a.The true proportion of times the racket lands face up

5 Section 1.1

1.1.1 C.

1.1.2 A.

1.1.3 C.

1.1.4 A.

1.1.5 B.

1.1.6 D.

1.1.7 a. The true proportion of times the racket lands face up b. Parameter c. 0.50 d. 48 out of 100 does not constitute strong evidence that the spin- 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 face up is a typical result.e.Plausible that the spinning process is fair 1.1.8 a. 24 out of 100 does constitute strong evidence that the spinning pro- 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 an atypical result.b. Statistically significant evidence that spinning is not fair 1.1.9 a.LeBron’s long-run proportion of making a field goal b. Statistic c. 0.50 d. Flip a coin 1,095 times and record the number of heads. Repeat this 1,000 times keeping track of the number of heads in each set of 1,095.e.Approximately ½ of 1,095 (547 or 548) will be one of the most likely values.1.1.10a. Parameter b. Statistic c. 1,383 d.  e.  The number of heads (or pr oportion of heads) out of 1,383 flips ½ of 1,383 or about 691 or 692 1.1.11a. Parameter b. Statistic

c. The correct matches are shown below:

Column A Column B Coin flip Kevin shoots a field goal Heads Kevin makes his field goal Tails Kevin misses his field goal Chance of headsLong-run proportion of field goals Kevin makes One repetition One set of 100 field goal shots by Kevin 1.1.12a.Parameter b. Statistic

c. The correct matches are shown below:

Column A Column B Coin flip Author plays a game of Minesweeper Heads Author wins a game Tails Author loses a game Chance of headsLong-run proportion of games that the author wins One repetitionOne set of 20 Minesweeper games played by author

1.1.13

a. 100 dots b. Each dot represents the number of times out of 20 attempts the author wins a game of Minesweeper when the probability that the author wins is 0.50.c. 10, because that is what will happen on average if the author plays 20 games and wins 50% of her games.d. No, we are not convinced that the author’s long-run proportion of winning at Minesweeper is above 0.50 because 12 is a fairly typical 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 value for the long-run proportion of games that the author wins Mine- sweeper based on the author getting 12 wins in 20 games.Solutions Manual for Introduction to Statistical Investigations, 2e Tintle, Chance, Cobb, Rossman, Roy, Swanson, Jill

CHAPTER 1

Significance: How

Strong Is the Evidence?

c01Solutions.indd 510/16/20 9:15 PM 1 / 4

6 CHAPTER 1 Significance: How Strong Is the ­Evidence

c. Yes, the simulation analysis gives strong evidence that the woman is not simply guessing. If she were guessing she’d rarely get 8 out of

• correct.

d. Statistically significance evidence she is not guessing

1.1.18

a. The conclusion you’ve drawn is incorrect, because 5 out of 8 is a likely result if someone is just guessing. In particular, when you do a simulation with probability of success = 0.5, sample size (n) = 8, getting 5 heads happens quite frequently.b. No, this does not prove that you cannot tell the difference. It’s plausi- ble (believable) you are not guessing, but we haven’t proven it.

c. Applet inputs are: probability of success (π) = 0.5, sample size

(n) = 16, number of samples = 1000. Applet output suggests that 14 out of 16 is a fairly unlikely result (~2 out of 1000 times). Thus, this result also provides strong evidence that the person actually has abil- ity better than random guessing. The applet value for π stays the same because 0.50 still represents guessing, and n = 16 now because there are 16 cups of tea.

1.1.19

a. The long-run proportion of times that Zwerg chooses the correct object b. Zwerg is just guessing or Zwerg is choosing the correct object be- cause she understands the cue.c. 37 out of 48 attempts seems fairly unlikely to happen by chance, be- cause 24 out of 48 is what we would expect to happen in the long run.d. 50%

1.1.20

a. 37 times out of 48 attempts

b. Applet input: probability of success is 0.5, sample size is 48, num-

ber of samples is 1,000

1210 14 16 18 20 22 24 26 28 30 32 34 36

40 80 120 160 Number of successes 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 picking the correct object.d. We have strong evidence that Zwerg can correctly follow this type of direction more than 50% of the time.e. The results are statistically significant because we have strong evi- dence that the chance model is incorrect.

1.1.21

a. Zwerg is just guessing or Zwerg is picking up on the experimenter cue to make a choice.b. 26 out of 48 seems like the kind of thing that could happen just by chance because 24 out of 48 is what we would expect on average in the long run.e. No, 0.50 is just a plausible (reasonable) explanation for the data.Other explanations are possible (e.g., the author’s long-run propor- tion of wins could be 0.55).f. Yes, it means that there were special circumstances when the author played these 20 games and so these 20 games may not be a good rep- resentation of the author’s long-run proportion of wins in Minesweeper.

