Solution Manual for Interactive Statistics, 3rd Edition

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Solutions to All Exercises275SectionIII.SolutionstoallChapter Exercises

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276Chapter 1Interactive Statistics3rdEdition: Chapter 1FullSolutions1.1In hypothesis testing, the purpose is to determine whether there is sufficient evidence with which to reject the nullhypothesis (H0), which generally reflects the prevailing viewpoint. The alternative hypothesis (H1) is often whatsomeoneis hopeful that the data will support.1.2(a)True.(b)True.(c)False.(d)False.1.3H0: The 5-year survival rate forallthose using the vaccine is equal to 10%.H1: The 5-year survival rate forallthose using the vaccine is greater than 10%.1.4H0: Using the new technique, the percentage ofallwhales leaving the area is 40%H1: Using the new technique, the percentage ofallwhales leaving the area is more than 40%1.5(a)Since a Type I error is rejectingH0whenH0is true, we would be concluding the gun is not loaded when it isloaded.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the gun is loadedwhen it is not. A Type I error may be more serious as one might accidentally shoot a loaded gun.(b)Since a Type I error is rejectingH0whenH0is true, we would be concluding the dog does not bite when it doesbite.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the dog bites when itdoes not. A Type I error may be more serious as we might approach a dog that could bite.(c)Since a Type I error is rejectingH0whenH0is true, we would be concluding the mall is closed when it is open.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the mall is open when it isclosed. A Type II error may be more serious as we might waste the time and gas to drive to the mall expectingit open when it is closed.(d)Since a Type I error is rejectingH0whenH0is true, we would be concluding the watch is waterproof when it isnot.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the watch is notwaterproof when it is waterproof. A Type I error may be more serious as we might ruin our watch if it gets wet.1.6(a)Since a Type I error is rejectingH0whenH0is true, we would be concluding the electricity is not turned onwhen it is.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the electricity isturned on when it is not. A Type I error may be more serious as one might accidentally be electrocuted.(b)Since a Type I error is rejectingH0whenH0is true, we would be concluding the brakes are operational whenthey are not.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the brakes arenot operational when they are operational. A Type I error may be more serious as it may result in an accident.(c)Since a Type I error is rejectingH0whenH0is true, we would be concluding the snake is not poisonous when itis poisonous.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking the snake ispoisonous when it is not poisonous.A Type I error may be more serious as it may result in getting bit by apoisonous snake.(d)Since a Type I error is rejectingH0whenH0is true, we would be concluding it is not safe to cross the streetwhen it is safe.Since a Type II error isfailing to rejectH0whenH1is true, we would be thinking it is safe tocross the street when it is not safe. A Type II error may be more serious as it may result in an accident.

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Solutions to All Exercises2771.7H0:The average tomato yield for Brand A fertilizer is the same as the averagetomatoyield for the more expensiveBrand B fertilizer.H1: The average tomato yield for the more expensive Brand B fertilizer is greater than the average tomato yield forthe Brand A fertilizer.Type I Error:Spend more money on the Brand B fertilizer when it really is not better than the Brand A fertilizerregarding the average tomato yield.Type II Error:Continue to use the Brand A fertilizer when the Brand Bfertilizer results in a higher tomato yield on average.1.8(a)Type I error: to classify a person as a drug user when he/she is not. Type II error: to classify a person as a non-drug user when he/she actually is.(b)Since the test classifies the person as a drug user (i.e. rejectsH0) 4% of the time when the person is not a druguser (i.e.H0is true), the chance of a Type I error is 4%.1.9A Type I error is rejecting the null hypothesis when it is true.So the owner would conclude the patrons are olderand the owner would spend the time and money to remodel, when the crowd is actually not older. The owner wouldhave spent money unnecessarily and the remodeling may not appeal to some of the patrons, but in general, it is not aserious error.1.10Based on the article "U.S. Health Improves but Rural Areas Lag,"from CNN,September10,2001. Summary of thearticle appears below.Contains many possible hypotheses.H0:People in Rural areas of the US are as healthy as their urban counterparts.H1:People in Rural areas of the US are less healthy than their urban counterparts.Type I error:concludethatrural people are less healthy than urban peoplewhen in fact they are not.Type II error:concludethatrural people are as healthy as urban people when in fact it isn't the case.Summary of article: "Americans overall are healthier today than they were 25 years ago. A new government reportoffers some reasons: longer life expectancy, better infant survival, fewer smokers, less hypertension and lowercholesterol levels. But in small-town America, the news on health is far from good, said an annual report releasedMonday by the Centers for Disease Control and Prevention. Rural residents tend to smoke more, lose more teeth asthey age and die sooner than suburban and many big-city counterparts, the government snapshot of the country'shealth shows. For instance:--10.6 percent of the wealthiest residents in rural areas and 10 percent of urban residentslacked health insurance in 1997 and 1998, compared with about 6.6 percent of suburbanites.--37.6 percent of ruralresidents over 65 had edentulism, a total loss of teeth, in 1997 and 1998, compared with about 25.7 percent in thesuburbs and 26.8 percent in cities.--18.9 percent of children age 12 to 17 in the most rural areas were regularsmokers in 1999, compared with 11 percent in urban areas and 15.9 percent in the suburbs. Rural adults also smokedat higher rates than urban or suburban adults.--46.5 of men and women in the most rural areas did not exercise, playsports or pursue active hobbies in 1997 and 1998, compared with 40.9 percent of urban dwellers and 31.1 percent ofsuburbanites who were not fitness-minded. The youth death rate from all causes was higher in rural areas from 1996to 1998, as was the adult death rate.1.11(a)The null hypothesiscould not be rejected.(b)No, a complaint was not registered.(c)Yes, a Type II error may have been made.The cans are thought to containthe stated sodium content whenactually they contain higher amounts of sodium on average.1.12(a)H0: Septaphine is not better than Cephaline for reducing blood pressure.H1: Septaphine is better than Cephaline for reducing blood pressure.(b)(i)Thenull hypothesis was rejected and thealternative hypothesis wassupported.(ii)A Type I error could have been made, namely, concluding Septaphine is better when it really is not better atreducing blood pressure.

