (if just guessing)
1.1.33 f. If Sarah doesn’t understand how to solve problems and is just guess-
ing at which picture to select, the probability she would get 7 or more
a. We would flip 40 coins where one side (say heads) represents
correct out of 8 is 0.04.
Donaghy’s foul calls favoring the team that received heavier betting
and the other side (say tails) represents Donaghy’s foul calls not favor- g. A single dot represents the number of times Sarah would choose the
ing the team that received heavier betting. correct picture (out of 8) if she were just guessing.
b. Because 28 out of 40 is way out in the upper tail of the chance 1.2.17
distribution, it is very unlikely that Donaghy’s foul calls would favor a. Null: The long-run proportion of times Hope will go to the correct object
the team that received heavier betting 28 out of 40 times if the chance is 0.50, Alternative: The long-run proportion of times that Hope will go to
model is true. the correct object is more than 0.50
1.1.34 b. H0: π = 0.50, Ha: π > 0.50
a. We are assuming that Buzz’s probability of choosing the cor- c. 0.23 (23 dots are 0.60 or larger)
rect button does not change and that previous trials don’t influence d. No, the approximate p-value is 0.23, which provides little to no ev-
future guesses. idence that Hope understands pointing.
b. The parameter is his actual probability of pushing the correct button. e. 0.70
f. i.
Section 1.2 1.2.18 Researcher A has stronger evidence against the null hypothe-
1.2.1 D. sis because his p-value is smaller.
1.2.2 C. 1.2.19
1.2.3 A. a. Roll a die 20 times, and keep track of how many times ‘one’ is
1.2.4 B. rolled. Repeat this many times.
1.2.5 A. b. Using a set of five black cards and one red card, shuffle the cards
and choose a card. Note the color of the card and return it to the deck.
1.2.6 D.
Shuffle and choose a card 20 times keeping track of how many times
1.2.7 C. the red card is selected. Repeat this many times.
1.2.8 B. c. Roll 30 times, then repeat.
1.2.9 B. d. Shuffle and choose a card 30 times, then repeat.
1.2.10 C. e. Roll a die 20 times, and keep track of how many times a ‘one, two,
1.2.11 B. three, or four’ is rolled. Repeat this many times.
1.2.12 A. f. Using a set of one black card and two red cards, shuffle the cards
1.2.13 and choose a card. Shuffle and choose a card 20 times keeping track
of how many times the red card is selected. Repeat this many times.
a. 0.25
1.2.20
b. 25 (because 0.25 × 100 = 25)
a. Observational units: 40 heterosexual couples who agreed on their
1.2.14
response to which person was the first to say “I love you,” Variable:
a. H0 = Null hypothesis Whether the man or woman said “I love you” first; this is a categorical
b. Ha = Alternative hypothesis variable
c. ˆ
p = sample proportion b. Null: The proportion of all couples where the male said “I love
d. π = long-run proportion (parameter) you” first is 0.50. Alternative: The proportion of all couples where
the male said “I love you” first greater than 0.50
e. n = sample size
c. π is the proportion of all couples where the male said “I love you”
1.2.15 ˆ p is the value of the observed statistic, while the p-value is
first
the probability that the observed statistic or more extreme occurs if
the null hypothesis is true; p-value is a measure of strength of the d. 28/40 = 0.7 is the sample proportion; we use the symbol
evidence.
1.2.16 e. Flip a coin 40 times and keep track of the number of heads. Repeat
the 40 coin flips, 1,000 times. Calculate the proportion of sets of 40
a. The long-run proportion of times that Sarah chooses the correct
coin flips where 28 or more heads were obtained. That proportion is
photo, π
the p-value.
b. 7/8 = 0.875.
f. Applet: π = 0.50, n = 40, number of samples = 1,000. To find the
c. Null: The long-run proportion of times Sarah chooses the correct pho- p-value, we find the proportion of times a value greater than or equal
to is 0.50. Alternative: The long-run proportion of times Sarah chooses to 28 is observed. The p-value is approximately 0.008.
the correct photo is more than 0.50.
g. The p-value is the probability of observing a value of 28 or greater,
H0: π = 0.50, Ha: π > 0.50 assuming that for 50% of couples the man said “I love you” first.
c01Solutions.indd 8 10/16/20 9:15
, Solutions to Problems 9
h. The small p-value gives us strong evidence that for more than 50% d. Null hypothesis: The long-run proportion of times that rhesus
of couples the man said “I love you” first. monkeys make the correct choice when the researcher looks to-
1.2.21 wards the correct box is 0.50 (just guessing). Alternative hypothesis: The
long-run proportion of times that rhesus monkeys make the correct
a. Obs units = university students, Variable = male or female said “I
choice is more than 0.50.
love you” first
H0: π = 0.50, Ha: π > 0.50
b. Null: The long-run proportion of university student relationships
in which the male says “I love you” first is 0.50, Alternative: The long- e. Flip a coin 40 times and record the number of heads. Repeat this
run proportion of university student relationships in which the male process 999 more times, yielding a set of 1,000 values of the number
says “I love you” first is more than 0.50, of heads received in 40 coin tosses. Compute the p-value as the pro-
portion of times 31 or larger was obtained by chance in the 1,000 sets
c. 59/96 = 0.61 is the sample proportion, we use the symbol ˆ
pto de-
of 40 coin tosses. If the p-value is small (indicating 31 or larger rarely
note this quantity.
occurs by chance), then this is convincing evidence that rhesus mon-
d. We could flip a coin 96 times and keep track of the number of keys can interpret human gestures better than by random chance.
heads. Then do many, many more sets of 96 coin flips, keeping track
f. The approximate p-value from the applet (using π = 0.50, n = 40,
of the number of heads each time.
number of samples = 1,000) is 0.001 (probability of 31 or greater). This
e. Probability of heads: 0.5, number of tosses: 96, number of repe- small p-value means that 31 out of 40 is strong evidence that the rhesus
titions: 1,000. To find the p-value, we find the proportion of times a monkeys are not guessing, which may lead us to believe that rhesus
value is greater than or equal to 0.61. The p-value is approximately monkeys can interpret gestures to indicate which box to choose.
0.016.
1.2.24 The p-value is approximately 0.25. We don’t have strong evidence
f. The p-value (0.016) is the probability of 0.61 or larger assuming the that the author’s long-run proportion of wins in Minesweeper is greater
null hypothesis is true. than 0.50. The null hypothesis (long-run proportion of wins is 0.50) is a
g. We have strong evidence that the long-run proportion of university plausible explanation for her winning 12 out of 20 games.
student relationships in which the male says “I love you” first is more 1.2.25 The p-value is approximately 0.134. We don’t have strong evi-
than 0.50. dence that the author’s long-run proportion of wins in Spider Solitaire
1.2.22 is greater than 0.50. The null hypothesis (long-run proportion of wins
a. Obs units: each of the 40 monkeys. Variable: correct choice or not is 0.50) is a plausible explanation for him winning 24 out of 40 games.
