(Test Bank) Introduction to Statistical Investigations

Tintle,Chance,Cobb, 2

Math & Statistics

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2nd Edition

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Test Bank for Introduction to Statistical Investigations, The downloadable file consists of 154 pages, and the total size of the file is 3.54 MB. Purchase now only for highly discounted price.

Author(s): Tintle, Chance, Cobb

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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 8 correct. d. Statistically significance evidence she is not guessing 1.1.18a. 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. 1.1.19a. The long-run proportion of times that Zwerg chooses the correct objectb. 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% 1.1.20a. 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 12 10 14 16 18 20 22 24 26 28 30 32 34 3604080120160 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.1.1.21a. 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.1.1.14a. 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). 1.1.15a. 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. 1.1.16a. 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 cupc. 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.  Get ting 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 6 10/16/20 9:15 PMSolutions 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. d. 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 timesc. 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 36060120180 Number of heads 260.3300 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 d. 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 7 10/16/20 9:15 PM8 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  R esearcher A has stronger evidence against the null hypothe- sis because his p-value is smaller.1.2.19a.  R oll 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. 1.2.20a. 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” firstd. 28/40 = 0.7 is the sample proportion; we use the symbol ˆ pt 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 0 = Null hypothesis b. Ha = Alternative hypothesis c.  ˆ p = sample proportion d. π = long-run proportion (parameter) e. n = sample size 1.2.15  ˆp is the value of the observed statistic, while the p-value is 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. 1.2.16a. 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.   H0: π = 0.50, H a: π > 0.50 c01Solutions.indd 8 10/16/20 9:15 PMSolutions to Problems 9 d. Null hypothesis: The long-run proportion of times that rhesus monkeys make the correct choice when the researcher looks to- wards the correct box is 0.50 (just guessing). Alternative hypothesis: The long-run proportion of times that rhesus monkeys make the correct choice is more than 0.50. H0: π = 0.50, H a: π > 0.50 e. Flip a coin 40 times and record the number of heads. Repeat 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 as the pro - portion of times 31 or larger was obtained by chance in the 1,000 sets of 40 coin tosses. If the p-value is small (indicating 31 or larger rarely occurs by chance), then this is convincing evidence that rhesus mon-keys can interpret human gestures better than by random chance. f. The approximate p-value from the applet (using π = 0.50, n = 40, number of samples = 1,000) is 0.001 (probability of 31 or greater). This small p-value means that 31 out of 40 is strong evidence that the rhesus monkeys are not guessing, which may lead us to believe that rhesus monkeys can interpret gestures to indicate which box to choose. 1.2.24  The p-value is approximately 0.25. We don’t have strong evidence that the author’ s long-run proportion of wins in Minesweeper is greater than 0.50. The null hypothesis (long-run proportion of wins is 0.50) is a plausible explanation for her winning 12 out of 20 games. 1.2.25  The p-value is approximately 0.134. We don’t have strong evi- dence that the author ’s long-run proportion of wins in Spider Solitaire is greater than 0.50. The null hypothesis (long-run proportion of wins is 0.50) is a plausible explanation for him winning 24 out of 40 games. 1.2.26a. The long-run proportion of times a spun penny lands heads b. The p-value = 0.16. There is little to no evidence that a spun penny lands heads less than 50% of the time.c. Null would be the same, Alternative would be > 0.50. To calculate the p-value, find the probability that 29 or larger (0.58 or larger) occurred.1.2.27a. Null: π = 0.50, Alternative: π > 0.50 b. The p-value = 0.31. There is little to no evidence that a coin that starts 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) is correct, just that it is a plausible value for the parameter.d. In the long run it will land heads 51% of the time, in any particular set of 100 flips it is likely the coin won’t land heads exactly 51 out of 100 times. 1.2.28  A simulation analysis using a null hypothesis probability of 0.75 yields a p-value of 0.10, meaning that the set of 20 free throws by your friend (and making 12/20 of them) provides little to no evidence that your friend’s long-run proportion of free throws made is worse than the NBA average. 1.2.29  A simulation analysis using a null hypothesis probability of 0.75 yields a p -value of 0.02, meaning that the set of 40 free throws by your friend (and making 24/20 of them) provides strong evidence that your friend’s long-run proportion of free throws made is worse than the NBA average. 1.2.30 a. Null: Mercury will randomly choose between two containers ver- sus Alternative: Mercury will choose the container with more banan- as more often than the other container. b. Null, H 0: π = 0.50 versus Alternative, H a: π > 0.50 c. 0.02 h. The small p-value gives us strong evidence that for more than 50% of couples the man said “I love you” first.1.2.21a. Obs units = university students, Variable = male or female said “I love you” firstb. Null: The long-run proportion of university student relationships in which the male says “I love you” first is 0.50, Alternative: The long- run proportion of university student relationships in which the male says “I love you” first is more than 0.50, c. 59/96 = 0.61 is the sample proportion, we use the symbol ˆ p t o d e - note this quantity.d. We could flip a coin 96 times and keep track of the number of heads. Then do many, many more sets of 96 coin flips, keeping track of the number of heads each time. e. Probability of heads: 0.5, number of tosses: 96, number of repe- titions: 1,000. To find the p-value, we find the proportion of times a value is greater than or equal to 0.61. The p-value is approximately 0.016. f. The p-value (0.016) is the probability of 0.61 or larger assuming the null hypothesis is true.g. We have strong evidence that the long-run proportion of university student relationships in which the male says “I love you” first is more than 0.50. 1.2.22a. Obs units: each of the 40 monkeys. Variable: correct choice or not (categorical)b. The long-run proportion of times that a monkey will make the cor- rect choice, πc. 30/40 = 0.75. Statistic. We use the symbol ˆ p to denote this quantity. d. Null hypothesis: The long-run proportion of times that rhesus monkeys make the correct choice when observing the researcher jerk their head is 0.50 (just guessing). Alternative hypothesis: The long-run proportion of times that rhesus monkeys make the correct choice is more than 0.50. H 0: π = 0.50, H a: π > 0.50 e. Flip a coin 40 times and record the number of heads. Repeat 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 as the proportion of times 30 or larger was obtained by chance in the 1,000 sets of 40 coin tosses. If the p-value is small (indicating 30 or larger rarely occurs by chance), then this is convincing evidence that rhesus monkeys can interpret human gestures better than by random chance. f. The approximate p-value from the applet (using π = 0.50, n = 40, number of samples = 1,000) is 0.001 (probability of 30 or greater). This small p-value means that 30 out of 40 is strong evidence that the rhesus monkeys are not guessing, which may lead us to believe that rhesus monkeys may be able to understand a head jerk to indi-cate which box to choose. 1.2.23a. Obs units: each of the 40 monkeys. Variable: correct choice or not (categorical)b. The long-run proportion of times that a monkey will make the cor- rect choice, πc. 31/40 = 0.775. Statistic. We use the symbol p ˆto denote this quantity. c01Solutions.indd 9 10/16/20 9:15 PM10 CHAPTER 1 Significance: How Strong Is the Evidence 1.3.11 a. –3.47 (100 out of 400; 25%), –3.80 (20 out of 120; 16.7%), –4.17 (65 out of 300; 21.7%). Because of the random nature of obtaining the SDs of the null distributions, these standardized statistics can vary by as much as about plus or minus 0.2. b. 65 out of 300 is the strongest evidence, 100 out of 400 is the least strong evidence.1.3.12a. Friend D because they played more games b.  F riend D because this is more evidence against the null hypothesis c. Friend D because this is more evidence against the null hypothesis d.  F riend D because a smaller standard deviation leads to a larger standardized statistic1.3.13  F riend G because the value of their statistic (30 out of 40) is larger than Friend F (15 out of 40) and thus will be farther in the tail of the distribution.1.3.14  Simulation yields a standard deviation of the null distribution of 0.112, and a standar dized statistic of approximately 0.89, which provides little or no evidence that the author’s long-run proportion of wins in Minesweeper is higher than 0.50. 1.3.15  Simulation yields a standard deviation of the null distribu- tion of 0.125, and a standardized statistic of approximately (0.9375 − 0.50)/0.125 = 3.5, which provides very strong evidence that the long-run proportion of Buzz pushing the correct button is higher than 0.50. 1.3.16   Simulation yields a standard deviation of the null distribution of approximately 0.094, and a standardized statistic of approximately 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. 1.3.17a. The long-run proportion of all couples that lean their heads to the right while kissing, π.b. Null: π = 0.50, Alternative: π > 0.50 c. 80/124 = 0.645 = ˆp d. (0.645 − 0.5)/0.045 = 3.22 = z e.  The observed pr oportion of couples leaning their heads to the right while kissing is 3.22 standard deviations away from the null hypothe- sized parameter value of 0.5. f. We have strong evidence that the proportion of couples that lean their heads to the right while kissing is more than 0.50. 1.3.18a.  (0.645 − 0.60)/0.044 = 1.02 b.  The standardized statistic is smaller. This makes sense because the null hypothesis is now closer to the observed statistic (less extreme).1.3.19a. Null: The long-run proportion of all couples that have the male say “I love you” first is 0.50. Alternative: The long-run proportion is more than 0.50. b. z = (0.70 − 0.50)/0.079 = 2.53 c. The observed proportion of couples where the males says “I love you” first is 2.53 standard deviations above the null hypothesized pa- rameter value of 0.50. d. We have strong evidence that the proportion of couples for which the male says ”I love you” first is more than 0.50.d. Yes, because very rarely (p-value = 0.02) would Mercury pick the container with more bananas at least 16 times in 20 trials by just ran- domly choosing. e. Answers may vary depending on what you consider “strong evi- dence.” For a p-value to be at most 0.05, the sample proportion would have to be at least 0.75; so, 15 or more out of 20. 