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278Chapter 11.13Ifis decreased thenwill increase, so the possible value is 0.30.1.14If you do not reject the null hypothesis, you may be making either a Type IIerror ora correct decision, so theanswer is (e).1.15(a)The significance levelis 2/30=0.067 and the level ofis 20/30=0.667.(b)Decision Rule #2: RejectH0if the selected voucher is$2 or is$9. The significance levelis 6/30=0.20and the level ofis 12/30=0.40. Enlarging the rejection region resulted in increasing the level offrom 0.067to 0.20 while decreasing the level offrom 0.667 to 0.40.1.16(a)= chance of a Type I error = chance of rejectingH0whenH0is true= chance of observing $1 or $10 from bag E= 2/30 = 1/15 = 0.0667(b)p-value = chance of observing $3 or more extreme underH0= chance of observing$3 or$8 from bag E= 12/30 = 6/15 = 0.40(c) (i)= chance of a Type II error = chance offailing to rejectH0whenH1is true= chance of observing $2-$9 from bag G= 23/30 ~ 0.767(ii)= chance of a Type II error = chance offailing to rejectH0whenH1is true= chance of observing $2-$9 from bag H= 23/30 ~ 0.7671.17No, we need a decision rule that states when we reject orfail to rejectH0.1.18(a)H0: The proportion of newborns that are girls equals 0.50.H1: The proportion of newborns that are girls does not equal 0.50.(b)Two-sided.1.19(a)False:+does not need to equal 1. The value ofis calculated underH0while the value ofis calculatedunderH1.(b)False: Type II error is the chance offailingto rejectH0whenH1is true.(c)True.(d)False:H0is rejected if the sample shows evidence against it.(e)False: The sample size does not influence the alternative hypothesis.The alternative hypothesiscan be one-sided no matter what the samplesize.1.20The 0.05 indicates: (b) ifH0is true, the chance of falsely rejecting it is 0.05.1.21(a)H0: The shown box is Box A.H1: The shown box is Box B.(b)The direction of extreme is one-sided to the left.(c)RejectH0if the selected token is $5 or less.(d)The significance level= 2/25 = 0.08 which is less than 0.10.(e)The chance of a Type II error is= 11/25 = 0.44.(f)Our decision is to rejectH0.

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Solutions to All Exercises2791.22(a)The frequency plots are given below. The direction of extreme is two-sided.Distribution in Bag AXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX24681012141618Token Value (in $)Distribution in Bag BXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX24681012141618Token Value (in $)(b)H0: The shown bag is Bag A.H1: The shown bag is Bag B.(c)Answers may vary. One reasonable rule is to rejectH0is the selected token is$4 or$16.(d)The significance level= 6/50 = 0.12.(e)The chance of a Type II error is= 16/50 = 0.32.(f)(i)Fail to rejectH0.(ii)A Type II error.(g)(i)RejectH0.(ii)A Type I error.1.23(a)False.+does not need to equal 1. The value ofis calculated underH0while the value ofis calculatedunderH1.(b)False.AType II error occurs ifH1is true and wefail torejectH0. Here we don’t know ifH1is true.(c)True.

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280Chapter 11.24(a)H0: The shown jar is Jar A.H1: The shown jar is Jar B.(b)No, the coins are of different sizes and weights. If we mixed up the contents of the shown jar and reached in topick one coin, the coins that we are more likely to pick might be a quarter as it has a larger size as compared tosay a dime. The larger coins may be more likely to be selected over the smaller coins. This will be referred tolater in Chapter 2 as a length-biased sampling method.(c)The responses are currently listed as P, N, D, and Q, but there is an ordering to them with respect to their value.So a better way to record the response would be using their worth of 1, 5, 10, or 25 cents.Correspondingly, abetter picture to depict the two models for the shown jar is given below. The direction of extreme would be tothe right, to the larger coin values.Distribution in Jar AXXXXXXXXXXXXXXXXXXXX12345678910111213141516171819202122232425Coin Value (in cents)Distribution in Jar BXXXXXXXXXXXXXXXXXXXX12345678910111213141516171819202122232425Coin Value (in cents)1.25(a)The frequency plots are provided below:Bag XBag YXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXBlueBrownYellowGreenRedBlueBrownYellowGreenRed(b)No, the response being recorded is the color, which has no particular ordering for the outcomes.If the colorshad been listed in the order of Blue, Red, Brown, Green, and Yellow, then the apparent direction of extremewould be one-sided to the left. So it is not appropriate to discuss a direction of extreme in this case.