(categorical) 1.2.26
b. The long-run proportion of times that a monkey will make the cor- a. The long-run proportion of times a spun penny lands heads
rect choice, π b. The p-value = 0.16. There is little to no evidence that a spun penny
c. 30/40 = 0.75. Statistic. We use the symbol ˆ
to denote this quantity.
p lands heads less than 50% of the time.
d. Null hypothesis: The long-run proportion of times that rhesus c. Null would be the same, Alternative would be > 0.50. To calculate the
monkeys make the correct choice when observing the researcher jerk p-value, find the probability that 29 or larger (0.58 or larger) occurred.
their head is 0.50 (just guessing). Alternative hypothesis: The long- 1.2.27
run proportion of times that rhesus monkeys make the correct choice
a. Null: π = 0.50, Alternative: π > 0.50
is more than 0.50.
b. The p-value = 0.31. There is little to no evidence that a coin that starts
H0: π = 0.50, Ha: π > 0.50 out heads will land heads more than 50% of the time.
c. No, the p-value does not prove the null hypothesized value (0.50)
e. Flip a coin 40 times and record the number of heads. Repeat
is correct, just that it is a plausible value for the parameter.
this process 999 more times, yielding a set of 1,000 values of the
number of heads received in 40 coin tosses. Compute the p-value d. In the long run it will land heads 51% of the time, in any particular
as the proportion of times 30 or larger was obtained by chance in set of 100 flips it is likely the coin won’t land heads exactly 51 out of
the 1,000 sets of 40 coin tosses. If the p-value is small (indicating 30 100 times.
or larger rarely occurs by chance), then this is convincing evidence 1.2.28 A simulation analysis using a null hypothesis probability of
that rhesus monkeys can interpret human gestures better than by 0.75 yields a p-value of 0.10, meaning that the set of 20 free throws by
random chance. your friend (and making 12/20 of them) provides little to no evidence
f. The approximate p-value from the applet (using π = 0.50, n = 40, that your friend’s long-run proportion of free throws made is worse
number of samples = 1,000) is 0.001 (probability of 30 or greater). than the NBA average.
This small p-value means that 30 out of 40 is strong evidence that 1.2.29 A simulation analysis using a null hypothesis probability of
the rhesus monkeys are not guessing, which may lead us to believe 0.75 yields a p-value of 0.02, meaning that the set of 40 free throws by
that rhesus monkeys may be able to understand a head jerk to indi- your friend (and making 24/20 of them) provides strong evidence that
cate which box to choose. your friend’s long-run proportion of free throws made is worse than
1.2.23 the NBA average.
a. Obs units: each of the 40 monkeys. Variable: correct choice or not 1.2.30
(categorical) a. Null: Mercury will randomly choose between two containers ver-
b. The long-run proportion of times that a monkey will make the cor- sus Alternative: Mercury will choose the container with more banan-
rect choice, π as more often than the other container.
c. 31/40 = 0.775. Statistic. We use the symbol ˆ
to denote this b. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
quantity. c. 0.02
c01Solutions.indd 9 10/16/20 9:15
, 10 C HA PTER 1 Significance: How Strong Is the Evidence
d. Yes, because very rarely (p-value = 0.02) would Mercury pick the 1.3.11
container with more bananas at least 16 times in 20 trials by just ran- a. –3.47 (100 out of 400; 25%), –3.80 (20 out of 120; 16.7%), –4.17
domly choosing. (65 out of 300; 21.7%). Because of the random nature of obtaining the
e. Answers may vary depending on what you consider “strong evi- SDs of the null distributions, these standardized statistics can vary by
dence.” For a p-value to be at most 0.05, the sample proportion would as much as about plus or minus 0.2.
have to be at least 0.75; so, 15 or more out of 20. b. 65 out of 300 is the strongest evidence, 100 out of 400 is the least
1.2.31 strong evidence.
a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50 1.3.12
b. About 0.06 a. Friend D because they played more games
c. We have only moderate evidence because the p-value is fairly b. Friend D because this is more evidence against the null hypothesis
small, but not small enough to be considered strong evidence.
c. Friend D because this is more evidence against the null hypothesis
1.2.32
d. Friend D because a smaller standard deviation leads to a larger
a. Null, H0: π = 0.25 versus Alternative, Ha: π > 0.25 standardized statistic
b. About 0.003 1.3.13 Friend G because the value of their statistic (30 out of 40) is
c. We have very strong evidence because the p-value is very small; larger than Friend F (15 out of 40) and thus will be farther in the tail of
it is unlikely that 42 correct matches out of 113 would happen by the distribution.
just guessing.
1.3.14 Simulation yields a standard deviation of the null distribution
1.2.33 of 0.112, and a standardized statistic of approximately 0.89, which
a. Null, H0: π = 0.25 versus Alternative, Ha: π > 0.25 provides little or no evidence that the author’s long-run proportion of
wins in Minesweeper is higher than 0.50.
b. About 0.09
c. We have only moderate evidence because the p-value is somewhat 1.3.15 Simulation yields a standard deviation of the null distribu-
small; it is somewhat unlikely that 35 correct matches out of 113 tion of 0.125, and a standardized statistic of approximately (0.9375
would happen by just guessing, but not unlikely enough to give us − 0.50)/0.125 = 3.5, which provides very strong evidence that the
strong evidence. long-run proportion of Buzz pushing the correct button is higher
than 0.50.
1.2.34 When there are many outcomes, the probability of a single out-
come, even if it is close to what we would expect, can be very small. 1.3.16 Simulation yields a standard deviation of the null distribution
We need to look at more outcomes to get a better understanding of the of approximately 0.094, and a standardized statistic of approximately
likeliness or unlikeliness of an outcome. 0.76, which provides little to no evidence that the long-run proportion
of times Buzz pushes the correct button is higher than 0.50.
1.2.35
1.3.17
a. He claims he can flip a coin and make it come up heads every
single time. a. The long-run proportion of all couples that lean their heads to the
right while kissing, π.
b. Five
b. Null: π = 0.50, Alternative: π > 0.50
Section 1.3
c. 80/124 = 0.645 = p̂
1.3.1 C.
d. (0.645 − 0.5)/0.045 = 3.22 = z
1.3.2
e. The observed proportion of couples leaning their heads to the right
1.3.3 because the standard deviation is smaller while kissing is 3.22 standard deviations away from the null hypothe-
1.3.4 sized parameter value of 0.5.
D. Even though C is farther away from 0 (z = 3), because a positive f. We have strong evidence that the proportion of couples that lean
standardized statistic puts the observed statistic in the right tail, this their heads to the right while kissing is more than 0.50.
means the observed statistic was a number (much) larger than 0.25,
1.3.18
which is not evidence for the alternative hypothesis that π < 0.25.
a. (0.645 − 0.60)/0.044 = 1.02
1.3.5
b. The standardized statistic is smaller. This makes sense because the null
a. False b. True c. False d. False
hypothesis is now closer to the observed statistic (less extreme).
1.3.6 B.
1.3.19
1.3.7 C. a. Null: The long-run proportion of all couples that have the male say
1.3.8 A. “I love you” first is 0.50. Alternative: The long-run proportion is more
1.3.9 D. than 0.50.
1.3.10 b. z = (0.70 − 0.50)/0.079 = 2.53
a. 0.06 c. The observed proportion of couples where the males says “I love
you” first is 2.53 standard deviations above the null hypothesized pa-
b. No, greater, 0.05
rameter value of 0.50.
c. 1.72
d. We have strong evidence that the proportion of couples for which
d. No, less, 2 the male says ”I love you” first is more than 0.50.
c01Solutions.indd 10 10/16/20 9:15
, Solutions to Problems 11
1.3.20 g. We have little to no evidence that the long-run proportion of times
a. Null: The long-run proportion of times that rhesus monkeys that Zwerg makes the correct choice when using a marker is greater
choose the correct box is 0.50. Alternative. The long-run proportion of than 0.50.
times that rhesus monkeys choose the correct box is greater than 0.50. 1.3.24
b. z = (0.75 – 0.5)/0.079 = 3.16 a. The long-run proportion of times that 10-month olds choose the
c. The observed proportion of rhesus monkeys that chose the box the helper toy, π
experimenter gestured towards is 3.16 standard deviations away from b. Null: π = 0.50, Alternative: π > 0.50
the null hypothesized parameter value of 0.5.