1.2.31 a. Null, H0: π = 0.50 versus Alternative, H a: π > 0.50 b. About 0.06 c. We have only moderate evidence because the p-value is fairly small, but not small enough to be considered strong evidence.1.2.32 a. Null, H 0: π = 0.25 versus Alternative, H a: π > 0.25 b. About 0.003 c. We have very strong evidence because the p-value is very small; it is unlikely that 42 correct matches out of 113 would happen by just guessing. 1.2.33 a. Null, H 0: π = 0.25 versus Alternative, H a: π > 0.25 b. About 0.09 c. We have only moderate evidence because the p-value is somewhat small; it is somewhat unlikely that 35 correct matches out of 113 would happen by just guessing, but not unlikely enough to give us strong evidence. 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. We need to look at more outcomes to get a better understanding of the likeliness or unlikeliness of an outcome. 1.2.35a. He claims he can flip a coin and make it come up heads every single time. b. Five Section 1.3 1.3.1  C. 1.3.2  (0.45 − 0.30)/0.091 = 1.65 1.3.3  A, because the standard deviation is smaller 1.3.4 D. Even though C is farther away from 0 (z = 3), because a positive standardized statistic puts the observed statistic in the right tail, this means the observed statistic was a number (much) larger than 0.25, which is not evidence for the alternative hypothesis that π < 0.25. 1.3.5 a.  F alse b. T rue c. False d. F alse 1.3.6 B. 1.3.7  C. 1.3.8  A. 1.3.9  D. 1.3.10 a. 0.06 b. No, greater, 0.05 c.  1.72 d. N o, less, 2 c01Solutions.indd 10 10/16/20 9:15 PMSolutions to Problems 11 g. We have little to no evidence that the long-run proportion of times that Zwerg makes the correct choice when using a marker is greater than 0.50. 1.3.24a. The long-run proportion of times that 10-month olds choose the helper toy, πb.  Null: π = 0.50, Alt ernative: π > 0.50 c. 14/16 = 0.875 = ˆp d. z = (0.875 − 0.50)/0.125 = 3 e.  The observed proportion of times the 10-month-old babies chose the helper toy is 3 standard deviations above the null hypothesized parame - ter value of 0.50. f. We have strong evidence that the long-run proportion of times that 10-minth-old babies choose the helper t oy is greater than 0.50. 1.3.25a. Y es, the p-value will be small because the standardized statistic is largeb. A p-value of approximately 0.002. The p-value is the probability observing 14/16 or larger assuming the null hypothesis is true.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.d. Yes, because both the p-value and the standardized statistic are meas - uring the strength of evidence (how far out in the tail the observed value is), and so should lead to the same conclusion. 1.3.26a. The long-run proportion of people that choose the number 3, π b. Null: π = 0.25, Alternative: π > 0.25 c. 14/33 = 0.42 = p ˆ d. The mean = 0.248 and SD = 0.076 e. (0.42 – 0.248)/0.076 = 2.26 f. The observed proportion of people that chose the number 3 is 2.26 standard deviations above the null hypothesized parameter value of 0.25. g. We have strong evidence that the long-run proportion of people that will choose the number 3 is greater than 0.25.1.3.27a. Yes, because the standardized statistic is far from zero. b. The p -value is approximately 0.02, and is the probability of observ- ing 14/33 or larger assuming the null hypothesis is true.c. We have strong evidence that the long-run proportion of people that will choose the number 3 is greater than 0.25.d. Yes, because both the p-value and the standardized statistic are measuring the strength of evidence (how far out in the tail the ob- served value is), and so should lead to the same conclusion. 1.3.28a. The long-run proportion of people that choose a big number, π b.  Null: π = 0.50, Alt ernative: π > 0.50 c. 19/33 = 0.58 = ˆp d.  The mean = 0.50 and the SD = 0.084 e. (0.58 − 0.5)/0.084 = 0.952 f. W e hav e little to no evidence that the long-run proportion of peo- ple that will choose a ”big number” is greater than 0.50.1.3.20 a. Null: The long-run proportion of times that rhesus monkeys choose the correct box is 0.50. Alternative. The long-run proportion of times that rhesus monkeys choose the correct box is greater than 0.50. b. z = (0.75 – 0.5)/0.079 = 3.16 c. The observed proportion of rhesus monkeys that chose the box the experimenter gestured towards is 3.16 standard deviations away from the null hypothesized parameter value of 0.5. d.  W e have strong evidence that the long-run proportion of times that rhesus monkeys choose the correct box is greater than 0.50.1.3.21a. The long-run proportion of times the lady correctly identifies which was poured first, πb. Null: π = 0.50, Alternative: π > 0.50 c.  8/8 = 1 = p ˆ d.  0.50, because that is the value of the parameter if the null hy- pothesis is true. The standard deviation will be positive because the 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 simulated statistics). e. z = (1 − 0.50)/0.177 = 2.82 f. The observed proportion of times the lady correctly identified which was poured first is 2.82 standard deviations away from the null hypothesized parameter value of 0.50. g. We have strong evidence that the long-run proportion of times that the lady makes the correct identification is greater than 0.50.1.3.22a. The long-run proportion of times that Zwerg makes the correct choice when the object is pointed at, πb. Null: π = 0.50, Alternative: π > 0.50 c. 37/48 = 0.77 = p ˆ d.  0.5, because that is the value of the parameter if the null hypothe- sis is true. The standar d deviation will be positive because the stand- ard 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 simulated statistics). e.  (0.77 – 0.5)/0.073 = 3.70 f. The observed proportion of times Zwerg made the correct choice is 3.70 standard deviations away from the null hypothesized parameter value of 0.5. g. We have strong evidence that the long-run proportion of times that Zwerg makes the correct choice is greater than 0.50.1.3.23a. The long-run proportion of times that Zwerg makes the correct choice when using a marker, πb.  Null: π = 0.50, Alt ernative: π > 0.50 c. 26/48 = 0.54 = pˆ d. 0.5, because that is the value of the parameter if the null hypothesis is true. The standard deviation will be positive because the standard deviation must be at least 0, and is only equal to zero if there is no vari- ability in the values (there will be variability in the simulated statistics). e. (0.54 – 0.5)/0.073 = 0.55 f. The observed proportion of times Zwerg made the correct choice is 0.55 standard deviations away from the null hypothesized parameter value of 0.5. c01Solutions.indd 11 10/16/20 9:15 PM12 CHAPTER 1 Significance: How Strong Is the Evidence 1.3.34  a. Null, H 0: π = 0.50 versus Alternative, H a: π > 0.50 b. z = (0.8667 – 0.50)/0.129 = 2.84; because the standardized statistic is greater than 2, we have strong evidence that dogs are more likely to approach the positive donor. c. p-value = 0.003; because this is less than 0.05, we again have strong evidence that dogs are more likely to approach the positive donor.1.3.35 a.  Null, H0: π = 0.50 versus Alternative, H a: π > 0.50 b. z = (0.6667 – 0.50)/0.129 = 1.29; because the standardized statistic is less than 2, we do not have strong evidence that dogs are more likely to approach the positive donor. 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- tive donor. Section 1.4 1.4.1  D. 1.4.2  A. 1.4.3  D. 1.4.4  B. 1.4.5 C. 1.4.6  A. 1.4.7 a. Larger, because the sample proportion is closer to the hypothe- sized value (0.50) of the long-run proportion.b. Smaller, because less likely to get extreme values of the statistic from a larger sample.c. Larger, because now including at least as extreme values of the sample proportion in both tails of the null distribution.1.4.8 a. Smaller, as in, closer to 0, because the sample proportion is closer to the hypothesized value (0.5) of the long-run proportion.b.  Larg er, as in, farther from 0; because less likely to get extreme val- ues of the statistic from a larger sample.c. Same, because the sample proportion is still the same distance away from 0 in the null distribution.1.4.9 a. Null, H 0: Students randomly choose between the two cookies, π = 0.50 versus Alternative, Ha: Students have a genuine preference for Chips Ahoy over Chipsters, π > 0.50, where π represents the long- run proportion of students that choose Chips Ahoy over Chipsters. b. 0.03 c. Null, H 0: π = 0.50 versus Alternative, H a: π ≠ 0.50 d.  0.08 1.4.10a.  0.05 b. 0.11 (0.30 or less and 0.70 or mor e) 1.4.11a. Smaller b.  Smaller c. Larg er1.3.29 a. No, because the standardized statistic is not far from zero . b. The p -value is approximately 0.153, and is the probability of ob - serving 19/33 or larger assuming the null hypothesis is truec. We have little to no evidence that the long-run proportion of peo- ple that will choose a big number is greater than 0.50.d. Yes, because both the p-value and the standardized statistic are measuring the strength of evidence (how far out in the tail the ob- served value is), and so should lead to the same conclusion. 1.3.30 a.  0.50 b. Null, H0: π = 0.50 versus Alternative, H a: π > 0.50 c. 0.144 d.  2.89 e. Because the standardized statistic is much larger than 2, we have convincing evidence that Milne’s answers are better than random guesses. f. Stronger evidence, because the standardized statistic is 3.46 which is farther from 0 than is 2.89.1.3.31 a. Null, H 0: GY just guesses, π = 0.50 versus Alternative, H a: GY is correct more often than just guessing, π > 0.50, where π represents the chance that GY guesses correctly on any trial. b. Sample proportion; SD of the sample proportion is about 0.034. c. 6.03 d. Very strong evidence that GY is more likely than random chance to correctly identify whether or not the square was present because the standardized statistic is so large. e. p-value is <0.001 that is very small; y es, leads to same conclusion as the standardized statistic.1.3.32 a. Null, H 0: GY just chooses between the two wagers randomly, π = 0.50 versus Alternative, Ha: GY chooses the higher wager more often than the lower wager, π > 0.50, where π represents the chance that GY chooses the higher wager. b.  Sample proportion; SD of the sample proportion is about 0.043. c. –0.58 d. N o evidence that GY is more likely to take the high wager than the low wager because the standardized statistic is close to zero.e. p-value is about 0.75 that is not small at all; yes, leads to the same conclusion as the standardized statistic. f. Because the 67/141 is 0.475 which less than 0.5, and in the oppo- site direction than that conjectured by the alternative hypothesis.1.3.33 a. Null, H 0: π = 0.50 versus Alternative, H a: π > 0.50 b. 0.003; there is about a 0.3% chance that Paul got all 8 predictions correct by just guessing. c. Based on the very small p-value, we have very strong evidence that Paul is doing better than just guessing. d. Based on the small p-value, it is surprising that an animal would correctly choose the winning team in 8 of 8 competitions. This is a fairly small sample size, however, and it would be interesting to see whether Paul could sustain this accuracy through a much larger num-ber of competitions. c01Solutions.indd 12 10/16/20 9:15 PMSolutions to Problems 13 c. The p-value (The probability of obtaining 0.60 or larger when the true chance Krieger chooses the correct object is 0.05) is approximate- ly 0.38. d. We do not have strong evidence that Krieger will choose the cor- rect object more than 50% of the time. It is plausible that Krieger will choose the correct object 50% of the time. e. Stronger 1.4.26a. Values of the long-run proportion less than 0.50 b. Increase, a two-sided p-value will be approximately twice as big as the corresponding one-sided p-value.c. The two-sided p-value will approximately double to 0.75. d. Stronger 1.4.27a. Decrease, larger sample size. b. The p-value decreased to 0.244, yes it did behave as predicted. c.  