Introduction to Statistical Investigations, 2nd Edition Tintle, Chance, Cobb, Rossman,
Roy, Swanson, Jill
Introduction to Statistical Investigations, 2nd Edition Tintle, Chance, Cobb, Rossman,
Roy, Swanson, Jill
,
, 6 CHAPTER 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, pˆ = 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.
Introduction to Statistical Investigations, 2nd Edition Tintle, Chance, Cobb, Rossman, Roy, Swanson, Jill
, 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
a. 0.50
one plausible explanation for Zwerg’s performance in this experiment.
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
Roy, Swanson, Jill
Introduction to Statistical Investigations, 2nd Edition Tintle, Chance, Cobb, Rossman,
Roy, Swanson, Jill
,
, 6 CHAPTER 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, pˆ = 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.
Introduction to Statistical Investigations, 2nd Edition Tintle, Chance, Cobb, Rossman, Roy, Swanson, Jill
, 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
a. 0.50
one plausible explanation for Zwerg’s performance in this experiment.
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