c. 14/16 = 0.875 = p̂
d. We have strong evidence that the long-run proportion of times that
d. z = (0.875 − 0.50)/0.125 = 3
rhesus monkeys choose the correct box is greater than 0.50.
e. The observed proportion of times the 10-month-old babies chose the
1.3.21
helper toy is 3 standard deviations above the null hypothesized parame-
a. The long-run proportion of times the lady correctly identifies ter value of 0.50.
which was poured first, π
f. We have strong evidence that the long-run proportion of times that
b. Null: π = 0.50, Alternative: π > 0.50 10-minth-old babies choose the helper toy is greater than 0.50.
c. 8/8 = 1 = ˆ
p 1.3.25
d. 0.50, because that is the value of the parameter if the null hy-
a. Yes, the p-value will be small because the standardized statistic is
pothesis is true. The standard deviation will be positive because the
large
standard deviation must be at least 0, and is only equal to zero if
there is no variability in the values (there will be variability in the b. A p-value of approximately 0.002. The p-value is the probability
simulated statistics). observing 14/16 or larger assuming the null hypothesis is true.
e. z = (1 − 0.50)/0.177 = 2.82 c. We have strong evidence that the long-run proportion of times that
10-month-old babies choose the helper toy is greater than 0.50.
f. The observed proportion of times the lady correctly identified
which was poured first is 2.82 standard deviations away from the null d. Yes, because both the p-value and the standardized statistic are meas-
hypothesized parameter value of 0.50. uring the strength of evidence (how far out in the tail the observed value
is), and so should lead to the same conclusion.
g. We have strong evidence that the long-run proportion of times that
the lady makes the correct identification is greater than 0.50. 1.3.26
1.3.22 a. The long-run proportion of people that choose the number 3, π
a. The long-run proportion of times that Zwerg makes the correct b. Null: π = 0.25, Alternative: π > 0.25
choice when the object is pointed at, π c. 14/33 = 0.42 = p̂
b. Null: π = 0.50, Alternative: π > 0.50 d. The mean = 0.248 and SD = 0.076
c. 37/48 = 0.77 = p̂ e. (0.42 – 0.248)/0.076 = 2.26
d. 0.5, because that is the value of the parameter if the null hypothe-
f. The observed proportion of people that chose the number 3 is 2.26
sis is true. The standard deviation will be positive because the stand-
standard deviations above the null hypothesized parameter value
ard deviation must be at least 0, and is only equal to zero if there is
of 0.25.
no variability in the values (there will be variability in the simulated
statistics). g. We have strong evidence that the long-run proportion of people
that will choose the number 3 is greater than 0.25.
e. (0.77 – 0.5)/0.073 = 3.70
1.3.27
f. The observed proportion of times Zwerg made the correct choice is
3.70 standard deviations away from the null hypothesized parameter a. Yes, because the standardized statistic is far from zero.
value of 0.5. b. The p-value is approximately 0.02, and is the probability of observ-
g. We have strong evidence that the long-run proportion of times that ing 14/33 or larger assuming the null hypothesis is true.
Zwerg makes the correct choice is greater than 0.50. c. We have strong evidence that the long-run proportion of people
1.3.23 that will choose the number 3 is greater than 0.25.
a. The long-run proportion of times that Zwerg makes the correct d. Yes, because both the p-value and the standardized statistic are
choice when using a marker, π measuring the strength of evidence (how far out in the tail the ob-
b. Null: π = 0.50, Alternative: π > 0.50 served value is), and so should lead to the same conclusion.
c. 26/48 = 0.54 = p̂ 1.3.28
d. 0.5, because that is the value of the parameter if the null hypothesis a. The long-run proportion of people that choose a big number, π
is true. The standard deviation will be positive because the standard b. Null: π = 0.50, Alternative: π > 0.50
deviation must be at least 0, and is only equal to zero if there is no vari-
c. 19/33 = 0.58 = p̂
ability in the values (there will be variability in the simulated statistics).
d. The mean = 0.50 and the SD = 0.084
e. (0.54 – 0.5)/0.073 = 0.55
e. (0.58 − 0.5)/0.084 = 0.952
f. The observed proportion of times Zwerg made the correct choice is
0.55 standard deviations away from the null hypothesized parameter f. We have little to no evidence that the long-run proportion of peo-
value of 0.5. ple that will choose a ”big number” is greater than 0.50.
c01Solutions.indd 11 10/16/20 9:15
, 12 C HA PTER 1 Significance: How Strong Is the Evidence
1.3.29 1.3.34
a. No, because the standardized statistic is not far from zero. a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
b. The p-value is approximately 0.153, and is the probability of ob- b. z = (0.8667 – 0.50)/0.129 = 2.84; because the standardized statistic
serving 19/33 or larger assuming the null hypothesis is true is greater than 2, we have strong evidence that dogs are more likely to
c. We have little to no evidence that the long-run proportion of peo- approach the positive donor.
ple that will choose a big number is greater than 0.50. c. p-value = 0.003; because this is less than 0.05, we again have strong
d. Yes, because both the p-value and the standardized statistic are evidence that dogs are more likely to approach the positive donor.
measuring the strength of evidence (how far out in the tail the ob- 1.3.35
served value is), and so should lead to the same conclusion. a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
1.3.30 b. z = (0.6667 – 0.50)/0.129 = 1.29; because the standardized statistic
a. 0.50 is less than 2, we do not have strong evidence that dogs are more likely
b. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50 to approach the positive donor.
c. 0.144 c. p-value = 0.150; because this is greater than 0.05, we again do not
have strong evidence that dogs are more likely to approach the posi-
d. 2.89
tive donor.
e. Because the standardized statistic is much larger than 2, we have
convincing evidence that Milne’s answers are better than random
Section 1.4
guesses.
1.4.1 D.
f. Stronger evidence, because the standardized statistic is 3.46 which
is farther from 0 than is 2.89. 1.4.2 A.
1.3.31 1.4.3 D.
a. Null, H0: GY just guesses, π = 0.50 versus Alternative, Ha: GY is 1.4.4 B.
correct more often than just guessing, π > 0.50, where π represents 1.4.5 C.
the chance that GY guesses correctly on any trial.
1.4.6 A.
b. Sample proportion; SD of the sample proportion is about 0.034.
1.4.7
c. 6.03
a. Larger, because the sample proportion is closer to the hypothe-
d. Very strong evidence that GY is more likely than random chance to sized value (0.50) of the long-run proportion.
correctly identify whether or not the square was present because the
b. Smaller, because less likely to get extreme values of the statistic
standardized statistic is so large.
from a larger sample.
e. p-value is <0.001 that is very small; yes, leads to same conclusion
c. Larger, because now including at least as extreme values of the
as the standardized statistic.
sample proportion in both tails of the null distribution.