Str onger 1.4.28a. The long-run proportion of times Krieger makes the correct choice when the experimenter leans towards the object, π.b. 50% c. Null: The long run proportion of times Krieger makes the correct choice when the experimenter leans towards the object is 0.50; Alterna- tive: The long-run proportion of times Krieger makes the correct choice when the experimenter leans towards the object is more than 0.50. d. Decrease, 9 out of 10 is farther out in the tail of the null distribution than 6 out of 10.1.4.29a. 9 out 10 (0.90) b.  Mean = 0.50, SD = 0.16 c. 0.01 d. Appro ximately 0.80 1.4.30a. z = (0.90 − 0.50)/0.16 = 2.5 b.  Using both the p-value and the standardized statistic, we have strong evidence that the long-run proportion of times that Krieger makes the correct choice is more than 0.50. c. Yes, the p-value got smaller (evidence got stronger). 1.4.31 a.  Analysis  method ExercisesSample  size nNull value  π0 Value  of    ˆ p A: 1.4.8 – 1.4.12 938,223 0.50 0.516 B: 1.4.25 82 0.50 1.00   b. Analysis Method A provides strong evidence, but Method B is overwhelming.1.4.32a. One-sided b.  Moderat e. 6 out of 7 is not that strong.1.4.12 a. True b. False 1.4.13a. Str onger. The statistic (18/ 20 = 0.90) is much farther away from the null hypothesized value (0.50) than before (12/20 = 0.60).b. Stronger. The statistic is the same (0.60) but the sample size is much larger (100 vs. 20).c. No, 12/30 = 0.40 is less than the null hypothesis value of 0.50: thus this is not evidence that the long-run proportion of wins in Minesweeper is more than 0.50. 1.4.14a. The long-run proportion of male births is greater than 0.50. b. The long-run proportion of male births c. Null: The long-run proportion of male births is 0.50. Null: π = 0.50.d. Alternative: One-sided. He wanted to demonstrate that male births outnumbered female births.1.4.15  Probably w eak because they are quite close (0.516 and 0.50) 1.4.16  Strong; the sample size is e xtremely large (n = 938,223). 1.4.17  Twice 1.4.18a. Tin y b. Huge c. One-sided 1.4.19a. Chang e: sample size. Stay the same: distance, one- or two-sided. b. Stronger 1.4.20a. Change: distance. Stay the same: sample size, one- or two-sided. b.  Weak er 1.4.21a. Double it. The alternative would now be two -sided. b. Weaker 1.4.22a. Sample size changes; distance and one-sided are the same. b.  Strong er 1.4.23a. Distance changes; sample size and one-sided are the same. b. W eaker 1.4.24a.  The long run proportion of times that Krieger chooses the correct object, π.b. 50% c. Null: The long-run proportion of times that Krieger chooses the correct object is 0.50; Alternative: The long-run proportion of times that Krieger chooses the correct object is more than 0.50. 1.4.25a. 6 out of 10 for a proportion of 0.60 b. Mean = 0.50, SD = 0.16 c01Solutions.indd 13 10/16/20 9:15 PM14 CHAPTER 1 Significance: How Strong Is the Evidence d.  Alternative  hypothesisStrongest  evidenceShape  of curve π > 0.50 ˆ p = 1 Decreasing π < 0.50 ˆ p = 0 Increasing π ≠ 0.50 ˆ p = 0 or ˆ p = 1 Up-down 1.4.40 a. Lower b. Larger c.  Alw ays lies above d. Less steep 1.4.41 Increasing Decreasing Up- Down Steeper B A F Flatter C D E 1.4.42 a. That would be considered cheating. The hypotheses are always statements about the parameter(s) of interest and are stated before we begin to explore the data. We use the data to test our hypotheses, not to form them. b. Two-sided tests always yield larger p-values than one-sided tests, making it harder to find strong evidence in support of your research conjecture. Section 1.5 1.5.1  C. 1.5.2  N o, the maximum standard deviation is always obtained at 0.5, regardless of sample size. 1.5.3  D. 1.5.4  C. 1.5.5  D. 1.5.6  A. 1.5.7  C. 1.5.8  The predicted value of the standard deviation of a null distri- bution of sample proportions. The prediction is accurate when the sample size is large. 1.5.9  The value of the standardized statistic, z. 1.5.10 a. Decrease b.  0.50 c. 0 or 1 1.5.11  The standardized statistic; a measur ement of how many standard deviations from the mean the observed statistic is on the null distribution 1.5.12a. The p-value from option 1 is more valid. b. The validity conditions are not met for this test because the light was green only 4 times (which is less than 10). We can also see this is a problem in the applet because the normal overlay does not match up nicely with the skewed null distribution.c. p-value = 0.0547 + 0.0078 = 0.0625. We have moderate evidence that individuals living in the country have healthier lungs than those of individuals living in cities. 1.4.33a. One-sided b.  Six out of 8 seems fairly likely to occur by chance; thus it is plau- sible that bees are just as likely to sting a target that has already been stung as they are to sting a target that is pristine.c. p-value = 0.1094 + 0.0313 + 0.0039 = 0.1446. We have little to no ev - idence that bees are more likely to sting a target that has already been stung compared to a pristine target. 1.4.34a. In a race for U.S. president, is the taller candidate more likely to win? Alternatively, is π > 0.5.b. Null: The long-run proportion of races where the taller candidate wins in U.S. presidential elections is 0.5. Alternative: The long-run pro - portion of races where the taller candidate wins in U.S. presidential elec- tions is larger than 0.5. Using symbols: H 0: π = 0.5, H a: π > 0.5; where π is the long-run proportion of races where the taller candidate won. c. Probability of heads: 0.5, number of tosses: 26, approximate p- value = 0.0047.d. We have very strong evidence against the null and in support of the taller candidate winning the race more often than would be predicted by random chance. e. It is somewhat arbitrary to only look at 20th and 21st century elections.1.4.35a. A: p-value = 0.3036, B: p-value = 0.0047, C: p-value = 0.9616, D: p-value = 0.1400b. Looking at the set of p-values suggests there is little evidence that taller candidates are more likely to win. In particular, looking at all presidential elections since 1796 yields a p-value of 0.1400. Looking at an arbitrary subset of presidential elections (Exercise 1.4.34) suggested a potentially significant result, but looking at more data suggested otherwise. 1.4.36  D = Sample size is n = 25, alternative hypothesis is right-sided: π > ½.  A = Sample size is n = 225, alternative hypothesis is right-sided: π > ½.  C = Sample size is n = 25, alternative hypothesis is left-sided: π < ½.  B = Sample size is n = 225, alternative hypothesis is left-sided: π < ½.  E = Sample size is n = 25, alternative hypothesis is two-sided: π ≠ ½.  F = Sample size is n = 225, alternative hypothesis is two-sided: π ≠ ½. 1.4.37 a. Incr easing: B,C b. Decr easing: A,D c. Up-down: E,F 1.4.38a.  100% b.  50% c. 0% 1.4.39a. Smaller, decreasing, A,D b.  Larger, incr easing, B,C c. Up-down, E,F c01Solutions.indd 14 10/16/20 9:15 PMSolutions to Problems 15 f. Using the Theory-based inference applet (which uses the normal approximation for the null distribution) yields a standardized statistic of 3.02 and p-value of 0.0012, again showing strong evidence that the proportion of correct guesses is larger than 0.25. It is not surprising that the two approaches give similar results as the sample size is very much larger (in particular, there are 106 successful guesses and 223 unsuccessful guesses. Both values are much larger than 10). 1.5.18a. 32/97 = 0.330 (Morris et al.) vs. 106/329 = 0.322 (Bern and Honorton). The proportions are about the same (Morris is just slightly more). b. The p-value for Morris will be larger because the sample size is smallerc. The simulation p-value is approximately 0.048; this is strong evi - dence that the Ganzfeld receivers choices are better than just chanced. The p-value = 0.0346. It is similar to what we got from simula- tion, which is not surprising because the validity conditions are met (32 successes and 65 failures). e. They are more similar in the previous study because the sample size is larger and, even though the validity conditions are met here, the p-values will continue to get closer and closer together as the sam-ple size increases. 1.5.19a. Null: The long-run proportion of times that the male says “I love you” first is 0.50.  Alternativ e: The long-run proportion of times that the male says “I love you” first is mor e than 0.50. b. 0.0057 c. We have very strong evidence that the long-run proportion of times that the male says “I love you” first is more than 0.50d. z = 2.53. That the proportion of males that say “I love you first” is 2.53 standard deviations above the mean of the null distribution. e. 0.0057 × 2 = 0.0114 1.5.20a. Null: The long-run proportion of rhesus monkeys that choose the correct box is 0.50  Alternativ e: The long-run proportion of rhesus monkeys that choose the correct box is more than 0.50  Null: π = 0.50  Alternativ e: π > 0.50 b. 0.0003 c. We ha ve very strong evidence that the long-run proportion of Rhesus monkeys that choose the correct box is more than 0.50d. Z = 3.48. That the observed proportion of rhesus monkeys that chose the correct box is 3.48 standard deviations above the mean of the null distribution. e. 0.0003 × 2 = 0.0006 1.5.21a. Null: The long-run proportion of times that a player starts with scissors is 0.333  Alternativ e: The long-run proportion of times that a player starts with scissors is different than 0.333  Null: π = 0.333  Alternativ e: π ≠ 0.333 b. p-v alue = 01.5.13a. Null: The long-run proportion of times that a penny lands heads when spun is 0.20 Alternative: The long-run proportion of times that a penny lands heads when spun is greater than 0.20 b. The simulation based p-value of 0.077, because the validity con- ditions are not met. There are not at least 10 times where the penny landed heads and at least 10 times where the penny landed tails in the sample. c. No, we do not have strong evidence that a penny will land heads more than 20% of the time in the long run, because the p-value is only 0.077 (but there is moderate evidence). 