1.1.14

a. 100 b. Each dot represents the number of times out of 10 attempts the toast lands buttered side down when the probability that the toast lands buttered side down is 0.50.c. 5, because that is what will happen on average if the toast is dropped 10 times and 50% of the drops it lands buttered side down.d. No, we are not convinced that the long-run proportion of times the toast lands buttered side down is above 0.50 because 7 is a fairly typical outcome for the number of times landing buttered side down out of 10 drops of toast when the long-run proportion of times it lands but- tered side down is 0.50. Stated another way, 0.50 is a plausible value for the long-run proportion of times that the toast lands buttered side down based on getting 7 times landing buttered side down in 10 drops.e. No, 0.50 is just a plausible (reasonable) explanation for the data.Other explanations are possible (e.g., the long-run proportion of times the toast lands buttered side down could be 0.60).

1.1.15

a. Statistic b. Parameter c. Yes, it is possible to get 17 out of 20 first serves in if Mark was just as likely to make his first serve as to miss it.d. Getting 17 out of 20 first serves in if Mark was just as likely to make the serve as to miss it is like flipping a coin 20 times and getting heads 17 times. This is fairly unlikely, so 17 out of 20 first serves in is not a very plausible outcome if Mark is just as likely to make his first serve as to miss it.

1.1.16

a. Observational unit is each cup, variable is whether the tea or milk was poured first.b. The long-run proportion of times the woman correctly identifies a cup c. 8, ˆp = 1.0 d. Yes, it’s possible she could get 8 out of 8 correct if she was just randomly guessing with each cup.e.Getting 8 out of 8 correct if she was randomly guessing is like flip- ping a coin 8 times and g etting heads every time—a fairly unlikely result. Thus, 8 out of 8 seems unlikely.1.1.17a. Toss a coin 8 times to represent the 8 cups of tea. Heads represents a correct identification of what was poured first, tea or milk, and tails represents an incorrect identification of what was poured first. Count the number of heads in the 8 tosses, this represents the number of correct identifications of what was poured first out of the 8 cups. Repeat this process many times (1,000). You will end up with a distribution of the number of correct identifications out of 8 cups when the chance of a cor- rect identification is 50%. If 8 correct out of 8 cups rarely occurs, then it is unlikely that the woman was just guessing as to what was poured first.b.Using the applet shows that 8 out of 8 occurs rarely by chance (~4 times out of 1,000), confirming the fact that 8 out of 8 is quite unlikely to occur just by chance.

c01Solutions.indd 610/16/20 9:15 PM 2 / 4

Solutions to Problems 7 b. 17 out of 30 seems like the kind of thing that could happen just by chance because 15 out of 30 is what we would expect on average in the long run.c. 50%

1.1.26

a. 17 times out of 30 attempts

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 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.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).f. This does not prove that Janine lands 50% of her right-handed 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 of serves in bounds as an explanation for Janine landing 17 out of 30 serves in bounds.

1.1.27

a. 0.50 b. 20 c. 1,000 (or some large number) d. 12 out 20 is a fairly likely value because it occurred frequently in the simulated data.

1.1.28

a. 0.50 b. 100 c. 1,000 (or some large number) d. 60 out of 100 is somewhat unlikely because it occurred somewhat infrequently in the simulated data.e. The sample size was different (20 serves vs. 100 serves).

1.1.29 B.

1.1.30 

a. 12 coin flips b. It is unlikely, because at least 11 correct (heads) happened very rarely on 12 coin flips.c. Yes, we have very strong evidence that Milne does better than just guess because 11 correct out of 12 is very unlikely to happen by just guessing.

• Even stronger evidence, because 12 correct is even more extreme

than 11 correct.e. Compared to 11, 12 is farther away in the tail and farther away from 6 (which is ½ of 12).

1.1.31 

a. 20 times b. Yes, because 16 heads out 20 coin flips is very rare.c. Yes, because 16 heads in 20 tosses is very rare, so we have strong evidence that Mercury’s choices are not random.

1.1.32 

a. 20 times c. 50%

1.1.22

a. 26 times out of 48 attempts

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.

12 15 18 21 24 27 30 33 36

60 120 180 Number of heads 26

0.3300

c. We cannot conclude the chance model is wrong because a value as large or larger than 26 is fairly likely.d. We do not have strong evidence that Zwerg can correctly follow this type of direction more than 50% of the time.e. The chance model (Zwerg guessing) is a plausible explanation for the observed data (26 out of 48), because the observed outcome was likely to occur under the chance model.f. Less convincing evidence that Zwerg can correctly follow this type of direction more than 50% of the time. We could have anticipated this because 26 out of 48 is closer to 24 out of 48 than is 37 out of 48.g. This does not prove that Zwerg is just guessing. Guessing is just one plausible explanation for Zwerg’s performance in this experiment.We cannot rule out guessing as an explanation for Zwerg getting 26 out of 48 correct.