1.3.32
1.4.8
a. Null, H0: GY just chooses between the two wagers randomly, π =
a. Smaller, as in, closer to 0, because the sample proportion is closer
0.50 versus Alternative, Ha: GY chooses the higher wager more often
to the hypothesized value (0.5) of the long-run proportion.
than the lower wager, π > 0.50, where π represents the chance that GY
chooses the higher wager. b. Larger, as in, farther from 0; because less likely to get extreme val-
ues of the statistic from a larger sample.
b. Sample proportion; SD of the sample proportion is about 0.043.
c. Same, because the sample proportion is still the same distance
c. –0.58
away from 0 in the null distribution.
d. No evidence that GY is more likely to take the high wager than the
low wager because the standardized statistic is close to zero. 1.4.9
e. p-value is about 0.75 that is not small at all; yes, leads to the same a. Null, H0: Students randomly choose between the two cookies,
conclusion as the standardized statistic. π = 0.50 versus Alternative, Ha: Students have a genuine preference
for Chips Ahoy over Chipsters, π > 0.50, where π represents the long-
f. Because the 67/141 is 0.475 which less than 0.5, and in the oppo- run proportion of students that choose Chips Ahoy over Chipsters.
site direction than that conjectured by the alternative hypothesis.
b. 0.03
1.3.33
c. Null, H0: π = 0.50 versus Alternative, Ha: π ≠ 0.50
a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
d. 0.08
b. 0.003; there is about a 0.3% chance that Paul got all 8 predictions
correct by just guessing. 1.4.10
c. Based on the very small p-value, we have very strong evidence that a. 0.05
Paul is doing better than just guessing. b. 0.11 (0.30 or less and 0.70 or more)
d. Based on the small p-value, it is surprising that an animal would 1.4.11
correctly choose the winning team in 8 of 8 competitions. This is a a. Smaller
fairly small sample size, however, and it would be interesting to see
whether Paul could sustain this accuracy through a much larger num- b. Smaller
ber of competitions. c. Larger
c01Solutions.indd 12 10/16/20 9:15
Investigations, 2e Tintle
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, CHAPTER 1
Significance: How
Strong Is the Evidence?
Section 1.1 e. ½ of 1,383 or about 691 or 692
1.1.1 C. 1.1.11
1.1.2 A. a. Parameter
1.1.3 C. b. Statistic
1.1.4 A. c. The correct matches are shown below:
1.1.5 B. Column A Column B
1.1.6 D.
Coin flip Kevin shoots a field goal
1.1.7
Heads Kevin makes his field goal
a. The true proportion of times the racket lands face up
Tails Kevin misses his field goal
b. Parameter
Chance of heads Long-run proportion of field goals Kevin makes
c. 0.50
One repetition One set of 100 field goal shots by Kevin
d. 48 out of 100 does not constitute strong evidence that the spin-
1.1.12
ning process is not fair, because if the spinning process was fair (50%
chance of racket landing face up), getting 48 out of 100 spins landing a. Parameter
face up is a typical result. b. Statistic
e. Plausible that the spinning process is fair c. The correct matches are shown below:
1.1.8
Column A Column B
a. 24 out of 100 does constitute strong evidence that the spinning pro-
Coin flip Author plays a game of Minesweeper
cess is not fair, because if the spinning process was fair (50% chance of
racket landing face up), getting 24 out of 100 spins landing face up is Heads Author wins a game
an atypical result. Tails Author loses a game
b. Statistically significant evidence that spinning is not fair Chance of heads Long-run proportion of games that the author wins
1.1.9 One repetition One set of 20 Minesweeper games played
a. LeBron’s long-run proportion of making a field goal by author
b. Statistic 1.1.13
c. 0.50 a. 100 dots
d. Flip a coin 1,095 times and record the number of heads. Repeat b. Each dot represents the number of times out of 20 attempts the
this 1,000 times keeping track of the number of heads in each set author wins a game of Minesweeper when the probability that the
of 1,095. author wins is 0.50.
e. Approximately ½ of 1,095 (547 or 548) will be one of the most c. 10, because that is what will happen on average if the author plays
likely values. 20 games and wins 50% of her games.
1.1.10 d. No, we are not convinced that the author’s long-run proportion of
a. Parameter winning at Minesweeper is above 0.50 because 12 is a fairly typical
b. Statistic outcome for the number of wins out of 20 games when the long-run
proportion of winning is 0.50. Stated another way, 0.50 is a plausible
c. 1,383
value for the long-run proportion of games that the author wins Mine-
d. The number of heads (or proportion of heads) out of 1,383 flips sweeper based on the author getting 12 wins in 20 games.
5
c01Solutions.indd 5 10/16/20 9:15
, 6 C HA PTER 1 Significance: How Strong Is the Evidence
e. No, 0.50 is just a plausible (reasonable) explanation for the data. c. Yes, the simulation analysis gives strong evidence that the woman
Other explanations are possible (e.g., the author’s long-run propor- is not simply guessing. If she were guessing she’d rarely get 8 out of
tion of wins could be 0.55). 8 correct.
f. Yes, it means that there were special circumstances when the author d. Statistically significance evidence she is not guessing
played these 20 games and so these 20 games may not be a good rep- 1.1.18
resentation of the author’s long-run proportion of wins in Minesweeper.
a. The conclusion you’ve drawn is incorrect, because 5 out of 8 is a
1.1.14 likely result if someone is just guessing. In particular, when you do
a. 100 a simulation with probability of success = 0.5, sample size (n) = 8,
b. Each dot represents the number of times out of 10 attempts the getting 5 heads happens quite frequently.
toast lands buttered side down when the probability that the toast b. No, this does not prove that you cannot tell the difference. It’s plausi-
lands buttered side down is 0.50. ble (believable) you are not guessing, but we haven’t proven it.
c. 5, because that is what will happen on average if the toast is c. Applet inputs are: probability of success (π) = 0.5, sample size
dropped 10 times and 50% of the drops it lands buttered side down. (n) = 16, number of samples = 1000. Applet output suggests that 14
d. No, we are not convinced that the long-run proportion of times the out of 16 is a fairly unlikely result (~2 out of 1000 times). Thus, this
toast lands buttered side down is above 0.50 because 7 is a fairly typical result also provides strong evidence that the person actually has abil-
outcome for the number of times landing buttered side down out of ity better than random guessing. The applet value for π stays the same
10 drops of toast when the long-run proportion of times it lands but- because 0.50 still represents guessing, and n = 16 now because there
tered side down is 0.50. Stated another way, 0.50 is a plausible value for are 16 cups of tea.
the long-run proportion of times that the toast lands buttered side down 1.1.19
based on getting 7 times landing buttered side down in 10 drops. a. The long-run proportion of times that Zwerg chooses the correct
e. No, 0.50 is just a plausible (reasonable) explanation for the data. object
Other explanations are possible (e.g., the long-run proportion of times b. Zwerg is just guessing or Zwerg is choosing the correct object be-
the toast lands buttered side down could be 0.60). cause she understands the cue.
1.1.15 c. 37 out of 48 attempts seems fairly unlikely to happen by chance, be-
a. Statistic cause 24 out of 48 is what we would expect to happen in the long run.
b. Parameter d. 50%
c. Yes, it is possible to get 17 out of 20 first serves in if Mark was just 1.1.20
as likely to make his first serve as to miss it. a. 37 times out of 48 attempts
d. Getting 17 out of 20 first serves in if Mark was just as likely to make b. Applet input: probability of success is 0.5, sample size is 48, num-
the serve as to miss it is like flipping a coin 20 times and getting heads ber of samples is 1,000
17 times. This is fairly unlikely, so 17 out of 20 first serves in is not a 160
very plausible outcome if Mark is just as likely to make his first serve
as to miss it.
1.1.16 120
a. Observational unit is each cup, variable is whether the tea or
milk was poured first.
80
b. The long-run proportion of times the woman correctly identifies a
cup
c. 8, p̂ = 1.0 40
d. Yes, it’s possible she could get 8 out of 8 correct if she was just
randomly guessing with each cup.
0
e. Getting 8 out of 8 correct if she was randomly guessing is like flip- 10 12 14 16 18 20 22 24 26 28 30 32 34 36
ping a coin 8 times and getting heads every time—a fairly unlikely
Number of successes
result. Thus, 8 out of 8 seems unlikely.