1.5.14 a. The symbol π represents the long-run proportion of times of the coin landing heads up. b. Null: π = 0.50. Alternative: π > 0.50 c. If the result was heads 52% of the time out 1000, then 520 must have been heads and 480 tails. Both of these are greater than 10.d. A standardized statistic of 1.26 means that our observed proportion of 0.52 is 1.26 standard deviations above 0.50 in the null distribution.e. We have little to no evidence that it is more likely for a coin to land the same side up as it started than not. 1.5.15 a. 0.152; this is very close to the hypothesized parameter value of 0.15. b. No, the validity conditions are not met. There are not at least 10 successes and 10 failures in the data (only 8 and 2).1.5.16 a. 0.15; this is the hypothesized parameter value of 0.15 b.  0.019; √ π(1 − π)/n =√ 0.15(1 − 0.15)/361 = 0.019_ ________________ c. Yes d. Because the validity conditions are met for this dataset (larger sample size)1.5.17a. The long-run pr oportion of times that a person identifies the cor- rect imageb. Null: The long-run proportion of times that a person identifies the correct image equals 0.25. Alternative: The long-run proportion of times that a person identifies the correct image greater than 0.25 c. Approximately 0.25 will identify the correct image if they have no psychic ability. This is the null hypothesis.d. See graph below 0.15260 40 20 0.170 0.188 0.207 0.225 0.243 0.261 0.280 0.298 0.316 0.334Mean = 0.248 SD = 0.023 0 Proportion of successes e. The p-value is approximately 0.002 because we only got a result 0.322 or larger 2 out of 1000 times by chance. Thus, there is strong evidence that proportion of correct guesses is larger than 0.25. c01Solutions.indd 15 10/16/20 9:15 PM16 CHAPTER 1 Significance: How Strong Is the Evidence 1.5.27 a. The long-run proportion of times a six is rolled b. Null: The long -run proportion of times a six is rolled is 0.167.  Alternativ e: The long-run proportion of times a six is rolled is more than 0.167. c. 0.1497 d. We hav e little to no evidence that the long-run proportion of times a six is rolled is more than 0.167.1.5.28a. The long-run proportion of times a one is rolled. b. Null: The long-run proportion of times a one is rolled is 0.167.  Alternativ e: The long-run proportion of times a one is rolled is less than 0.167.c. 0.0840 d. We hav e moderate evidence that the long-run proportion of times a one is rolled is less than 0.167.1.5.29a. The long-run proportion of times that Mario wins b.  Null: The long -run proportion of times Mario wins is 0.50.  Alternativ e: The long-run proportion of times Mario wins is not 0.50. c. p-value = 0.2733 d. W e have little to no evidence that the long-run proportion of times Mario wins is not 0.50.e. 100 games gives z = 2 ( p = 0.0455). So, if out of 100 games Mario wins 60, this would be strong evidence that Mario’s long-run propor- tion of times he wins is not 0.50. 1.5.30 a.  π = 0.50 Null, H 0: Students randomly choose a number between 1 and 10, versus Alternative, H a: Students have a genuine preference for numbers greater than 5, π > 0.50, where π represents the long-run proportion of students that a number greater than 5, when asked to choose a number between 1 and 10. b. 0.61 c. Yes, mor e than 10 successes (62) and 10 failures (39) d. 0.011 e. We hav e strong evidence (based on the small p-value = 0.011) that students have a genuine preference for numbers greater than 5. 1.5.31  a. Null, H 0: Cardiac arrests are just as likely on Mondays as on either Wednesdays or Fridays, π = 0.33 versus Alternative, Ha: Cardiac ar- rests are more likely on Mondays than on either Wednesdays or Fri- days, π > 0.33, where π represents the long-run proportion of cardiac arrests among dialysis patients that happen on Mondays. b. 0.454 c. Yes, mor e than 10 successes (93) and 10 failures (112) d. < 0.001 e. 0.0001 f. Very strong evidence that of cardiac arrests among dialysis patients more than a third happen on Mondays (compared to Wednesdays or Fridays)1.5.32 a.  Null, H0: Cardiac arrests are just as likely on Tuesdays as on ei- ther Thursdays or Saturdays, π = 0.33 versus Alternative, Ha: Cardiac c. We have very strong evidence that the long-run proportion of times that a player starts with scissors is different than 0.333. 1.5.22a.  Null: The long-run proportion of times that a player starts with rock is 0.333.  Alternative: The long-run proportion of times that a player starts with rock is different than 0.333.  Null: π = 0.333  Alternativ e: π ≠ 0.333 b. p-v alue = 0 c. We have very strong evidence that the long-run proportion of times that a player starts with rock is different than 0.333.1.5.23a. Null: The long-run proportion of times that people assign the name Tim to the face on the left is 0.50.  Alternativ e: The long-run proportion of times that people assign the name Tim to the face on the left is more than 0.50.b.  0.0721 c. We have moderate evidence that the long-run proportion of times that people assign the name Tim to the face on the left is more than 0.50. d. z = 1.46. That the proportion of the sample that chose Tim is 1.46 standard deviations above the mean of the null distribution1.5.24a. Null: The long-run proportion of times that the most compe- tent-looking candidate wins is 0.50.  Alternativ e: The long-run proportion of times that the most compe- tent-looking candidate wins is mor e 0.50. b. n = 279, ˆ p = 0.677 c. 0 d. We have very strong evidence that the most competent-looking candidate wins more than 50% of the time.1.5.25a. Null: The long-run proportion of matches that the red uniform wins is 0.50.  Alternativ e: The long-run proportion of matches that the red uniform wins is not 0.50.b. 0.0681 c. We ha ve moderate evidence that the long-run proportion of match- es the red uniform wins is not 0.50.d. z = 1.82. That the proportion of the sample in which red won is 1.82 standard deviations above the mean of the null distribution1.5.26a. Null: The long-run proportion of times that the red uniform wins a boxing match is 0.50.  Alt ernative: The long-run proportion of times that the red uniform wins a boxing match is not 0.50.b. 0.0896 c. W e have moderate evidence that the proportion of times the red uniform wins a boxing match is not 0.50.d. z = 1.70. That the proportion of the sample in which red won a boxing match is 1.70 standard deviations above the mean of the null distribution c01Solutions.indd 16 10/16/20 9:15 PMSolutions to Problems 17 1 0Mean 2.44 SD 1.37 2 3 4 7080160240320 5 6 Number of successes d. The p-value is the probability of observing 7 or more successes out of 12 attempts when each attempt has a 20% chance of being correct. e. The theory-based approach is not appropriate here because the resulting simulated distribution of statistics is not normal. This is be- cause the sample size is not large enough. In particular, there are only 7 successes and 5 failures, instead of at least 10 of each. 1.CE.4 Step 1: Ask a research question Jamie and Adam wanted to investigate which side buttered toast pre- fers to land on when it falls through the air. Step 2: Design a study and collect data They set up a specially designed rig and dropped 48 pieces of toast from the roof of the Mythbusters’ headquarters. They wish to test the following null and alternative hypotheses: Null: There is no preference for which side the buttered toast lands on; both sided are equally likely (have a 50% chance of landing face down) Alternative: One of the sides tends to land face down more than the other. They recorded which sided landed down (buttered or not buttered side) for each of the 48 attempts. Step 3: Explore the data In 19 out of 48 attempts the buttered side landed down. Step 4: Draw inferences Statistic: 19/48 = 0.396 Simulation: Used applet to generate a distribution (using probability = 0.5, sample size = n = 48, number of samples = 1000), to generate a two-sided p-value of 0.19 Strength of evidence: We do not have strong evidence that one side of buttered toast tends to fall face down more often than the other. Step 5: Formulate conclusions We don’t know if the results can be generalized to other situations (different bread? inside vs. outside? device used to drop bread?) Step 6: Look back and aheadarrests are more likely on Tuesdays than on either Thursdays or Satur- days, While no evidence was found that one side falls to the ground more than the other , further studies are needed to ensure that the results apply to all bread, inside, and when a person drops it instead of a machine. π > 0.33, wher e π represents the long-run proportion of cardiac arrests among dialysis patients that happen on Tuesdays. b. 0.342 c. Yes, more than 10 successes (65) and 10 failures (125) d. 0.384 e. 0.3613 f. No evidence that of cardiac arrests among dialysis patients more than a third happen on Tuesdays (compared to Thursdays or Saturdays) 1.5.33  a. Null, H0: The probability a spun coin will land on heads is 0.50, π = 0.50 versus Alternative, Ha: The probability a spun coin will land on heads is not 0.50, π ≠ 0.50, where π represents the probability the spun coin will land on heads. b. The sample statistic symbol is ˆ p . Its value will vary. c. Answers will vary. d. Answ ers will vary. e. Answ ers will vary. 1.5.34  a. 0.05525 b. 0.0289 c. The simulation-based p- value is more valid because the validity conditions for the normal approximation are not met here. d. 0.0547, closer t o the simulation-based p-value 1.5.35  a. Simulation-based p-value = 0.168, normal-approx.-based p-value = 0.103, exact binomial p-value = 0.1719; the simulation-based and exact binomial p-values are close to each other. b. Simulation-based p-value = 0.378, normal-approx.-based p-value = 0.2635, exact binomial p-value = 0.377; the simulation-based and exact binomial p-values are close to each other. End of Chapter 1 Exercises 1.CE.1  Null 1.CE.2  Probability value in the null hypothesis 1.CE.3 a. Null: The probability the statistics professor wins is 0.02, Alterna- tive: The probability the statistics professor wins is larger than 0.02. b. Start with 5 playing cards—one red and four black. Shuffle the cards. Randomly choose one card, record if it is red or not, and then place the card back in the deck. Shuffle and randomly choose cards until 12 cards have been selected. Record the number of red cards selected out of the 12 selections. Repeat this entire process 999 more times to generate a distribution of counts of red cards. If 7 out of 12 red cards chosen rarely happened in the 1,000 repetitions, then this would be convincing evidence that 7 out of 12 was unlikely to have occurred by chance. c. The observed data (7 wins out of 12 attempts) provide convincing evidence that the statistics professor’s probability of winning in one week was larger than would be expected if the 5 competitor’s were equally likely to win because 7 out of 12 rarely happens by chance, when everyone is equally likely to win; in particular the p-value is 0.005. c01Solutions.indd 17 10/16/20 9:15 PM18 CHAPTER 1 Significance: How Strong Is the Evidence  Null: π = 0.50  Alternative: π < 0.50 b. The p-value is approximately 0, with a standardized statistic of –6.45. This is extremely strong evidence that the long-run proportion of times that New Zealand students associate the name Bob with the face on the left is less than 0.50. c. Yes, because there are at least 10 successes (30 Bob on left) and at least 10 failures (105 Tim on left).d. p-value = 0; z = –6.45; This is extremely strong evidence that the long-run proportion of times that New Zealand students associate the name Bob with the face on the left is less than 0.50. 1.CE.9  Not necessarily. A larger sample size yields a smaller p-value if the value of the statistic is the same; there is no guarantee the value of the statistic (proportion of heads) will be the same in Jose and Ro-berto’s separate samples 1.CE.10  A is the only correct answer . 1.CE.11  Roll a die. If it comes up 1 or 2 then call it a success, other- wise failure. Repeat the process 50 times keeping track of the total number of successes out of 50. Then, repeat sets of 50 rolls keeping track of the number of successes within the 50 rolls. 