1.1.23

a. The long-run proportion of times that Janine’s short serve lands in bounds when serving left-handed b. Janine has a 50-50 chance of landing in- bounds and so 23 out of 30 happened by chance; Janine’s chance of landing her serve in bounds is greater than 50%.c. 23 out of 30 seems somewhat unlikely to occur if she has a 50-50 chance of landing the serve in-bounds

• 50%

1.1.24

a. 23 out of 30 attempts

b. Applet input: probability of success is 0.5, sample size is 30, num-

ber of samples is 1,000. Centered ~15 c. Yes, it appears as if the chance model is wrong, as it is highly un- likely to obtain a value as large as 23 when there is a 50% chance of getting the serve in-bounds.d. We have strong evidence that Janine can land the majority of her serves in bounds.e. The results are statistically significant because we have strong evi- dence that the chance model is incorrect.

1.1.25

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 is greater than 50%.

c01Solutions.indd 710/16/20 9:15 PM 3 / 4

8 CHAPTER 1 Significance: How Strong Is the ­Evidence

d. Because 4 of the 100 simulated outcomes gave a result of 7 or more, the p-value is 0.04.e. We have strong evidence that Sarah is not simply guessing, be- cause 7 out 8 rarely occurs by chance (if just guessing) 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 correct out of 8 is 0.04.g. A single dot represents the number of times Sarah would choose the correct picture (out of 8) if she were just guessing.

1.2.17

a. Null: The long-run proportion of times Hope will go to the correct object

is 0.50, Alternative: The long-run proportion of times that Hope will go to

the correct object is more than 0.50 b. H

0: π = 0.50, H

a: π > 0.50

c. 0.23 (23 dots are 0.60 or larger) d. No, the approximate p-value is 0.23, which provides little to no ev- idence that Hope understands pointing.e. 0.70 f. i.

1.2.18 

Researcher A has stronger evidence against the null hypothe- sis because his p-value is smaller.1.2.19a.Roll a die 20 times, and keep track of how many times ‘one’ is rolled. Repeat this many times.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.Shuffle and choose a card 20 times keeping track of how many times the red card is selected. Repeat this many times.c. Roll 30 times, then repeat.d. Shuffle and choose a card 30 times, then repeat.e. Roll a die 20 times, and keep track of how many times a ‘one, two, three, or four’ is rolled. Repeat this many times.f. Using a set of one black card and two red cards, shuffle the cards 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.

1.2.20

a. Observational units: 40 heterosexual couples who agreed on their

response to which person was the first to say “I love you,” Variable:

Whether the man or woman said “I love you” first; this is a categorical variable

b. Null: The proportion of all couples where the male said “I love

you” first is 0.50. Alternative: The proportion of all couples where

the male said “I love you” first greater than 0.50 c. π is the proportion of all couples where the male said “I love you” first d. 28/40 = 0.7 is the sample proportion; we use the symbol t o d e- note this quantity.

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 coin flips where 28 or more heads were obtained. That proportion is the p-value.

f. Applet:

π = 0.50, n = 40, number of samples = 1,000. To find the p-value, we find the proportion of times a value greater than or equal to 28 is observed. The p-value is approximately 0.008.g. The p-value is the probability of observing a value of 28 or greater, assuming that for 50% of couples the man said “I love you” first.b. No, because 11 heads in 20 coin flips is not unusual.c. No, we do not have evidence that Panzee’s choices are that dif- ferent from what could have happened if Panzee was just randomly choosing containers.

1.1.33 

a. We would flip 40 coins where one side (say heads) represents Donaghy’s foul calls favoring the team that received heavier betting and the other side (say tails) represents Donaghy’s foul calls not favor- ing the team that received heavier betting.b. Because 28 out of 40 is way out in the upper tail of the chance distribution, it is very unlikely that Donaghy’s foul calls would favor the team that received heavier betting 28 out of 40 times if the chance model is true.

1.1.34 

a. We are assuming that Buzz’s probability of choosing the cor- rect button does not change and that previous trials don’t influence future guesses.b. The parameter is his actual probability of pushing the correct button.Section 1.2

1.2.1 D.

1.2.2 C.

1.2.3  A.

1.2.4 B.

1.2.5 A.

1.2.6 D.

1.2.7 C.

1.2.8 B.

1.2.9 B.

1.2.10 C.

1.2.11 B.

1.2.12 A.

1.2.13a. 0.25 b. 25 (because 0.25 × 100 = 25) 1.2.14a. H

• =

Null hypothesis b. H a = Alternative hypothesis c.  ˆp = sample proportion d. π = long-run proportion (parameter) e. n = sample size 1.2.15  ˆp 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 evidence.

1.2.16

a. The long-run proportion of times that Sarah chooses the correct photo, π b. 7/8 = 0.875.

c. Null: The long-run proportion of times Sarah chooses the correct pho-

to is 0.50. Alternative: The long-run proportion of times Sarah chooses

the correct photo is more than 0.50.H

0: π = 0.50, H

a: π > 0.50

c01Solutions.indd 810/16/20 9:15 PM

• / 4