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Solutions to All Exercises2811.26Thep-value is a number between 0 and 1 which measures how likely the observed result is, or a result that is evenmore extreme (in the direction ofH0), assuming the null hypothesisH0is true.1.27Thep-value should be small in order to reject the null hypothesisH0.A smallp-value indicates that the observeddata or data even more extreme is very unlikely or unusual if the null hypothesis is true. In general, we rejectH0ifp-value is less than or equal to, the significance level.1.28(a)It is statedthere was no significant difference in the caesarean delivery rates between the two groups, so thedata supported the null hypothesisH0.(b) Thep-value would have been "large".Since the null hypothesis wasnot rejected, the observed data were notconsidered unlikely under the null hypothesis.1.29(a) Frequency plots for the two completing hypotheses.(b)= chance of a Type I error = chance of rejectingH0whenH0is true= chance of observing $1 from the winning bag= 2/10 = 0.20(c)= chance of a Type II error = chance offailing to rejectH0whenH1is true= chance of observing $10 or $100 fromthe losing bag= 3/8 =0.375(d)No, we did not actually observe a voucher.1.30(a)The direction of extreme is one-sided to the left, since the smaller values are more likely underH1and lesslikely underH0.(b)The chance of a Type I error is=3/15 = 0.20 (found under the Machine A model for 2 or fewer flaws). Thechance of a Type II error is= 6/15 = 0.40 (found under the Machine B model for more than 2 flaws).(c)Thep-value is the chance of getting the observed 4 flaws or something evenmore extreme (less than 4 flaws),assuming the null hypothesis is true and the machine is Machine A. Thep-value = 10/15 = 0.667.(d)Since the data were not usually assuming the null hypothesis is true, that is, thep-value is larger thanthe dataarenot statistically significant at the level

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282Chapter 11.31(a)The direction of extreme is one-sided to the left (to the smaller values).(b)The chance of a Type I error is= 1/6 = 0.1667 (found under the Die A model for 1 or less). The chance of aType II error is= 7/10 = 0.70 (found under the Die B model for2 ormore).(c)Thep-value is the chance of getting the observed value of 2 or less, assuming the die is Die A, so thep-value=2/6 = 0.333.Since thisp-value is greater than the significance levelof 0.1667, wefail torejectH0andconclude thattheselected die appears to be Die A.1.32(a)The direction of extreme is one-sided to the left. Note that the smallervalues show the most support forH1and the least support forH0.(b)See pictures at the right for the circling and labeling.i.= 3/35 = 0.0857.ii.=14/35 = 0.40.(c)p-value = 7/35 = 0.20(d)The observed result is not statistically significant because thep-valueis more than.(e)A new decision rule is rejectH0if the observed voucher value is $6 ormore extreme, that is less than or equal to $6. Using a cut-off value of$6 or more will yield a larger rejection region and thus a larger valuefor. The new= 7/35 = 0.20 is larger than= 3/35 = 0.0857.1.33(a)H0: The shownbag is Bag A.H1: The shown bag is Bag B.(b)The direction of extreme is two-sided.(c)(i)Thep-value is10.0404=.(ii)Yes, since thep-value is.(iii)No, since thep-value is >.(d)(i)Thep-value is 1.(ii)No, since thep-value is >.(iii)No, since thep-value is >.1.34(a)Two possiblep-values that are statistically significant at 0.01are 0.002 and 0.004. Note that any two values 0 ≤p-value ≤ 0.01 will work.(b)Two possiblep-values thatare statistically significant at 0.05, but not statistically significant at 0.01 are0.03 and 0.04. Note that any two values 0.01 <p-value ≤ 0.05 will work.(c)Two possiblep-values that are not statistically significant at 0.10 are 0.20 and 0.30. Notethat any two valuesp-value > 0.10 but not >1 will work.1.35(a)All new model 100-watt light bulbs produced at Claude's plant.(b)H0: The population of all new model 100-watt light bulbs (produced at Claude's plant) has an average lifetimeequal to 40 hours.H1: The population of all new model 100-watt light bulbs (produced atClaude’splant) hasan average lifetime greater than 40 hours.(c)10%(d)Thep-value can be any value between 0 and 0.10.(e)Yes, if thep-value is less than or equal to 0.10, then thep-value is also less than or equal to 0.15 so it issignificant at the 0.15 level. However, if we only know that thep-value is less than or equal to 0.10, we cannotbe sure whether thep-value is also less than or equal to 0.05. Without further information about the value of thep-value we cannot determine if the data would also be significant at the 0.05 level.Box IXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX24681012$Box IIXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX24681012$p-value

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Solutions to All Exercises2831.36(a)H0: The shown bag is Bag A.H1: The shown bag is Bag B.(b)The direction of extreme is one-sided to the right, to the larger values.(c)Thep-value is06.0503=.(d)Yes, the $14 voucher is statistically significant at the 10% level as thep-value is less than or equal to 0.10.(e)No, the $14 voucher is not statistically significant at the 5% level as thep-value is more than 0.05.1.37(a)H0(b)p-value > 0.10(c)Wefailed to rejectH0, so we could have made a Type II error.(d)One-sided to the right. We want to see if the numbers have increased.1.38(a)Since the data were statistically significant, the null hypothesis was rejected and the alternative hypothesis wassupported.(b)The direction of extreme is two-sided.(c)Answers will vary. Two possible values are 0.02 and 0.03.Note that any two values between 0 and 0.05 willwork.1.39(a)We rejectedH0, so we could have made a Type I error.(b)We decide that the average cost is higher than $350 while it is really not. Maybe you decide that the cost is toohigh and decide to attend a different college, while in reality you could have attended this college after all.(c)Thep-value is0.10.(d)Yes.1.40(a)H0: The population of all goats born to mother goats that were trained to walk on a treadmill has a mean birthweight of 1600 grams.H1: The population of all goats born to mother goats that were trained to walk on atreadmill has a mean birth weight different from 1600 grams.(b)The null hypothesis was supported at the 1% level.(c)Thep-value was more than 0.01 (but of course less than 1).(d)Answers will vary. One possiblep-value is 0.08.(e)Yes, as now thep-value must be larger than 0.01 but less than or equal to 0.05.A value that will satisfy thisstatement on significance is 0.04.(f)A Type II error could have been made.1.41(a)For study A,a possiblep-value is 0.001;for study B, a possiblep-value is 0.11;and for study C, a possiblep-value is 0.03.(b)RejectH0if thep-value is small, so support forH0is shown if thep-value is large, the largestp-value is forStudy B.(c)We rejectedH0, but it was true, so a Type I errorwasmade.(d)For Study A: one-sided to the right, Study B: two-sided, Study C: one-sided to the left.1.42(a)One-sided to the right.(b)H0(c)Thep-value is0.05.(d)Type I error.