1.CE.12  Flip two coins. If they both come up heads call it a “success,” otherwise ‘failure.’ Repeat the process 25 times keeping track of the total number of successes out of 25. Then, repeat sets of 25 pairs of flips keeping track of the number of successes within the 25 paired flips. 1.CE.13a.  The long -run proportion of times that Rick makes a free throw un- derhandedb. Null: The long -run proportion of times that Rick makes a free throw underhanded is 0.90.  Alternative: The long-run proportion of times that Rick makes a free throw underhanded is more than 0.90.  Null: π = 0.90  Alternative: π > 0.90 1.CE.14 a. The long-run proportion of times that Lorena makes a 10-foot putt b. a) Null: The long-run proportion of times that Lorena makes a 10- foot putt is 0.60.  Alternativ e: The long-run proportion of times that Lorena makes a 10-foot putt is more than 0.60.  Null: π = 0.60  Alternativ e: π > 0.60 1.CE.15a. The long-run pr oportion of times someone chooses an odd number b. Null: The long-run proportion of times that someone chooses an odd number is 0.50.  Alternativ e: The long-run proportion of times that someone chooses an odd-number is more than 0.50.c. 1029/1770 = 0.581 d. Yes, the validity conditions are met. There are 1029 successes and 741 failures, both well above the minimum of 10.e.  6.85 f. p- value = 0 g.  We have very strong evidence that the long-run proportion of times someone chooses an odd number is more than 0.50.1.CE.5a. Even though 34.6% (the percentage of players suspended for PED use who were from the United States) is less than the percentage of all baseball players born in the United States (57.3%), it’s possible that this could have happened by chance (just like it’s possible to flip a coin and get heads 8 times out of 8); the question is how likely this (34.6%) would happen just by chance. If it is quite unlikely then we say that the result is statistically significant, meaning that there is something about U.S. baseball players which make them less likely to be suspended for PED use. b. The likelihood of having only 34.6% of 595 suspended baseball players be from the United States when the percentage of all baseball players who are from the United States is 57.3% is extremely unlikely; in other words, 34.6% is (statistically) significantly less than 57.3%. c. z = –11.19. This tells us that the observed proportion (0.346) is 11.19 SDs less than the mean of the null distribution, confirming that we have extremely strong evidence that the null hypothesized value of the parameter is incorrect. 1.CE.6a. Even though 61.8% (the percentage of players suspended for PED use who were from Latin America) is more than the percentage of all baseball players born in Latin America (34.6%), it’s possible that this could have happened by chance (just like it’s possible to flip a coin and get heads 8 times out of 8); the question is how likely is it that this (61.8%) would happen just by chance. If it is quite unlikely then we say that the result is statistically significant, meaning that there is something about Latin American baseball players which make them more likely to be suspended for PED use. b. The likelihood of having 61.8% of 595 suspended baseball players be from Latin America when the percentage of all baseball players who are from the United States is 34.6% is extremely unlikely; in other words, 61.8% is (statistically) significantly more than 34.6%. c.  z = 13.97. This tells us that the observed percentage (61.8%) is more than 13.95 SDs less than the mean of the null distribution, confirming that we have extremely strong evidence that the null hypothesized value of the parameter is incorrect. 1.CE.7a. Null: The long-run proportion of times that New Zealand students associate the name Tim with the face on the left is 0.50.  Alternativ e: The long-run proportion of times that New Zealand stu- dents associate the name Tim with the face on the left is more than 0.50.  Null: π = 0.50  Alternativ e: π > 0.50 b. The p- value is approximately 0, with a standardized statistic of 6.45. This is extremely strong evidence that the long-run proportion of times that New Zealand students associate the name Tim with the face on the left is more than 0.50. c. Yes, because there are at least 10 successes (105 Tim on left) and at least 10 failures (30 Bob on left).d. p-value = 0; Z = 6.45; This is extremely strong evidence that the long-run proportion of times that New Zealand students associate the name Tim with the face on the left is more than 0.50. 1.CE.8a. Null: The long-run proportion of times that New Zealand students associate the name Bob with the face on the left is 0.50.  Alternativ e: The long-run proportion of times that New Zealand students associate the name Bob with the face on the left is less than 0.50. c01Solutions.indd 18 10/16/20 9:15 PMSolutions to Problems 19 4. Null: The probability that students choose the right front tire is 0.25 (π = 0.25). Alternative: The probability that students choose the right front tire is more than 0.25 (π > 0.25). 5. In this sample,14 out of 28 or 50% of the students selected the right front tire. This is more than we would expect if students choose ran- domly and equally among the four tires (25% of the time in the long run). 6. Yes, anything is possible, although some outcomes will be less like- ly or believable when the null hypothesis is true. 7. Our statistic is p = 0.50.ˆ 8. Probability of success (π) = 0.25, sample size ( n) = 28, number of samples = 1,000 9. The cent er is located at 0.25. Yes, it makes sense that this is the center because 0.25 is the specified null hypothesis proportion. 10.  a. The p-value from the simulation should be around 0.004. In other words, in a large number of samples (from a process with = 0.25), roughly .4% of those samples should have a sample proportion of 0.50 or larger. b. The standard deviation of the simulated sample proportions should be around 0.082, so the standardized statistic is approximately z = (0.50 – 0.25)/0.082 = 3.05. (This number may vary a bit because the SD of the simulated null distribution will vary a bit.) Therefore, a sample proportion of 0.50 is more than 3 standard deviations above the hypothesized process probability of 0.25. c. The One Proportion applet reports a theory-based p-value of 0.0011. The validity conditions are met because there are 14 “successes” (choose right front) and 14 “failures” (choose something else), both of which exceed 10. 11. Yes, all three methods in question 10 give strong evidence against the null h ypothesis; the p-values are quite small and the standardized statistic is large (e.g., above 2). 12. We have strong evidence that students pick the right front tire more than 25% of the time because we would rarely have 50% of a sample of 28 students choose the right front tire if the long-run pro- portion of students choosing the right front tire is 0.25. 13. Hard to say. The question is: “Who are these students?” Will they perform similarly to other people in the same situation? It’s hard to say that these students will necessarily act like people in general. We can probably say that we can infer these results to people similar to those that were in the study. 14. Answers will vary. Some things to consider include selecting a broader representation of students to participate in the study, exam - ining exactly how the question is posed to students and whether that impacts their choices, considering whether there might be gender differences or a tendency for different responses among individuals who have recently had a flat tire, and whether this tendency is similar across cultures (including countries where motorists drive on other side of the road). 15. We would expect to find weaker evidence because the sample size is smaller and the statistic (0.50) has stayed the same. 16. The p-value should be around 0.04 which is larger than we got before and hence weak er evidence, as expected. 17. Null: The probability that students choose the right front tire is 0.25 (π = 0.25). Alternative: The probability that students choose the right-front tire is different than 0.25 (π ≠ 0.25). 18. The p-value is 0.0023 for the two-sided test—about twice as large as before.1.CE.16 a. The long-run proportion of times someone chooses 7 b. Null: The long-run proportion of times that someone chooses 7 is 0.10.  Alternative: The long-run proportion of times that someone chooses 7 is more than 0.10. c. 503/1770 = 0.284 d. Yes, the validity conditions are met. There are 503 successes and 1267 failures, both well abov e the minimum of 10. e. 25.83 f. p-value ≈ 0 g. We have very strong evidence that the long-run proportion of times someone chooses 7 is more than 0.10. 1.CE.17 a. We don’t have strong evidence that the probability a spun tennis racket lands with the label up is differ ent from 0.50. b. If the probability that a spun tennis racket lands with the label is actually 0.50, then it is quite likely to get 46 spins out of 100 with the label up; thus, a reasonable (plausible) explanation for the author’s data (46 out 100) is that the tennis racket is fair (0.50 chance of label landing up). 1.CE.18  Null h ypothesis probability = 0.50  Statistic = 0.46  p-value = 0.484 1.CE.19  Answers will vary . 0.05000 00.0050 0 0.07 0.14 0.21180 120 60 0 0.29 Proportion of successes0.36 0.43 0.50 0.57Mean = 0.243 SD= 0.081 1.CE.20  It would be more surprising for the sample proportion to be 0.70 in a sample of 60 compared to a sample of 30 students, if the population proportion is 0.60. This is because it is more unusual to see extreme results in a larger sample (p-value = 0.0719) rather than a small sample (p-value = 0.1763). Chapter 1 Investigation 1. The observational units are the 28 students. 2. The variable recorded is which tire each student indicates (Right front, right rear, left front, left rear). This is a categorical, non- bina- ry variable. We could also define the variable to be “right front” or “not right front” as that is the primary outcome of interest in our research question. 3. The parameter is the long-term proportion of students who will pick the right front tire. c01Solutions.indd 19 10/16/20 9:15 PM20 CHAPTER 1 Significance: How Strong Is the Evidence 19. Yes, we have strong evidence the probability is different than one- fourth because the p-value of 0.0023 is still small enough to be consid- ered strong evidence against the null hypothesis.  It is int eresting to note here that even though we pass the technical con- ditions, the sample size of 28 is still pretty small and the theory-based method is not in close agreement with the simulation approach. Chapter 1 Research Article 1. The researchers are examining the nature and development of at- titudes toward similar and dissimilar others in human infancy 2. (two options, among others) (a) Dissimilar others are perceived as unkind trustworthy and unintelligent (Brewer 1979, etc.) (b) Humans may engage, support or ignore violence directed towards individuals who differ from themselves (Prentice and Miller, 1999) 3.  16 4. To figure out what kind of food the babies preferred (graham crackers or green beans) so that information could be used later in the study 5. Food preference: green beans or graham crackers, categorical with 2 outcomes (green beans or graham crackers)6. To have babies establish which rabbit is similar to them, and which is dissimilar7. No 8. The researchers needed to make sure that the babies understood what they were seeing (one puppy be nice to the rabbit, and one puppy be mean to the rabbit). 