1.1.17 c. Yes, it appears as if the chance model is wrong, as it is highly un-
likely to obtain a value as large as 37 when there is a 50% chance of
a. Toss a coin 8 times to represent the 8 cups of tea. Heads represents
picking the correct object.
a correct identification of what was poured first, tea or milk, and tails
represents an incorrect identification of what was poured first. Count the d. We have strong evidence that Zwerg can correctly follow this type
number of heads in the 8 tosses, this represents the number of correct of direction more than 50% of the time.
identifications of what was poured first out of the 8 cups. Repeat this e. The results are statistically significant because we have strong evi-
process many times (1,000). You will end up with a distribution of the dence that the chance model is incorrect.
number of correct identifications out of 8 cups when the chance of a cor- 1.1.21
rect identification is 50%. If 8 correct out of 8 cups rarely occurs, then it is
a. Zwerg is just guessing or Zwerg is picking up on the experimenter
unlikely that the woman was just guessing as to what was poured first.
cue to make a choice.
b. Using the applet shows that 8 out of 8 occurs rarely by chance (~4
b. 26 out of 48 seems like the kind of thing that could happen just by
times out of 1,000), confirming the fact that 8 out of 8 is quite unlikely
chance because 24 out of 48 is what we would expect on average in the
to occur just by chance.
long run.
c01Solutions.indd 6 10/16/20 9:15
, Solutions to Problems 7
c. 50% b. 17 out of 30 seems like the kind of thing that could happen just by
1.1.22 chance because 15 out of 30 is what we would expect on average in the
long run.
a. 26 times out of 48 attempts
c. 50%
b. Applet input: probability of success is 0.5, sample size is 48, num-
ber of samples is 1,000. This distribution is centered at 24. 1.1.26
a. 17 times out of 30 attempts
180 0.3300 b. Applet input: probability of success is 0.5, sample size is 30, num-
ber of samples is 1,000. The distribution is centered at 15.
c. We cannot conclude the chance model is wrong because a value as
120
large or larger than 17 is fairly likely.
d. We do not have strong evidence that Janine can land the majority
of her serves in-bounds when serving right-handed.
60
e. The chance model (Janine lands 50% of her serves in-bounds when
serving right-handed) is a plausible explanation for the observed data
(17 out of 30).
0
12 15 18 21 24 27 30 33 36
26 f. This does not prove that Janine lands 50% of her right-handed
Number of heads short-serves in bounds. This is just one plausible explanation for Ja-
nine’s performance. We cannot rule out a 0.50 long-run proportion
c. We cannot conclude the chance model is wrong because a value as of serves in bounds as an explanation for Janine landing 17 out of 30
large or larger than 26 is fairly likely. serves in bounds.
d. We do not have strong evidence that Zwerg can correctly follow 1.1.27
this type of direction more than 50% of the time.
a. 0.50
e. The chance model (Zwerg guessing) is a plausible explanation for
the observed data (26 out of 48), because the observed outcome was b. 20
likely to occur under the chance model. c. 1,000 (or some large number)
f. Less convincing evidence that Zwerg can correctly follow this type d. 12 out 20 is a fairly likely value because it occurred frequently in
of direction more than 50% of the time. We could have anticipated this the simulated data.
because 26 out of 48 is closer to 24 out of 48 than is 37 out of 48.
1.1.28
g. This does not prove that Zwerg is just guessing. Guessing is just
one plausible explanation for Zwerg’s performance in this experiment. a. 0.50
We cannot rule out guessing as an explanation for Zwerg getting 26 b. 100
out of 48 correct. c. 1,000 (or some large number)
1.1.23 d. 60 out of 100 is somewhat unlikely because it occurred somewhat
a. The long-run proportion of times that Janine’s short serve lands in infrequently in the simulated data.
bounds when serving left-handed
e. The sample size was different (20 serves vs. 100 serves).
b. Janine has a 50-50 chance of landing in- bounds and so 23 out of 30
1.1.29 B.
happened by chance; Janine’s chance of landing her serve in bounds
is greater than 50%. 1.1.30
c. 23 out of 30 seems somewhat unlikely to occur if she has a 50-50 a. 12 coin flips
chance of landing the serve in-bounds b. It is unlikely, because at least 11 correct (heads) happened very
d. 50% rarely on 12 coin flips.
1.1.24 c. Yes, we have very strong evidence that Milne does better than
a. 23 out of 30 attempts just guess because 11 correct out of 12 is very unlikely to happen by
b. Applet input: probability of success is 0.5, sample size is 30, num- just guessing.
ber of samples is 1,000. Centered ~15 d. Even stronger evidence, because 12 correct is even more extreme
c. Yes, it appears as if the chance model is wrong, as it is highly un- than 11 correct.
likely to obtain a value as large as 23 when there is a 50% chance of e. Compared to 11, 12 is farther away in the tail and farther away
getting the serve in-bounds. from 6 (which is ½ of 12).
d. We have strong evidence that Janine can land the majority of her 1.1.31
serves in bounds.
a. 20 times
e. The results are statistically significant because we have strong evi-
dence that the chance model is incorrect. b. Yes, because 16 heads out 20 coin flips is very rare.
1.1.25 c. Yes, because 16 heads in 20 tosses is very rare, so we have strong
evidence that Mercury’s choices are not random.
a. Janine has a 50-50 chance of landing in- bounds and so 17 out of 30
happened by chance; Janine’s chance of landing her serve in-bounds 1.1.32
is greater than 50%. a. 20 times
c01Solutions.indd 7 10/16/20 9:15
, 8 C HA PTER 1 Significance: How Strong Is the Evidence
b. No, because 11 heads in 20 coin flips is not unusual. d. Because 4 of the 100 simulated outcomes gave a result of 7 or more,
c. No, we do not have evidence that Panzee’s choices are that dif- the p-value is 0.04.
ferent from what could have happened if Panzee was just randomly e. We have strong evidence that Sarah is not simply guessing, be-
choosing containers. cause 7 out 8 rarely occurs by chance (if just guessing)
1.1.33 f. If Sarah doesn’t understand how to solve problems and is just guess-
ing at which picture to select, the probability she would get 7 or more
a. We would flip 40 coins where one side (say heads) represents
correct out of 8 is 0.04.
Donaghy’s foul calls favoring the team that received heavier betting
and the other side (say tails) represents Donaghy’s foul calls not favor- g. A single dot represents the number of times Sarah would choose the
ing the team that received heavier betting. correct picture (out of 8) if she were just guessing.
b. Because 28 out of 40 is way out in the upper tail of the chance 1.2.17
distribution, it is very unlikely that Donaghy’s foul calls would favor a. Null: The long-run proportion of times Hope will go to the correct object
the team that received heavier betting 28 out of 40 times if the chance is 0.50, Alternative: The long-run proportion of times that Hope will go to
model is true. the correct object is more than 0.50
1.1.34 b. H0: π = 0.50, Ha: π > 0.50
a. We are assuming that Buzz’s probability of choosing the cor- c. 0.23 (23 dots are 0.60 or larger)
rect button does not change and that previous trials don’t influence d. No, the approximate p-value is 0.23, which provides little to no ev-
future guesses. idence that Hope understands pointing.
b. The parameter is his actual probability of pushing the correct button. e. 0.70
f. i.