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284Chapter 11.43(a)See the chart below for the alternative hypotheses.(b)See the chart below for the possiblep-values.(c)The results for Study C had the most support for the null hypothesis since thep-value was the largest.(d)This would be called a Type I error.1.44(a)The alternative hypothesis could be stated in words asH1: The proportion of all convicted persons who havebeen through the program and later reconvicted is lower than the national proportion of convicted persons whoare released and later reconvicted.(b)Thep-value must be greater than 0.05 but less than or equal to 0.10. Thus there is one possible value of 0.06.1.45(a)True.(b)False.1.46(a)False. Thep-value is the chance of getting the observed value or values more extremeassumingthatH0is true.(b)True. Thep-value of 0.04 was less than the significance level of 0.05, so the decision is to rejectH0.(c)True. When we rejectH0, we say that the results are statistically significant.(d)True.If the results are statistically significant at the 0.05 level (i.e. thep-value was less than 0.05), then theresults will also be significant at the 0.10 level (as thep-value will also be less than 0.10).1.47(c) to be statistically significant at the 5% level means thep-value is less than or equal to 0.05. However, we do notknow if thep-value is less than or equal to 0.01 or if it is between 0.01 and 0.05, so the answer is "sometimes yes"(if thep-value is also0.01) and "sometimes no" (if thep-value is > 0.01).

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Solutions to All Exercises2851.48(a)Type I error: The subject does not haveESP but you conclude he does.(b)Type II error: The subject has ESP but you conclude he does not.(c)Level of significance is the chance of rejecting Ho when Ho is true. If the subject does not have ESP (i.e.,H0istrue) then the chance that he correctly identifies the card (and we rejectH0) is 1 in 52, 1/52 = 0.019.(d)The chance of a Type II error is the chance wefail to rejectH0whenH0is true. If the subject has perfect ESP,then the chance he correctly identifies the card is 1 or 100%.He will correctly identify it, so the chance of aType II error is 0.(e)Thep-value calculates the chance of getting the observed result (here 0 correct answers) or something moreextreme, in the direction ofH0(namely, 1 correct answer), under the assumption thatH0is true (the person doesnot have ESP). If the subject does not have ESP, the chance of incorrectly identifying the card is 51/52 = 0.981and the chance of correctly identifying it is 1/52 = 0.019. So thep-value is 51/52 + 1/52 = 1.(f)No, the null hypothesis is either true or false (correct or not).The individual either has ESP or does not.Thep-value depends on the sample.1.49(a)You observed a yellow ball, which is the most extreme result that you could get.With only one observationthere is no more extreme than observing a yellow. So thep-value is the chance of observing a yellow ball underthe null hypothesis, which is 1/5 = 0.20.Since thep-value is larger than 10%, the result is not statisticallysignificant.(b)The data consists of selecting two balls with replacement (and the order is not important).The possibleoutcomes are shown in the picture below:We observed one yellow and one blue ball. Results that are even more extreme would be observing two yellowballs.So thep-value is (1+8)/25 = 0.36.Since thep-value is larger than 10%, the result is not statisticallysignificant.1.50(a)H0: The percentage of Republicans who are in favor of the death penalty is equal to the percentage ofDemocrats who are in favor of the death penalty.H1: The percentage of Republicans who are in favor of the death penalty is larger than the percentage ofDemocrats who are in favor of the death penalty.(b)It would be a one-sided rejection region since we are looking for a particular direction (larger than), and not justany difference.(c)The result could be statistically significant if thep-value were.(d)The difference of 2% would probably not be practically significant.

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286Chapter 11.51Statement (a) since the effect is small so we would need a larger sample size to detect it.1.52The significance level is set at 0.05.Forn=1: the decision rule is rejectH0if the result is$60 and thecorrespondingis 12/20 = 0.60. Forn=2: the decision rule is rejectH0if the result is$45 and the correspondingis 53/190 = 0.28. As the sample size is increased, the level ofdecreased from 0.60 to 0.28.1.53(a)(i)= chance of observing an average45 from bag A = 4+2+1/190 = 0.0368(ii)45(iii)= 0.2789(iv)The chance that we decide that the bag shown is bag A, while it really is bag B is 0.28 (or 28%).(b)(i)First note that the observed average voucher value is $35, so we have:p-value= chance of observing anaverage of $35 or more extreme underH0= 17+9+4+2+1/190 = 0.1737(ii)p-valueso we rejectH0(iii)20+17+9+4+2+1/190 = 0.2789(iv)p-value >so wefail to rejectH0(v)The cut-off value is the average of $35, since we rejectedH0for $35 and wefailed to rejectH0for $30.1.54(a)The first table shows the possible samples, possible averages, and their frequencies of occurring.Possible SamplesAverageFrequency if Bag CFrequency if Bag D1,110101,21.52201,323151,42.54101,53552,22162,32.56122,43882,53.51043,33333,43.51263,541534,44614,54.52025,55100The next table combines the entries in the first table according to the different possible averages.AverageFrequency if Bag CFrequency if Bag D10101.522024212.51022316163.5221042144.52025100Total105105The corresponding frequency plots are given next.