9. Puppy preference: Harmful or Helpful 10. 12/16 chose helper when viewing activities involving the rabbit similar to them, compared to 4/16 who chose the harmful puppy when viewing activities involving the rabbit similar to them.11. 100% similar chose helper; 0% dissimilar chose helper 12. Fifty-three percent is fairly close to 50% (the null hypothesis) and the sample size (36) is not large.13. To modify the experiment so that babies have a “neutral” option to provide strong comparisons between groups14. Null hypothesis: The long-run proportion of times an infant chooses the harmer dog is 0.50 when the dog interacts with the dis- similar rabbit; Alternative hypothesis: The long-run proportion of times an infant chooses the harmer dog is not 0.50 when the dog in-teracts with the dissimilar puppet.. 15. Answers will vary. Sixty-three percent of 14-month-olds in the study chose graham crackers over green beans when given a choice between the two. 16. Answers will vary. Null hypothesis: The long-run proportion of times that a 14-month-old chooses helper character instead of the neutral character is 0.50 when the dog interacts with the similar rab-bit; Alternative hypothesis: The long-run proportion of times that a 14-month-old chooses the helper character instead of the neutral character is not 0.50 when the dog interacts with the similar rabbit. The p-value is 0.08, meaning that there is moderate evidence that the long run proportion of times that a 14-month-old chooses the helper characters instead of the neutral character is not 0.50. 17. If the infants in the study are special in some way (e.g., par - ticularly developmentally advanced or not; different ethnicity, socio- economic status, etc.) then the results from this study may not generalize to all infants. 18. Answers will vary. (a) Select babies to represent different ethnici- ties/SES in order to improve the ability to generalize the results. (b) Give babies different foods to choose from initially to ensure that there is no impact of green beans/graham crackers in particular. 19. If babies generally liked the graham cracker rabbit/dog and didn’t like the green bean rabbit/dog, then the researchers’ conclu- sions about similarity/dissimilarity would be invalidated, because the differences between the similar and dissimilar conditions would be better explained by a different variable (green beans/graham crackers). 20. The researchers refer to prior research that links adult and child similarity preferences and group psychology, which, they argue, sug- gests that their results are more inborn (nature) rather than the result of accumulated experiences (nurture). c01Solutions.indd 20 10/16/20 9:15 PM21 Section 2.1 2.1.1  B. 2.1.2  A. 2.1.3  A. 2.1.4  C. 2.1.5  C. 2.1.6  C. 2.1.7  D. 2.1.8  D. 2.1.9  A. 2.1.10  A. 2.1.11  B. 2.1.12  A. False B. True C. True 2.1.13  A. 2.1.14  B. 2.1.15a. F. b. C. c. A. 2.1.16a.  _____________ SD = √ 0.41 ( 0.59 ) / 30 = 0.090 b. The SD should be about 0.085 which is a bit smaller than that from part (a). It is different because the population is less than 10 times the sample size. It should be at least 20 times the sample size.c. The SD should be about 0.090 which is what we got in part (a). The larger population size helped.2.1.17  The graph of the most recent sample represents whether or not a word was short, a categorical variable. However, in the graph of the proportions, the horizontal axis represents the proportion of short words in a sample, a quantitative variable.2.1.18 a. Mean = 0.25, and SD = 0.068 b. Mean = 0.25, and SD = 0.022 2.1.19a.  ______________ Mean = 0.20, SD = √ 0.20 ( 0.80 ) / 1,414 = 0.0106 b.  ( 271 / 1,414 − 0.200 ) / 0.0106 = − 0.7873 . So the sample proportion is 0.7873 standard deviations below the mean.c. It is not very unlikely to get a sample proportion of 271/1,414 be- cause it is within 2 standard deviations of the mean. 2.1.20  Using a phone call as the method of asking this question is probably a biased method. Those answ ering a person on a phone call were much more unlikely to say that they exercise less than once per week. Having an interaction with a person probably makes some peo-ple not give the socially undesirable answer. 2.1.21a. Likely representative, because the distribution of blood type is probably not different among those that eat at the cafeteria compared to that of the U.S. population.b. Likely not representative because those in the cafeteria may eat most of their meals in the cafeteria and not regularly eat fast foodc. Perhaps repr esentative because the proportion with brown hair is probably not too different among those that eat at the cafeteria com- pared to that of all the students at the school.d. This may not be representative. Those in the cafeteria (as well as those at the school) could differ quit e a bit racially from the U.S. pop- ulation and thus would differ in the proportion that has brown hair.2.1.22  Although some of these could be argued the other way, all of the samples would likely not be representative. a. If you have food that finches like and other birds do not, you would overestimate the proportion of finches. b.  You could have food that finches do not like, and you would rarely see more than one eating at a time. c. The proportion of male birds could be species-dependent and depending on the type of food you have could affect the type of spe- cies and hence affect the proportion of male birds that come to your feeder. CHAPTER 2 Generalization: How Broadly Do the Results Apply? c02Solutions.indd 21 10/19/20 6:22 PM22 CHAPTER 2 Generalization: How Broadly Do the Results Apply? c. Not comfortable generalizing to all U.S. adults. Instead, comfort- able generalizing to a population like the one that participated in the survey—watchers of the TV news program who were motivated enough to participate. d. Theory-based is appropriate because at least 10 successes and 10 failures in the data, z = 3.83 and p-value = 0.0001.2.1.33a. The population is all the sharks at the zoo. b. The pr oportion should be similar to the population because it came from a random sample.2.1.34a. The parameter is the proportion of sharks in the zoo that have the disease.b. The statistic is the proportion of sharks in the sample (0.20) that have the disease.2.1.35a. The variable measured is whether or not they have the disease. b. Categorical c. Pr oportion d. Bar graph 2.1.36a. Null: The proportion of all sharks at the zoo that have the disease is 0.25. Alternative: The proportion of all sharks at the zoo that have the disease is less than 0.25.b. 3/15 = 0.20 2.1.37a. p-value = 0.46 b. We do not have any evidence that the proportion of sharks in the zoo that have the disease is less than 0.25.c. The sharks at the zoo d. A theory-based approach using the normal distribution is not rea- sonable to use because there were only three sharks with the disease. We need at least 10.2.1.38a. All the customers of the store b. The 100 people asked to fill out the survey c. The proportion of all customers that visit the store because of the sale on coatsd. The proportion of the sample that said they visited the store be- cause of the sale on coats (0.40)2.1.39 It may not be representative because it was not a random sample.2.1.40a. The new pr oportion of 1,523/2,216 = 0.687 is closer to the propor- tion that actually voted.b. ________________ SD = √ 0.592 ( 0.408 ) / 2,216 = 0.0104 c. (0.687 – 0.592)/0.0104 = 9.13. Because the sample proportion is 9.13 SD above 0.592, it is significantly larger. Results like this are very unlikely to happen by chance.2.1.41  The question is awkwardly phrased. “Does it seem possible or impossible … it never happened?” The latt er part (“impossible it never happened”) involves a double negative.2.1.42  This question is more clearly phrased; they got rid of the double negative.d. This proportion could then be differ ent than the proportion of males in your area as well as those that typically visit feeders. 2.1.23 a. The population is all the likely voters in the city. b.  Because it is a random sample, I would think the proportion that favor the incumbent in the sample is similar to that for the population. 2.1.24a. The proportion of all city voters that plan to vote for the incumbent b. The proportion of those in the sample that plan to vote for the incumbent (0.65). 2.1.25a. The variable is who they plan to vote for or whether or not they plan to vote for the incumbent.b. Categorical c. Proportion d. Bar graph 2.1.26a. Null: 50% of the all the city voters plan to vote for the incumbent. Al- ternative: A majority of all the city voters plan to vote for the incumbent. b. The proportion of all city voters that plan to vote for the incum- bent, 0.65.2.1.27a. p-value ≈ 0 b. We have strong evidence that a majority of all the city voters plan to vote for the incumbent.c. We can infer these results to all likely city voters because they came from a random sample. d. Using theory-based methods is appropriate and we also obtain a p-value of approximately 0.2.1.28a. The population is all adults in the United States b.  It is perhaps greater than the population. People were allowed to self-select themselves to be part of the sample. This method will of- ten overestimate the population proportion because people that really care about the issue will be the ones to respond. 2.1.29a. The proportion of U.S. adults that are unhappy with the verdict. b. The proportion of the sample (0.82) that are unhappy with the verdict. 2.1.30a. Whether or not someone is unhappy with the verdict b. Categorical c. Proportion d. Bar graph 2.1.31a. Null: The proportion of U.S. adults that are unhappy with the ver- dict is 0.75. Alternative: The proportion of U.S. adults that are unhap- py with the verdict is greater than 0.75.b. 0.82 2.1.32a. p-value ≈ 0 b. We have very strong evidence that the proportion of all U.S. adults that are opposed to the verdict is greater than 0.75. c02Solutions.indd 22 10/19/20 6:22 PMSolutions to Problems 23 d. 7.5 e. The mean would be larger because the distribution is skewed to the right. f. The mean would be smaller, the median would stay the same, and the standard deviation would be smaller.2.2.20a. The distribution is skewed t o the right. b. Because the distribution is skewed to the right, we should expect the mean to be higher than the median.c. The median is $35, and the mean is $45.68. The mean is higher, as expected.d. If a $150 haircut is changed to $300 we should expect the median to stay the same (because $150 or $300 is just another larger value), but the mean should increase (because the total of all haircut costs will increase). When the change is made, we can see the median does stay the same and the mean increases to $48.68. 2.2.21a. E. b. C. c. B. 2.2.22a.  _____2.119 SD = _ = 0.387 √ 30 b. The SD is about 0.366. It is smaller than what is predicted because the population size is too small. The population size should be at least 20 times the sample size and it is less than 10 times. c. The SD is about 0.386 or 0.387, much closer to what was predicted in (a).2.2.23a. Mean = 10, and SD = 0.8 b. Mean = 10, and SD = 0.4 2.2.24 a. Decreases b. Increases c. 100 2.2.25  The sample size must be four times as large. 