Section 1.2 1.2.18 Researcher A has stronger evidence against the null hypothe-
1.2.1 D. sis because his p-value is smaller.
1.2.2 C. 1.2.19
1.2.3 A. a. Roll a die 20 times, and keep track of how many times ‘one’ is
1.2.4 B. rolled. Repeat this many times.
1.2.5 A. b. Using a set of five black cards and one red card, shuffle the cards
and choose a card. Note the color of the card and return it to the deck.
1.2.6 D.
Shuffle and choose a card 20 times keeping track of how many times
1.2.7 C. the red card is selected. Repeat this many times.
1.2.8 B. c. Roll 30 times, then repeat.
1.2.9 B. d. Shuffle and choose a card 30 times, then repeat.
1.2.10 C. e. Roll a die 20 times, and keep track of how many times a ‘one, two,
1.2.11 B. three, or four’ is rolled. Repeat this many times.
1.2.12 A. f. Using a set of one black card and two red cards, shuffle the cards
1.2.13 and choose a card. Shuffle and choose a card 20 times keeping track
of how many times the red card is selected. Repeat this many times.
a. 0.25
1.2.20
b. 25 (because 0.25 × 100 = 25)
a. Observational units: 40 heterosexual couples who agreed on their
1.2.14
response to which person was the first to say “I love you,” Variable:
a. H0 = Null hypothesis Whether the man or woman said “I love you” first; this is a categorical
b. Ha = Alternative hypothesis variable
c. ˆ
p = sample proportion b. Null: The proportion of all couples where the male said “I love
d. π = long-run proportion (parameter) you” first is 0.50. Alternative: The proportion of all couples where
the male said “I love you” first greater than 0.50
e. n = sample size
c. π is the proportion of all couples where the male said “I love you”
1.2.15 ˆ p is the value of the observed statistic, while the p-value is
first
the probability that the observed statistic or more extreme occurs if
the null hypothesis is true; p-value is a measure of strength of the d. 28/40 = 0.7 is the sample proportion; we use the symbol
evidence.
1.2.16 e. Flip a coin 40 times and keep track of the number of heads. Repeat
the 40 coin flips, 1,000 times. Calculate the proportion of sets of 40
a. The long-run proportion of times that Sarah chooses the correct
coin flips where 28 or more heads were obtained. That proportion is
photo, π
the p-value.
b. 7/8 = 0.875.
f. Applet: π = 0.50, n = 40, number of samples = 1,000. To find the
c. Null: The long-run proportion of times Sarah chooses the correct pho- p-value, we find the proportion of times a value greater than or equal
to is 0.50. Alternative: The long-run proportion of times Sarah chooses to 28 is observed. The p-value is approximately 0.008.
the correct photo is more than 0.50.
g. The p-value is the probability of observing a value of 28 or greater,
H0: π = 0.50, Ha: π > 0.50 assuming that for 50% of couples the man said “I love you” first.
c01Solutions.indd 8 10/16/20 9:15
, Solutions to Problems 9
h. The small p-value gives us strong evidence that for more than 50% d. Null hypothesis: The long-run proportion of times that rhesus
of couples the man said “I love you” first. monkeys make the correct choice when the researcher looks to-
1.2.21 wards the correct box is 0.50 (just guessing). Alternative hypothesis: The
long-run proportion of times that rhesus monkeys make the correct
a. Obs units = university students, Variable = male or female said “I
choice is more than 0.50.
love you” first
H0: π = 0.50, Ha: π > 0.50
b. Null: The long-run proportion of university student relationships
in which the male says “I love you” first is 0.50, Alternative: The long- e. Flip a coin 40 times and record the number of heads. Repeat this
run proportion of university student relationships in which the male process 999 more times, yielding a set of 1,000 values of the number
says “I love you” first is more than 0.50, of heads received in 40 coin tosses. Compute the p-value as the pro-
portion of times 31 or larger was obtained by chance in the 1,000 sets
c. 59/96 = 0.61 is the sample proportion, we use the symbol ˆ
pto de-
of 40 coin tosses. If the p-value is small (indicating 31 or larger rarely
note this quantity.
occurs by chance), then this is convincing evidence that rhesus mon-
d. We could flip a coin 96 times and keep track of the number of keys can interpret human gestures better than by random chance.
heads. Then do many, many more sets of 96 coin flips, keeping track
f. The approximate p-value from the applet (using π = 0.50, n = 40,
of the number of heads each time.
number of samples = 1,000) is 0.001 (probability of 31 or greater). This
e. Probability of heads: 0.5, number of tosses: 96, number of repe- small p-value means that 31 out of 40 is strong evidence that the rhesus
titions: 1,000. To find the p-value, we find the proportion of times a monkeys are not guessing, which may lead us to believe that rhesus
value is greater than or equal to 0.61. The p-value is approximately monkeys can interpret gestures to indicate which box to choose.
0.016.
1.2.24 The p-value is approximately 0.25. We don’t have strong evidence
f. The p-value (0.016) is the probability of 0.61 or larger assuming the that the author’s long-run proportion of wins in Minesweeper is greater
null hypothesis is true. than 0.50. The null hypothesis (long-run proportion of wins is 0.50) is a
g. We have strong evidence that the long-run proportion of university plausible explanation for her winning 12 out of 20 games.
student relationships in which the male says “I love you” first is more 1.2.25 The p-value is approximately 0.134. We don’t have strong evi-
than 0.50. dence that the author’s long-run proportion of wins in Spider Solitaire
1.2.22 is greater than 0.50. The null hypothesis (long-run proportion of wins
a. Obs units: each of the 40 monkeys. Variable: correct choice or not is 0.50) is a plausible explanation for him winning 24 out of 40 games.
(categorical) 1.2.26
b. The long-run proportion of times that a monkey will make the cor- a. The long-run proportion of times a spun penny lands heads
rect choice, π b. The p-value = 0.16. There is little to no evidence that a spun penny
c. 30/40 = 0.75. Statistic. We use the symbol ˆ
to denote this quantity.
p lands heads less than 50% of the time.
d. Null hypothesis: The long-run proportion of times that rhesus c. Null would be the same, Alternative would be > 0.50. To calculate the
monkeys make the correct choice when observing the researcher jerk p-value, find the probability that 29 or larger (0.58 or larger) occurred.
their head is 0.50 (just guessing). Alternative hypothesis: The long- 1.2.27
run proportion of times that rhesus monkeys make the correct choice
a. Null: π = 0.50, Alternative: π > 0.50
is more than 0.50.
b. The p-value = 0.31. There is little to no evidence that a coin that starts
H0: π = 0.50, Ha: π > 0.50 out heads will land heads more than 50% of the time.
c. No, the p-value does not prove the null hypothesized value (0.50)
e. Flip a coin 40 times and record the number of heads. Repeat
is correct, just that it is a plausible value for the parameter.
this process 999 more times, yielding a set of 1,000 values of the
number of heads received in 40 coin tosses. Compute the p-value d. In the long run it will land heads 51% of the time, in any particular
as the proportion of times 30 or larger was obtained by chance in set of 100 flips it is likely the coin won’t land heads exactly 51 out of
the 1,000 sets of 40 coin tosses. If the p-value is small (indicating 30 100 times.
or larger rarely occurs by chance), then this is convincing evidence 1.2.28 A simulation analysis using a null hypothesis probability of
that rhesus monkeys can interpret human gestures better than by 0.75 yields a p-value of 0.10, meaning that the set of 20 free throws by
random chance. your friend (and making 12/20 of them) provides little to no evidence
f. The approximate p-value from the applet (using π = 0.50, n = 40, that your friend’s long-run proportion of free throws made is worse
number of samples = 1,000) is 0.001 (probability of 30 or greater). than the NBA average.