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Solutions to All Exercises287xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1.54.53.52.52543Possible AveragesBag Axxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1.54.53.52.52543Possible AveragesBag B11Bag CBag D(b)For the decision rule to reject the null hypothesis if the average is$2, we have the significance levelis6/105=0.057 and the level ofis 54/105=0.514.(c)If you observe an average of $3, thep-value would be the chance of observing $3 or less, which isp-value =32/105 = 0.305.1.55(a)20 nCr 2 = 190(b)20 nCr 3 = 11401.56(a)The height must be 1/(base) = 1/7 or 0.1429.(b)(i)= (1)(1/7) =1/7 = 0.1429, as represented by the light grey shaded rectangle from 1 and to the left undertheH0density.(ii)=0.25 + 0.15 + 0.08 + 0.05 + 0.04 + 0.03 = 0.60 (or from 1–0.40), as represented by the area to the rightof 1 under theH1density.(iii)Power = 1–= 1–0.60 = 0.40, as represented by the area to the left of 1 under theH0density.(c)Thep-value is (0.5)(1/7) = 0.0714, as represented by the black striped rectangle from 0.5 and to the left undertheH0density.(d)(ii) Increasing the sample size would lead to a reduction in the chance of committing a Type II error.(e)False.−=powerp-value

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288Chapter 11.57(a)The sketches are provided below.1_41_4p-value = area to the left of 4.1(b)(i)See the shadedarea marked asabove.(ii)= (0.4)(1/4) = 0.1(iii)See the shadedarea marked asabove.(iv)Power = 1–= 1–(0.6)(1/4) = 1–(6/10)(1/4) = 1-0.15 = 0.85.(c)(i)See the shadedarea marked asthep-valueabove.(ii)p-value =(1.1)(1/4) = 0.275.(d)No, thep-value > 0.1 =.1_41_4

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Solutions to All Exercises289Interactive Statistics3rdEdition: Chapter 2FullSolutions2.135% is a parameter–itis a numerical summary of the population.28% is a statistic–itis a numerical summary of a sample from the population.2.2(a)The value of 9% is a parameter–it is a numerical summary of the population.(b)The value of 12.5% is a statistics–it is a numerical summary of a sample from the population.2.3The proportion of register voters in Ann Arbor who would vote “Yes” on Proposal A is an example of (c) aparameter.2.4(a)The population consists of (the planned vote for) the 100 U.S. Senators.(b)N= 100.(c)The sample consists of (the planned vote for) the ten selected U.S. Senators.(d)n= 10.2.5For a simple random sample ofn= 200 and the number of items that were defective = 5. All we can say is that (c)the percent of defective items in the sample is 5/200 = 2.5%.p=5/200 = 0.025 or 2.5%.2.6(a)22.4.(b)19. No.(c)22.5. No. No.(d)18 and 21. The sample mean age is 19.5.18 and 26. The sample mean age is 22.20 and 27. The sample mean age is 23.5.20 and 21. The sample mean age is 20.5.20 and 26. The sample mean age is 23.27 and 21. The sample mean age is 24.27 and 26. The sample mean age is 26.5.2.7(a)The value of 34 is a statistic.(b)Response bias.2.8(a)Statistic since the 54% is a sample percentage, not of the population.(b)Nonresponse bias.2.9Response bias.2.10Nonresponse bias.2.11Nonresponse bias.

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290Chapter 22.12False.When only 272 out of 1000 people respond, for a 27.2% response rate, thisresults innonresponse bias.2.13The best answer is (c).2.14Results will vary.(a)Using the calculator with a seed value of 291, the selected persons are 39 (79 years old), person 3 (75 years old)and person 24 (70 years old). The average age is 74.67 years.(b)22 (36 years old), 34 (89 years old) and 29 (89 years old). The average age is 71.33 years, different from themean calculated in part (a).(c)69.19 years.(d)74.67 and 71.33 are statistics, while the mean in part (c) is a parameter.2.15(a)Yes, each sample of size 20 has the same chance as any other sample of size 20 to be selected (assuming all tagsare exactly the same and the box is thoroughly mixed).(b)It is drawn without replacement.2.16(a)12/100.(b)With the graphing calculator the selected employees are: 77, 51, 72, 40, 71, 42, 17, 34, 62, 23, 35,and 12.From the random table the selected employees are:7, 5, 69, 76, 28, 33, 78, 70, 99, 98, 42,and 80.2.17(a)H0: Thepopulationproportion of dissatisfied customers equals 0.10.H1:Thepopulationproportion of dissatisfied customers is less than 0.10.(b)With the calculator the selected customers are: 34318, 15553, 8461,and614.With the random table theselected customers are: 15409, 23336, 29490,and 30414.(c)Type II error.(d)0.21.(e)No, thep-value is > 0.05.(f)(ii)Statistic.2.18Using the TI: Label the sites from 1 to 80. With aseed value of 29the five labels, and thus sites, selected at randomare as follows:Sites #50, #66, #43, #49,and #74.Using the Random Number Table:Since there are more than 10,but less than 100, sites, we can use double-digit labels, such as01,02,03,through80.WithRow 10 of the table,starting at Column 1,thesimple random sample of sites consists of sites #47, #53, #68, #57, and #34.2.19Stratified random sampling.You are dividing up your population by class rank then selecting 100 students fromwithin eachstratumat random.2.20(a)Stratifying by declared major (field of study) would be a good stratification variable.The cost of textbooks islikely to vary from one field of study to the next, but not vary as much within a particular field of study (manystudents have to take the same basic classes for their field of study and hence may be paying similar textbookcosts.) You would also learn about the costs for each field of major as well as costs overall.(b)Stratifying by gender may not be as useful as a stratification variable. The cost of textbooks may vary a lot forfemales and vary a lot for males, and there may not be much variation between males and females. However, ifyou wanted to learn about the costs for each gender as well as overall, you might wish to stratify by gender.(c)Stratifying by class rank may be useful as a stratification variable.The cost of textbooks may vary from oneclass-rank to another.However, there may be quite a lot of variation within each class-rank as well.If youwanted to learn about the costs for each class-rank as well as overall, you might wish to stratify by class-rank.2.21