2.2.26a. Mean = 100 points, SD = 3.35 points b. 80 c. 320 2.2.27a. i. Mean = 8 hours, SD = 0.47 hours, and bell-shaped;  ii. Yes b.  i.  Mean = 8 hours, SD = 0.47 hours, and slight skewed to the right;  ii. yes c. i. Mean = 8 hours, SD = 0.47 hours, and bell-shaped;  ii. Yes d. R egardless of the shape of the population distribution, the mean and SD were about 8 and 0.47, respectively. When the population had a symmetric shape, the sampling distribution had a bell shape. But when the population had a skewed distribution, the sampling distri-bution was also skewed. 2.2.28a. Students at her school2.1.43  Yes, assistance to the poor likely elicits a more favorable re- sponse toward progr ams than the term w elfare. 2.1.44  Probably the question that did not give two options will yield a higher percentage of affirmative responses. 2.1.45  A sample of 1,500 might not have many or any members of a rare subpopulation, but that does not matter because members of a rare subpopulation are such a small part of the entire population and hence do not really affect an overall population proportion. 2.1.46  False. Increasing the sample size will not affect bias but will only affect sampling variability.2.1.47  In the Doris and Buzz example, Dr. Bastian randomly deter- mine if the light would flash or if it would be shining steadily.2.1.48  Randomness occurs in the chance model by flipping a coin to determine if Buzz would push the correct button if he was just guess- ing (a constant probability). In reality, Buzz may not have a constant probability of choosing the correct button. He may be learning along the way, he may get tired, or his stomach may get full of fish. Section 2.2 2.2.1  B. 2.2.2  B. 2.2.3  A. 2.2.4  D. 2.2.5  B. 2.2.6  C. 2.2.7  B. 2.2.8  B. 2.2.9  D. 2.2.10  C. 2.2.11  A. 2.2.12  A, D. 2.2.13  E. 2.2.14  Graph (a) is a distribution of sample means from samples of size 30; we know this because it is the distribution with the smaller SD. 2.2.15 a. False b. False c. False 2.2.16  The horizontal axis of the graph of the most recent sample is the length of the wor ds, but the horizontal axis of the graph of the statistic is the mean length of the words in a sample of 10 words. Both of these are quantitative. 2.2.17a. Because the distribution is skewed to the left, the mean will be to the left of the median; hence, 65.86°F is the mean and 67.50°F is the median.b. Mean: Larger; Median: Larger; Standard deviation: Smaller 2.2.18  Because your score of 84 is below the median of 87, more stu- dents had exam scores higher than yours than below.2.2.19a. The students in the class b.  The variable is the number of states visited and it is quantitativ e. c. The graph is centered about 8, most of the data is between 2 and 16 (and skewed to the right a bit) with outliers 25, 30, and 43. c02Solutions.indd 23 10/19/20 6:22 PM24 CHAPTER 2 Generalization: How Broadly Do the Results Apply? 2.2.37  VoterTime spent reading/ learning about local politicsVoting for  incumbent? #1 0 Yes #2 0 Yes #3 60 No … … … 2.2.38  The study was done using a random sample. It could have been done by obtaining a list of all the voters in the city and then as- signing every voter a number. Then have a random number generator give 267 random numbers. The voter’s names that match the numbers chosen will be the random sample. 2.2.39a. The variable is the time spent reading or watching news coverage about the trial in the last 3 days.b. Quantitative c. Mean or median d. Dotplot 2.2.40 RespondentTime spent reading/ watching about the trialNot happy  with verdict? #1 240 Not happy #2 90 Not happy #3 30 Happy … … … 2.2.41  The study was not done using a random sample. Because national polls like this don’ t have available lists of all U.S. adults from which to sample, typically random-digit dialing is done. Phone numbers are randomly generated. This could have been done for this poll. 2.2.42 a. The shark’s blood oxygen content b. Quantitative c. Mean or median d. Dotplot 2.2.43 Shark Has disease? Blood oxygen content #1 Yes 1.2% #2 No 5.6% #3 No 6.2% … … … 2.2.44  The study was done using a random sample. It could have been done by obtaining a list of all the sharks at the zoo and assigning each a number. Then have a random number generator give 15 ran- dom numbers. The sharks that match the numbers chosen will be the random sample. 2.2.45a. All full-time students at the school b. The 150 students in the sample c. The average daily study time for all full-time students at the schoolb. I would guess that the average number of hours students in a statistics class watch TV per da y is similar to that of students in the entire school. 2.2.29a. The parameter is the average number of hours students at the school watch TV per day.b. The statistic is the average number of hours students in the sample watch TV per day (1.2 hours).2.2.30a. The variable is the number of hours of TV is watched per day. b. Quantitative c. Mean or median d. Dotplot 2.2.31a. If a student watched more than 10 minutes of TV yesterday b. Categorical c. Proportion d. Bar graph 2.2.32a. Null: 50% of the students at the school watched at least 10 minutes of TV yesterday . Alternative: More than 50% of the students at the school watched at least 10 minutes of TV yesterday. b. The proportion (or percentage) of all students that watched at least 10 minutes of TV yesterdayc. 21/30 = 0.70 2.2.33a. 0.0214 b.  We have strong evidence that students watched more than 10 min- utes of TV yesterday . c. Concerned because not a random sample of students at the school, although that is the population of inter est; generalize to this popula- tion with caution. d. A theory-based approach is not reasonable because there are not at least 10 students in the sample who wat ched less than 10 minutes of TV yesterday.2.2.34  Student Hours per day Watched TV yesterday Alejandra 2 no Ben 4 yes Cassie 0.5 no … … … 2.2.35  The study was not done using a random sample. To take a random sample of 30 students at the school, we first need to obtain a list of all the students at the school and randomly choose from that list. One way to do this is to assign every student a number and then have a random number generator give you 30 random numbers. The students’ names that match the numbers chosen will be your ran-dom sample. 2.2.36a. How long people spend reading or learning about local politics b. Quantitative c. Mean or median d. Dotplot c02Solutions.indd 24 10/19/20 6:22 PMSolutions to Problems 25 2.3.8 a. Statistic b. _x = 3.01 c. You would have to fabricate a large dataset to represent the popu- lation of times for all Cal Poly students with the variability similar to that of the sample data and a mean of 2.84. From that data you would take a sample of 100 and find its mean. Repeat this at least 1,000 times to develop a null distribution. To determine the p-value, determine the proportion of simulated statistics that are at least as far away from 2.84 as that of 3.01. d. Simulation Real study One repetition = A random sample of 100 students Null model = Population mean hours of TV watching is 2.84 hours Statistic = The average TV watching time in the sample 2.3.9 a. If the mean daily TV watching time for Cal Poly students is 2.84 hours, the probability we would g et a sample mean as extreme as 3.01 from a random sample of 100 students is 0.16. b. Because this is a one-sided test, Dr. Elliot’s p-value should be about half that of Dr. Sameer’s. 2.3.10 a. s = 1.97 is a statistic. b. _____ (3.01 − 2.84)/(1.97/ √ 100 ) = 0.86 c. Because the standardized statistic is 0.86 (and that is less than 2) we do not have strong evidence that the averag e time Cal P oly students watch TV is different than 2.84 hours. 2.3.11 a. Yes, because the sample size is large b. The standardized statistic is 0.86 and the p-value is 0.3903. c. See Solution 2.3.11c. Because the p-value is greater than 0.05 we do not have strong evidence that the average time Cal Poly students watch TV is different than 2.84 hours. Sample data SD SD Mean 1 2 3 4 5100 3.01 1.97 CalculatePaste dtataScenario: n: mean, x: sample sd, s: ResetOne mean CalculateTheory-based inference t = 0.86 Standardized statistic df = 99Mean = 2.84 SD = 0.197 t= − 4 t= − 2 t= 0 t= 4 t= 2Ho: = Ha: 2.84 2.84 2 2.4 2.8 3.1 3.5 p-value 0.3903 x" " -------- -------- Solution 2.3.11 cd. The average daily study time for the students in the sample (2.23 hours) 2.2.46  Because it is a random sample it should be representative of the population. 2.2.47 a. If one coffee bar opens earlier than the other, the lines may be different at the bars when they first open (if one opens before first class starts it may have a longer line than one that opens after first class starts). b. A more representative sample of students might be observed at different times of the day and on different days of the week. 2.2.48 a. Maybe ov erstate, as students may tend to think they are getting more sleep than they are actually getting b. Maybe overstate, as students may tend to think they are volunteer- ing more than they are actually volunteering c. Maybe overstate, as students may tend to think they are attending church more often than they actually are attending d. Maybe overstat e, as students may tend to think they are studying more than they actually are studying e. Maybe overstate, as students ma y tend to think they are wearing a seat belt more often than they actually wear a seat belt Section 2.3 2.3.1 B. 2.3.2 C. 2.3.3 D. 2.3.4 B  . 2.3.5  C. 2.3.6 C. 2.3.7 a. The number of hours Cal Poly students watch TV per day and it is quantitative b. μ = the average number of hours Cal Poly students watch TV per day c. H0: μ = 2.84 hrs. H a: μ≠ 2.84 hrs c02Solutions.indd 25 10/19/20 6:22 PM26 CHAPTER 2 Generalization: How Broadly Do the Results Apply? 2.3.15 a. Yes, because the sample size is larger than 20 b. The standardized statistic: t = 2.13 and the p-v alue = 0.0191 (ap- plet output is shown in Solution 2.3.15b).c. If the mean SPF for all students at the school is 30, the probabil- ity we would get a sample mean as large or larger than 35.29 from a random sample of 48 is 0.0191. d. We have strong evidence that the average SPF used by students at the school is more than 30. 2.3.16 a. The diameter of needles and it is quantitative b. H 0: μ = 1.65 mm H a: μ≠ 1.65 mm c. _n = 35, x     = 1.64, s = 0.07 d. See Solution 2.3.16d. The mean should be about 1.65 and the standard deviation of the distribution should be about 0.07_____ √___ 35 ≈ 0.01. 2.3.17a. Y ou would have to fabricate a large dataset to represent the popu- lation of needles with the variability similar to that of the sample data and a mean of 1.65. From that data you would take a sample of 35 and find its mean. Repeat this at least 1000 times to develop a null distribu-tion. To determine the p-value, determine the proportion of simulated statistics that are at least as extreme or more extreme than 1.64. b. Simulation Real study One repetition = A random sample of 35 needles Null model = Population mean is 1.65mm Statistic = Average diameter of needles in the sample t−4 t−2 t0 t2 t4One mean Reverse (30.299, 40.281)Calculaten: mean, x: sample sd, s:48 17.1935.29¯ Ho: µ=30 Ha: µ>30test statistic t =2.13 95 %p-value 0.0191 <----------“x”----------> ¯20.1 25 30 35 39.9Mean = 30.00 SD = 2.481 Solution 2.3.15b 1.65 1.64 1.63 1.62 1.67 1.66 1.68Mean = 1.65 SD = 0.01 Sample mean diameter (mm) Solution 2.3.16d2.3.12  The standardized statistic should be more than 2 because a t-distribution has more area (probability) in the tail, we would have to move the standardized statistic out farther to reduce the probability to what you would find beyond 2 in the tail of a normal distribution. 