This small p-value means that 30 out of 40 is strong evidence that 1.2.29 A simulation analysis using a null hypothesis probability of
the rhesus monkeys are not guessing, which may lead us to believe 0.75 yields a p-value of 0.02, meaning that the set of 40 free throws by
that rhesus monkeys may be able to understand a head jerk to indi- your friend (and making 24/20 of them) provides strong evidence that
cate which box to choose. your friend’s long-run proportion of free throws made is worse than
1.2.23 the NBA average.
a. Obs units: each of the 40 monkeys. Variable: correct choice or not 1.2.30
(categorical) a. Null: Mercury will randomly choose between two containers ver-
b. The long-run proportion of times that a monkey will make the cor- sus Alternative: Mercury will choose the container with more banan-
rect choice, π as more often than the other container.
c. 31/40 = 0.775. Statistic. We use the symbol ˆ
to denote this b. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
quantity. c. 0.02
c01Solutions.indd 9 10/16/20 9:15
, 10 C HA PTER 1 Significance: How Strong Is the Evidence
d. Yes, because very rarely (p-value = 0.02) would Mercury pick the 1.3.11
container with more bananas at least 16 times in 20 trials by just ran- a. –3.47 (100 out of 400; 25%), –3.80 (20 out of 120; 16.7%), –4.17
domly choosing. (65 out of 300; 21.7%). Because of the random nature of obtaining the
e. Answers may vary depending on what you consider “strong evi- SDs of the null distributions, these standardized statistics can vary by
dence.” For a p-value to be at most 0.05, the sample proportion would as much as about plus or minus 0.2.
have to be at least 0.75; so, 15 or more out of 20. b. 65 out of 300 is the strongest evidence, 100 out of 400 is the least
1.2.31 strong evidence.
a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50 1.3.12
b. About 0.06 a. Friend D because they played more games
c. We have only moderate evidence because the p-value is fairly b. Friend D because this is more evidence against the null hypothesis
small, but not small enough to be considered strong evidence.
c. Friend D because this is more evidence against the null hypothesis
1.2.32
d. Friend D because a smaller standard deviation leads to a larger
a. Null, H0: π = 0.25 versus Alternative, Ha: π > 0.25 standardized statistic
b. About 0.003 1.3.13 Friend G because the value of their statistic (30 out of 40) is
c. We have very strong evidence because the p-value is very small; larger than Friend F (15 out of 40) and thus will be farther in the tail of
it is unlikely that 42 correct matches out of 113 would happen by the distribution.
just guessing.
1.3.14 Simulation yields a standard deviation of the null distribution
1.2.33 of 0.112, and a standardized statistic of approximately 0.89, which
a. Null, H0: π = 0.25 versus Alternative, Ha: π > 0.25 provides little or no evidence that the author’s long-run proportion of
wins in Minesweeper is higher than 0.50.
b. About 0.09
c. We have only moderate evidence because the p-value is somewhat 1.3.15 Simulation yields a standard deviation of the null distribu-
small; it is somewhat unlikely that 35 correct matches out of 113 tion of 0.125, and a standardized statistic of approximately (0.9375
would happen by just guessing, but not unlikely enough to give us − 0.50)/0.125 = 3.5, which provides very strong evidence that the
strong evidence. long-run proportion of Buzz pushing the correct button is higher
than 0.50.
1.2.34 When there are many outcomes, the probability of a single out-
come, even if it is close to what we would expect, can be very small. 1.3.16 Simulation yields a standard deviation of the null distribution
We need to look at more outcomes to get a better understanding of the of approximately 0.094, and a standardized statistic of approximately
likeliness or unlikeliness of an outcome. 0.76, which provides little to no evidence that the long-run proportion
of times Buzz pushes the correct button is higher than 0.50.
1.2.35
1.3.17
a. He claims he can flip a coin and make it come up heads every
single time. a. The long-run proportion of all couples that lean their heads to the
right while kissing, π.
b. Five
b. Null: π = 0.50, Alternative: π > 0.50
Section 1.3
c. 80/124 = 0.645 = p̂
1.3.1 C.
d. (0.645 − 0.5)/0.045 = 3.22 = z
1.3.2
e. The observed proportion of couples leaning their heads to the right
1.3.3 because the standard deviation is smaller while kissing is 3.22 standard deviations away from the null hypothe-
1.3.4 sized parameter value of 0.5.
D. Even though C is farther away from 0 (z = 3), because a positive f. We have strong evidence that the proportion of couples that lean
standardized statistic puts the observed statistic in the right tail, this their heads to the right while kissing is more than 0.50.
means the observed statistic was a number (much) larger than 0.25,
1.3.18
which is not evidence for the alternative hypothesis that π < 0.25.
a. (0.645 − 0.60)/0.044 = 1.02
1.3.5
b. The standardized statistic is smaller. This makes sense because the null
a. False b. True c. False d. False
hypothesis is now closer to the observed statistic (less extreme).
1.3.6 B.
1.3.19
1.3.7 C. a. Null: The long-run proportion of all couples that have the male say
1.3.8 A. “I love you” first is 0.50. Alternative: The long-run proportion is more
1.3.9 D. than 0.50.
1.3.10 b. z = (0.70 − 0.50)/0.079 = 2.53
a. 0.06 c. The observed proportion of couples where the males says “I love
you” first is 2.53 standard deviations above the null hypothesized pa-
b. No, greater, 0.05
rameter value of 0.50.
c. 1.72
d. We have strong evidence that the proportion of couples for which
d. No, less, 2 the male says ”I love you” first is more than 0.50.
c01Solutions.indd 10 10/16/20 9:15
, Solutions to Problems 11
1.3.20 g. We have little to no evidence that the long-run proportion of times
a. Null: The long-run proportion of times that rhesus monkeys that Zwerg makes the correct choice when using a marker is greater
choose the correct box is 0.50. Alternative. The long-run proportion of than 0.50.
times that rhesus monkeys choose the correct box is greater than 0.50. 1.3.24
b. z = (0.75 – 0.5)/0.079 = 3.16 a. The long-run proportion of times that 10-month olds choose the
c. The observed proportion of rhesus monkeys that chose the box the helper toy, π
experimenter gestured towards is 3.16 standard deviations away from b. Null: π = 0.50, Alternative: π > 0.50
the null hypothesized parameter value of 0.5.
c. 14/16 = 0.875 = p̂
d. We have strong evidence that the long-run proportion of times that
d. z = (0.875 − 0.50)/0.125 = 3
rhesus monkeys choose the correct box is greater than 0.50.
e. The observed proportion of times the 10-month-old babies chose the
1.3.21
helper toy is 3 standard deviations above the null hypothesized parame-
a. The long-run proportion of times the lady correctly identifies ter value of 0.50.
which was poured first, π
f. We have strong evidence that the long-run proportion of times that
b. Null: π = 0.50, Alternative: π > 0.50 10-minth-old babies choose the helper toy is greater than 0.50.
c. 8/8 = 1 = ˆ
p 1.3.25
d. 0.50, because that is the value of the parameter if the null hy-
a. Yes, the p-value will be small because the standardized statistic is
pothesis is true. The standard deviation will be positive because the
large
standard deviation must be at least 0, and is only equal to zero if
there is no variability in the values (there will be variability in the b. A p-value of approximately 0.002. The p-value is the probability
simulated statistics). observing 14/16 or larger assuming the null hypothesis is true.
e. z = (1 − 0.50)/0.177 = 2.82 c. We have strong evidence that the long-run proportion of times that
10-month-old babies choose the helper toy is greater than 0.50.