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Solutions to All Exercises291(a)45 8th-grade students; 25 10th-grade students; 48 12th-grade students. For a total in the sample of 118 students.(b)Withthe calculator the selected students are: 193, 127, 430, 100,and 427.With the random number table theselected students are: 241, 304, 22, 364,and 151.2.22(a)100/200 = 1/2 = 0.50 or 50%.(b)100/1000 = 1/10 = 0.10 or 10%.(c)The chance of being chosen is 0.10. This is NOT a simple random sample. For this stratified random samplingplan each possible sample would contain exactly 100 males and 20 females.All samples of size 120 are notequally likely (as it should be for simple random sample). Some samples of size 120 are not even possible, forexample, having 120 males.2.23We should use a weighted average:4010070()+6010063()=65.8 inches2.24(a)20/200 = 0.10.(b)With the calculator the first five females selected in the sample are: 132, 195, 147, 171,and 85.With therandom number table: To each woman we assigned 5 numbers, for example 1,201,401,601 and 801.Thenumbers from the table were 521, 625, 391, 646,and 369.So the first five selected females are: 121, 25, 191,46,and 169.(c)The weighted average is: (0.75)8 + (0.25)5 = 7.25.2.25(a)Stratified random sampling.(b)With the calculator the first five selected homes are: 386, 81, 379, 211,and 156. With the random number tablethe first five selected homes are: 94, 299, 396, 378,and 363.(c)(0.60)(2100) + (0.40)(2600) =2300 square feet.2.26(0.20)(16) + (0.50)(43) + (0.30)(71) = 46 years.2.27(a)Stratified random sampling.(b)With the calculator the first six homes selected from the 1000 homes in County I are: 918, 193, 902, 502, 370,and5.Withtherandom number table the first six homes selected from the 1000 homes in County I are:963, 19,197, 705, 463,and79where the homes are labeled 0 to 999.(c)(0.50)(175) + (0.30)(200) + (0.20)(195) = 186.5 thousands of dollars or $186,500.2.28(a)With the calculator, the label of the first student selected is 3.With the random number table, the label of thefirst student selected is 3.(b)The students in the sample are those with ID numbers 3, (3 + 4 =) 7, (7 + 4 =) 11, and (11 + 4=) 15, for a totalof 4 students.2.29(a)Withthe calculator with the first 100 addresses labeled 1 through 100, the sample is: Addresses =#79, #179,#279, #379, #479.Usingthe random number tablewith the first 100 addresses labeled 01, 02, ... , 98, 99, 00,the sample is: Addresses = #30, #130, #230, #330, #430.(b)1/100 = 0.01or 1% There are 100 possible systematic samples of size 5, each equally likely.

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292Chapter 22.30(a)Stratified random sampling.(b)No, if there are more people whose family name begins with say A and fewer people whose family name beginswith Z, then those with Z will have a higher chance of being selected.(c)Selection bias,a systematic tendency to exclude those with unlisted phone numbers.2.31Yes, the chance is 1/5 or 0.20 or 20%. There are 5 possible systematic samples, each equally likely.2.32(a)1/14 = 0.0714.(b)Note that there are555 members so555/14→39.64or39groups of14and onelast group of9.For the 17thmemberto be selected, the starting point must have been a 3, that is, the 3rdmember in each group of 14 isselected. This will result in a sample of40 members.2.33(a)1/8 = 0.125.(b)Note that there are350 members so350/8→43.75or43groups of8andonelast group of6.For the 300thmember to be selected, the starting point must have been a 4, that is, the 4thmember in each group of 8 isselected. This will result in a sample of44 members.2.34(a)1/40 = 0.025.(b)Using the TI calculator with seed of 19, and with the first 40 members labeled 1 through 40, the first fivemembers are 7, 47, 87, 127,and 167. Using row 15, column 1, andlabelingthe first 40 membersas01, 02, ... ,39, 40, the first 5 members are 7, 47, 87, 127,and167.(c)Note that there are2220 members so2220/40→55.5or55groups of40andonelast group of20.Since thestarting pointwasa7, that is, the7thmember in each group of 40was selected. Thiswouldresult in acompletesample of56members.2.35False, the chance depends on the number of clusters.2.36(a)With the calculator, the two departments selected were 2 and 4; Chemistry and Mathematics.Using the randomnumber table, the two department selected were 2 and 3; Chemistry and Biology.(b)With calculator, the sample size is 80. With random number table, the sample size is 90.2.37(a)Cluster sampling.The faculty are grouped into departments which serve as clusters.Sixof the clusters areselected at random. All of the units in the cluster are in the sample.(b)Label the list of the 60 departments 1, 2, 3, ..., 60. The numbers generated and thus the departmentsselected using the TI calculator with seed = 79 are:54, 37, 49, 5, 15,and 43.If you are using the randomnumber table, you might label the list of the 60 departments 01, 02, 03, ... , 60. The numbers generated and thusthe departments selected starting at row 60, column 1 are: 23, 22, 47, 40, 25,and 37.(c)Yes we can determine the chance that any specific professor will be selected.The chance of selecting aprofessor is the same chance that his/her department or cluster will be selected. The reason being if a cluster isselected, every element in the cluster is selected.Therefore the probability that a cluster is selected is 6/60=0.10, thechance that any specific professor will be selected.