2.3.13 a. The SPF value for sunscreen used by students at a certain school b. μ = mean SPF of sunscreens used by all students at this school. c. H0: μ = 30 versus H a: μ > 30 d. _n = 48, x    = 35.29, s = 17.19 e. No, it just came from students in her class f. They probably don’t differ much from the students as a whole on this issue. g. It may not be representative for students at a Midwestern college where it is very cloudy.2.3.14 a. You would ha v e to fabricate a large dataset to represent the pop- ulation of SPF numbers of sunscr eens for all students at the school with the variability similar to that of the sample data and a mean of 30. From that data you would take a sample of 48 and find its mean. Repeat this at least 1,000 times to develop a null distribution. To de-termine the p-value, determine the proportion of simulated statistics that are at or more than 35.29. b. Simulation Real study One repetition = A sample of 48 students Null model = Population mean is 30 Statistic = Average SPF number in the sample c02Solutions.indd 26 10/19/20 6:22 PMSolutions to Problems 27 c. Although both mistakes are problematic, telling someone they are healthy when they are not could hav e serious, negative, long-term personal and community health impacts (e.g., they could give others the disease). 2.3.25 a. We have strong evidence the subject is not telling the truth. b. It is plausible the subject is telling the truth (we cannot rule out the fact that they are telling the truth). c. We have strong evidence the subject is not telling the truth, when in fact they are telling the truth.d. We do not have str ong evidence the subject is not telling the truth (it’s plausible they are telling the truth), when in fact they are lying. 2.3.26 a. We have strong evidence the incoming messag e is not legitimate. b. It is plausible the incoming message is legitimate. c. We have strong evidence the incoming message is not legitimate, but it is legitimate. d. It is plausible the incoming message is legitimat e, when it is actu- ally not legitimate. 2.3.27 a. We have strong evidence the new treatment is better, when it actu- ally is not. b. We do not have evidence the new treatment is better, when it actu- ally is. 2.3.28 a. We find strong evidence that Buzz is not guessing, but he is guessing. b. We do not have evidence that Buzz is not guessing (guessing is plausible), when Buzz is actually not guessing. 2.3.29  A Type I error is possible her e, which means that we conclude there is strong evidence Buzz is not guessing, when in fact Buzz is guessing.2.3.30 a. The true, long-run, average needle diameter ( μ ). b. Null: The long-run av erage needle diameter is 1.65 mm ( μ = 1.65 mm). Alternative: The long-run average needle diameter is different than 1.65 mm ( μ ≠ 1.65 mm). c. We find evidence that the long-run average needle diameter is dif- ferent than 1.65 mm, when it is actually 1.65 mm. The consequence is that production may stop when it should not have.d. We do not have evidence that the long-run average needle diameter is different from 1.65 mm, when it is actually different than 1.65 mm. The consequence is producing needles with the wrong diameter and selling them in the marketplace, which ultimately may be bad for business. 2.3.31  Type I error would be the producer’s risk because they risk shutting down manufacturing when it should not ha ve and Type II error would be the consumer’s risk because they are risking using needles which (purportedly) have an average diameter of 1.65 mm when the true average may not be 1.65 mm. 2.3.32  If a dataset is symmetric, the mean and median will be about the same and (mean – median) will be about 0. As the dataset gets more skewed to the high numbers the mean will get larger than the median and (mean – median) will get larger. Similarly as the dataset 2.3.18 a. The standardized statistic is –0.85 and the p-value is 0.4039. b. Because the p-value is much greater than 0.05, we do not have strong evidence that the average diameter of the population of nee- dles is different than 1.65mm. 2.3.19 a. The distribution is fairly symmetric. b. The mean and the median will be about the same because the dis- tribution of temperatures is fairly symmetric. c. The mean is 98.105 and the median is 98.100. Yes, they are very close.d. The actual standard deviation is 0.699. 2.3.20 a.  Null: The average body temper ature for males is 98.6°F (  μ = 98.6°F), Alt ernative: The average body temperature for males is not 98.6°F ( μ ≠ 98.6°F). b. The standardized statistic is −5.71 and the p-value is 0. c. Because the p-value is less than 0.05 we have strong evidence that the average male body temperature is not 98.6°F. d. Any gener alization should be done with caution, but we can probably generalize it to healthy male adults similar t o those that were in the study. 2.3.21 a. Null hypothesis: The average body temperature for females is 98.6 °F ( μ = 98.6°F). Alternative: hypothesis: the average body tem- perature for females is not 98.6°F ( μ ≠ 98.6°F). b. Hard to tell, there is a lot of variability in the data c. The standardized statistic is −2.24 and the p-value = 0.0289. d. Because the p-value is less than 0.05 we have strong evidence that the average body temperature for females is different than 98.6°F. 2.3.22 a. Null h ypothesis: The average commute time in the California city is 27.5 minutes. Alternative hypothesis: The a verage commute time in the California city is different than 27.5 minutes. b. The standardized statistic is –2.23 and the p-value is 0.0308. c. Because the p-value is less than 0.05 we have strong evidence that the average commute time in the California city is different than 27.5 minutes. d. Perhaps we can generalize just to the people like those that she and her parents know. They could easily not be r epresentative of the city residents as a whole. 2.3.23 a. Rejecting a true null hypothesis is finding an innocent person guilty. b. Not rejecting a false null hypothesis is finding a guilty person not guilty. c. Our judicial syst em is supposed to be set up to make it difficult to find an innocent person guilty, the one described in part (a). 2.3.24 a. We hav e strong evidence the patient has the disease, when in fact they are healthy. The consequence is pot entially frightening a patient into thinking they are diseased when, in fact, they are not. b. We do not have evidence the patient has the disease, when in fact they are diseased. The consequence is giving a patient a false/untrue sense of security. c02Solutions.indd 27 10/19/20 6:22 PM28 CHAPTER 2 Generalization: How Broadly Do the Results Apply? c. The mean would increase because 5 would be added to the total of the scores. We cannot really say for certain how or whether the median or standard deviation would change. It depends on the orig- inal distribution of scores. For example, if adding the 5 points to the lowest score could still keep it the lowest score or it could change it to the highest score. 2.4.19  No, the distribution of means is not symmetric but is skewed right. Therefore, a t-distribution will not model this well.2.4.20  Randomly choose a coin out of your collection, note its year, and put it back. Do this same thing 99 mor e times so you have a sam- ple of 100 years. Find the mean of those 100 years. Repeat this process many, many times to develop a bootstrap sampling distribution. 2.4.21 a. (3, 3, 3), (3, 3, 6), (3, 6, 6), (3, 3, 9), (3, 9, 9), (6, 6, 6), (6, 6, 9), (6, 9, 9), (9, 9, 9), (3, 6, 9)b. There are seven possible means: 3, 4, 5, 6, 7, 8, 9. c. There are only three possible medians: 3, 6, 9. 2.4.22a. H 0: The mean body temperature for males is 98.6°F ( μ = 98.6°F). Ha: The mean body temperature for males is not 98.6°F ( μ ≠ 98.6°F). b.  _x = 98.105°F, s = 0.699 c. SD ≈ 0.086 d. (98.105 – 98.6)/0.086 = –5.76. Because the sample mean 98.105°F is more than 5 SDs below the hypothesized mean of 98.6°F, we have very strong evidence that the mean body temperature for males is dif- ferent (less) than 98.6°F. e. As the p-value < 0.0001, we come to the same conclusion. f. We can probably generalize to all healthy adult males in the United States between the ages of 18 and 40. 2.4.23  98.105 ± 2(0.086) = 97.933°F to 98.277°F 2.4.24a. The standar dized statistic is t = –5.71. It is very similar to what was found using a bootstrap sampling distribution.b. The p-value is < 0.001. It is very similar to what was found using a bootstrap sampling distribution.2.4.25a. H 0: The mean body temperature for females is 98.6°F ( μ = 98.6 °F). Ha: The mean body temperature for females is not 98.6°F ( μ ≠ 98.6°F).b.  _x = 98.394°F, s = 0.743 c. SD ≈ 0.092 d. (98.394 – 98.6)/0.092 = –2.24. Because the sample mean 98.394°F is more than 2 SDs below the hypothesized mean of 98.6°F, we have strong evidence that the mean body temperature for females is differ- ent (less) than 98.6°F. e. As the p-value ≈ 0.022, yes, we come to the same conclusion. 2.4.26a. The standardized statistic is t = –2.24. It is exactly what was found using a bootstrap sampling distribution.b. The p-value is 0.0289. It is very similar to what was found using a bootstrap sampling distribution.2.4.27a. H 0: The mean increase in the laugh rating is 0 ( μ = 0). H a: The mean increase in the laugh rating greater than 0 ( μ > 0). b.  _x = 0.295, s = 0.427 c. SD ≈ 0.067gets more skewed to the low numbers the mean will get smaller than the median and (mean – median) will get smaller. When we divide by the SD we standardize this difference and thus this makes this statistic a good measure of skewedness. 2.3.33  Deviations fr om the average always sum to be zero. So if you know n – 1 deviations, you can figure out the other one. This means that only n – 1 of the numbers of a dataset are free. The last one is de- termined by the rest. The more degrees of freedom for a t-distribution, the more it looks like a normal distribution. 2.3.34  A p-value of 0.04999 or 0.050001 are very similar and came from very similar results. It is not like something magical happens when a p-value drops below 0.05. p-values around 0.05 are really all about the same, just as p-values close to any other number are all about the same. Therefore, we should not have dramatically different conclusions for p-values of 0.04999 and 0.050001. Section 2.4 2.4.1  B. 2.4.2  A. 2.4.3  B. 2.4.4  B. 2.4.5  C. 2.4.6  B. 2.4.7  C. 2.4.8  D. 2.4.9  B. 2.4.10  A. 2.4.11  B. 2.4.12  A. 2.4.13  A. 2.4.14  B. 2.4.15  a. Get larger b. Stay the same c. Get larger d. Get larger 2.4.16 a. False b.  True c. False d. True 2.4.17a. True b. F alse c. True d. False 2.4.18a. The mean and median would increase by 5 points; the standard deviation would stay the same because the entire distribution would move up but the variability stays the same.b. The mean would increase because 5 would be added to the total of the scores. The median would stay the same because the high score re- mains the high score. The standard deviation would increase because there would be more variability with the larger number. c02Solutions.indd 28 10/19/20 6:22 PM