f. The observed proportion of times the lady correctly identified
which was poured first is 2.82 standard deviations away from the null d. Yes, because both the p-value and the standardized statistic are meas-
hypothesized parameter value of 0.50. uring the strength of evidence (how far out in the tail the observed value
is), and so should lead to the same conclusion.
g. We have strong evidence that the long-run proportion of times that
the lady makes the correct identification is greater than 0.50. 1.3.26
1.3.22 a. The long-run proportion of people that choose the number 3, π
a. The long-run proportion of times that Zwerg makes the correct b. Null: π = 0.25, Alternative: π > 0.25
choice when the object is pointed at, π c. 14/33 = 0.42 = p̂
b. Null: π = 0.50, Alternative: π > 0.50 d. The mean = 0.248 and SD = 0.076
c. 37/48 = 0.77 = p̂ e. (0.42 – 0.248)/0.076 = 2.26
d. 0.5, because that is the value of the parameter if the null hypothe-
f. The observed proportion of people that chose the number 3 is 2.26
sis is true. The standard deviation will be positive because the stand-
standard deviations above the null hypothesized parameter value
ard deviation must be at least 0, and is only equal to zero if there is
of 0.25.
no variability in the values (there will be variability in the simulated
statistics). g. We have strong evidence that the long-run proportion of people
that will choose the number 3 is greater than 0.25.
e. (0.77 – 0.5)/0.073 = 3.70
1.3.27
f. The observed proportion of times Zwerg made the correct choice is
3.70 standard deviations away from the null hypothesized parameter a. Yes, because the standardized statistic is far from zero.
value of 0.5. b. The p-value is approximately 0.02, and is the probability of observ-
g. We have strong evidence that the long-run proportion of times that ing 14/33 or larger assuming the null hypothesis is true.
Zwerg makes the correct choice is greater than 0.50. c. We have strong evidence that the long-run proportion of people
1.3.23 that will choose the number 3 is greater than 0.25.
a. The long-run proportion of times that Zwerg makes the correct d. Yes, because both the p-value and the standardized statistic are
choice when using a marker, π measuring the strength of evidence (how far out in the tail the ob-
b. Null: π = 0.50, Alternative: π > 0.50 served value is), and so should lead to the same conclusion.
c. 26/48 = 0.54 = p̂ 1.3.28
d. 0.5, because that is the value of the parameter if the null hypothesis a. The long-run proportion of people that choose a big number, π
is true. The standard deviation will be positive because the standard b. Null: π = 0.50, Alternative: π > 0.50
deviation must be at least 0, and is only equal to zero if there is no vari-
c. 19/33 = 0.58 = p̂
ability in the values (there will be variability in the simulated statistics).
d. The mean = 0.50 and the SD = 0.084
e. (0.54 – 0.5)/0.073 = 0.55
e. (0.58 − 0.5)/0.084 = 0.952
f. The observed proportion of times Zwerg made the correct choice is
0.55 standard deviations away from the null hypothesized parameter f. We have little to no evidence that the long-run proportion of peo-
value of 0.5. ple that will choose a ”big number” is greater than 0.50.
c01Solutions.indd 11 10/16/20 9:15
, 12 C HA PTER 1 Significance: How Strong Is the Evidence
1.3.29 1.3.34
a. No, because the standardized statistic is not far from zero. a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
b. The p-value is approximately 0.153, and is the probability of ob- b. z = (0.8667 – 0.50)/0.129 = 2.84; because the standardized statistic
serving 19/33 or larger assuming the null hypothesis is true is greater than 2, we have strong evidence that dogs are more likely to
c. We have little to no evidence that the long-run proportion of peo- approach the positive donor.
ple that will choose a big number is greater than 0.50. c. p-value = 0.003; because this is less than 0.05, we again have strong
d. Yes, because both the p-value and the standardized statistic are evidence that dogs are more likely to approach the positive donor.
measuring the strength of evidence (how far out in the tail the ob- 1.3.35
served value is), and so should lead to the same conclusion. a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
1.3.30 b. z = (0.6667 – 0.50)/0.129 = 1.29; because the standardized statistic
a. 0.50 is less than 2, we do not have strong evidence that dogs are more likely
b. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50 to approach the positive donor.
c. 0.144 c. p-value = 0.150; because this is greater than 0.05, we again do not
have strong evidence that dogs are more likely to approach the posi-
d. 2.89
tive donor.
e. Because the standardized statistic is much larger than 2, we have
convincing evidence that Milne’s answers are better than random
Section 1.4
guesses.
1.4.1 D.
f. Stronger evidence, because the standardized statistic is 3.46 which
is farther from 0 than is 2.89. 1.4.2 A.
1.3.31 1.4.3 D.
a. Null, H0: GY just guesses, π = 0.50 versus Alternative, Ha: GY is 1.4.4 B.
correct more often than just guessing, π > 0.50, where π represents 1.4.5 C.
the chance that GY guesses correctly on any trial.
1.4.6 A.
b. Sample proportion; SD of the sample proportion is about 0.034.
1.4.7
c. 6.03
a. Larger, because the sample proportion is closer to the hypothe-
d. Very strong evidence that GY is more likely than random chance to sized value (0.50) of the long-run proportion.
correctly identify whether or not the square was present because the
b. Smaller, because less likely to get extreme values of the statistic
standardized statistic is so large.
from a larger sample.
e. p-value is <0.001 that is very small; yes, leads to same conclusion
c. Larger, because now including at least as extreme values of the
as the standardized statistic.
sample proportion in both tails of the null distribution.
1.3.32
1.4.8
a. Null, H0: GY just chooses between the two wagers randomly, π =
a. Smaller, as in, closer to 0, because the sample proportion is closer
0.50 versus Alternative, Ha: GY chooses the higher wager more often
to the hypothesized value (0.5) of the long-run proportion.
than the lower wager, π > 0.50, where π represents the chance that GY
chooses the higher wager. b. Larger, as in, farther from 0; because less likely to get extreme val-
ues of the statistic from a larger sample.
b. Sample proportion; SD of the sample proportion is about 0.043.
c. Same, because the sample proportion is still the same distance
c. –0.58
away from 0 in the null distribution.
d. No evidence that GY is more likely to take the high wager than the
low wager because the standardized statistic is close to zero. 1.4.9
e. p-value is about 0.75 that is not small at all; yes, leads to the same a. Null, H0: Students randomly choose between the two cookies,
conclusion as the standardized statistic. π = 0.50 versus Alternative, Ha: Students have a genuine preference
for Chips Ahoy over Chipsters, π > 0.50, where π represents the long-
f. Because the 67/141 is 0.475 which less than 0.5, and in the oppo- run proportion of students that choose Chips Ahoy over Chipsters.
site direction than that conjectured by the alternative hypothesis.
b. 0.03
1.3.33
c. Null, H0: π = 0.50 versus Alternative, Ha: π ≠ 0.50
a. Null, H0: π = 0.50 versus Alternative, Ha: π > 0.50
d. 0.08
b. 0.003; there is about a 0.3% chance that Paul got all 8 predictions
correct by just guessing. 1.4.10
c. Based on the very small p-value, we have very strong evidence that a. 0.05
Paul is doing better than just guessing. b. 0.11 (0.30 or less and 0.70 or more)
d. Based on the small p-value, it is surprising that an animal would 1.4.11
correctly choose the winning team in 8 of 8 competitions. This is a a. Smaller
fairly small sample size, however, and it would be interesting to see
whether Paul could sustain this accuracy through a much larger num- b. Smaller
ber of competitions. c. Larger
c01Solutions.indd 12 10/16/20 9:15