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Solutions to All Exercises2932.38(a)This is cluster sampling since the students are first divided into clusters (undergraduate classes). A class/clusteris then selected using a simple random sample and all students from that class/cluster are sampled.(b)No, since not all students take the same number of classes.Students who attend more classes have a greaterchance of being selected.(c)No it is not biased since the cluster was selected at random. It is a case of poor design together with bad luck.When clustering the variability between clusters should not be more important than the variability withinclusters. Here we have poor design because there might be more variability between a class with many studentson athletic scholarship and a class without any, than variability within each of those classes.2.39(a)The type of sampling performed in each dorm is cluster sampling, with the rooms forming the clusters and 3clusters were selected at randomfrom each of the four dorms.(b)If each cluster selected has one student, we would have 3 students from each of the four dorms for a minimumsample size of 12 students.(c)If each cluster selected has three students, we would have 9 students from each of the four dorms for amaximum sample size of 36 students.(d)We do not know, it will depend on how many clusters selected are rooms with women. The number of womencould be as low as 0 and as high as 36.(e)No, there will be anywhere from 3 to 9 freshmen students sampled from the freshmen dorm, as well as from 3to 9 sophomores, from 3 and 9 juniors, and from 3 to 9 seniors.2.40(a)Cluster sampling.(b)No, if the number of adults per city block is unknown.Yes, if you have the seed value and you know howmany adults are in each block.2.41(a)With the calculator or the random number table, the selected region is 3 = Southwest.(b)Stratified random sampling.(c)(i)1-in-10 systematic sampling.(ii)0.10.(iii)With the calculator, the first five selected cans are 7, 17, 27, 37, and 47. With the random number table wemight label the first can 1, the second can 2, …, and the 10thcan 0. Then the first five selected cans are 7,17,27,37, and47.(iv)Note that 125/10 = 12.5→12 or 13 cans. However, there will not be a 7thcan to select in last group. Thusthe total number of cans in the sample will be 12.(d)(i)No.(ii)Two possible values are 0.12 and 0.15.(iii)Yes, a Type II error.2.42(a)(i)Convenience sampling.(ii)Yes, a selection bias.(iii)The calculated average is expected to be higher than the true average, as all of the books in the sample havealready been checked out at least once, and may include some of the more popular books.(b)Cluster sampling.(c)(i)Stratified random sampling.(ii)Overall estimate:(40/1200)(20) + (200/1200)(15) + (600/1200)(10) = 14.2 times checked out.(d)For each of the three categories of books the following stages are followed.Stage 1:Divide the books into clusters according to the last digit of the call number (0 through 9).Take asimple random sample of 3 digits from the list of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.The clusters of books (in that category) with call numbers ending with those selected digits are selected.Stage 2:Within each of the selected clusters of books from Stage 2, select a simple random sample of 7 books.Note that with this multistage sampling plan, we will have a total of 3 categories x 3 clusters x 7 books = 63books.2.43

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294Chapter 2(a)A cluster sampling of blocks.(b)Yes, the chance is 5/50 = 0.10.(c)No, we do not know how many households are in each block.Additionally,if the number of households ineach block was not the same across all blocks, then we would also need to know which of the 5 blocks wereselected.(d)With the calculator, the five selectedblocks are8, 44, 33, 6, and 38.With the random number table we mightlabel theblocks 01 to 50, then the five selectedblocks are 12, 18, 11, 43, and 05.(e)Response bias because the interviewers are college students.People may not feel comfortable telling thesestudents they want to forbid loud music at parties in the college dorms.2.44Answer is (c)statisticis to a sample.2.45The proportion 153/200 is a parameter since the instructor polled her entire class and that was the group that she wasinterested in learning about.2.46(a)Population: Adults U.S. residents. The sample sizen= 1500.(b)Population: Today’s shipment of 1-gallon milk cartons. The sample sizen= 5.(c)Population: The 740 members of the local women’s business association. The sample sizen= 100.2.47Nonresponse bias.2.48(a)Selection bias, where the sampling frame is either incomplete (such as sampling only one dealership) orincorrect.(b)Nonresponse bias, where persons who do not respond to a survey (such as the lazy car owners) may havedifferent opinions to those who do.(c)Response bias, where respondents may have a tendency to lie (such as car dealerships that believe in lowermiles per gallon) or refuse to answer.2.49This survey may be subject to nonresponse bias.Only those alumni who respond and report their income will beincluded. Alumni who perhaps are currently unemployed or in a low paying position may elect not to respond.Therefore, the reported average income based on such a survey may be biased upwards--the average may be largerthan the actual average for all alumni.2.50(a)The sampling design is a simple random sample of size 227 taken from the cocaine users who called.(b)The population this sample is drawn from is cocaine users who called the National help line between Februaryand March.(c)The sample is not from the population of workers, but only from those who called the hotline, so the surveyresults do not generalize to the population of workers. Also, "more people" implies a comparison for which nodata were given.2.51(a)Withthe calculator the selected ID numbers are: 179, 2274,and 3327.With the random number table theselected ID numbers are: 2398, 2258,and 3540.(b)A statistic.2.52(a)True. If wefail to rejectthe null hypothesis, them the population consists of all males.(b)False.(c)True.2.53

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Solutions to All Exercises295(a)Based on her decision rule,Jane rejectsH0.(b)Since Jane rejected the null hypothesis, the data are statistically significant.(c)No, based on a sample of two $1 Jane can not be certain of which purse she has since both purses contain atleast two $1.(d)Type I error: rejectH0whenH0is true.(e)Simple random samples of size 2 from null purse:$11and $12,$11and $51,$11and $52$12and $51,$12and $52,$51and $52(f)Simple random samples of size 2 from alternative purse:$11and $12,$11and $13,$11and $14$12and $13,$12and $14,$13and $14(g)Thep-value is the chance of getting two $1 bills or more extreme (in the direction ofH0, but in this case, thereis no “more extreme”) ifH0is true. Thep-value is the chance of getting two $1 bills ifH0is true, i.e., 1/6.2.54You first need to label the 4000 signatures.Hopefully no one signed the petition more than one time. If you use acalculator or computer, the labels can simply be 1 to 4000. If you use the random number table, you could label thesignatures from 0001 to 4000.Using a seed value of say 29 with the calculator or row 120, column 11 of therandom number table, you can proceed to take a simple random sample of 400 signatures.2.55Stratified random sampling.You are dividing up your population by gender then selecting your sample within thestrata at random.2.56(a)Stratified random sampling.(b)75/416.(c)(0.20)(15/25) + (0.80)(60/75) = 0.76.(d)Statistic.2.57(a)All former university graduate students.(b)Stratified random sampling.(c)False.2.58(a)Stratified random sampling.(b)Selection bias.2.59(a)Stratified random sampling(b)The chance is 0, only 1 of the two fiction books (A, B) will be selected.(c)Two books that could be selected are Book A and Book C.(d)Another pair of books that could be selectedisBook A and Book D.(e)The chance that the total number of pages exceeds 800 is the same as the chance that the two selected books areBook B and Book D. There are 4 possible pairs of books that could be selected of which the (B, D) pair is 1, sothe chance is 1/4 = 0.25.2.60(a)Stratified random sampling.(b)High: 20 clients, Moderate: 125 clients, Low: 45 clients.(c)With the calculator the selected clients are: 163, 2196, 214, 2462, 740.With the random number table the selected clients are: 1887, 1209, 2294, 954,and 1869.

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296Chapter 22.61(a)With the calculator the first student selected is 1. With the random number table the first student selected is 1.(b)The sample size is 6.2.62(a)Systematic sampling (1-in-30).Once you select the sample you can determine the number of freshmen youselected. Also knowing the population list you can calculate the number of freshmen you would select for eachof the 30 possible samples (depending on the starting point.)(b)Stratified random sampling. Yes, you would have 25 freshmen.2.63(a)(i)0.02(500)+0.03(1200)+0.05(18000) = 10+36+900 = 946.(ii)Stratified Random Sampling.(iii)With the calculator the selected labels are 432, 232, 304, 412,and 372. With the random number table, wemight assign the first High category driver the labels 001 and 501.We would assign the second Highcategory driver 002 and 502. This assignment pattern would continue until the 500thHigh category driverwho would be assigned the labels 000 and 500.Reading off labels from row 60, column 1, we have: 789,191, 947, 423, and 632. This would correspond to selecting the 289th, 191st, 447th, 423rd, and132ndHighcategory drivers in the list of 500 High category drivers.(b)(i)With the calculator or the random number table, the first selected label is 15. Thus the selected labels willbe 15, 35, 55, 75, 95, and so on.(ii)Since the 500 High category drivers divide evenly into groups of 20 (500/20 = 25), there will be a total of25 High category drivers in the systematic 1-in-20 sample.2.64(a)907.(b)Cluster sample.(c)(iii) Selection bias.2.65These results are based on a study of 125 aerobic classes in five health clubs, not all aerobic classes in all healthclubs. Thus, the 60% figure is a statistic and the sample size isn= 125.2.66(a)It is a systematic 1-in-45 sampling resulting in 50 students (1 from each of the 50 sections, the 33rdin each ofthe 50 lists of 45 students).(b)It is a cluster sample and you cannot know the sample size (number of students selected) because we do notknow how many students are in the various major clusters.2.67(a)Cluster sampling.(b)1/5.(c)Response bias.2.68(a)False.(b)False.(c)True.2.69(a)A 1-in-40 systematic sample.(b)1/40 is the chance that any specific address is chosen since one of the first 40 addresses is picked at random.The other addresses are directly linked to that first random pick.

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Solutions to All Exercises2972.70(a)H0: The population mean increase in the GMAT score is 40 points.H1:The population mean increase in the GMAT score is less than 40 points.(d)With the calculator thefirst fourselectedstudents are: 209, 218, 7, and 750.Withthe random numbertable thefirst four selected students are: 070, 569, 762, and 833.(c)(i)A possiblep-value is 0.03.(ii)RejectH0.(iii)Type I error.(iv)Yes, since thep-value was≤ 0.05, it would also be ≤ 0.10.(d)(ii) Statistic.2.71(a)Multistage, with the first stage being a cluster sample of 3 lab sections and the second stage being a simplerandom sample of 25% of the students in the 3 selected labs.(b)0.0625.(c)The78% is a statistic since it was based on the sample of students surveyed.2.72Nonresponse bias is the distortion that can arise because a large number of units selectedfor the sample do notrespond or refuse to respond, and these nonresponders have atendency to be different from the responders.Sononresponse bias has to do withwhoresponds.Response bias is the distortion that can arise because the wording ofa question and thebehavior of the interviewer can affect the responses received.So response bias has to do withhowthe responders answer.2.73(a)Since 15 patients were selected with a 1-in-9 systematic sample, there were at least 15 groups of 9 patients eachor 135 patients in all.(b)(ii)Statistic.2.74(a)Sincethere is no prior information,we willsample in proportion to the size of the stratum relative to the size ofthe population.So the sample size from Stratum I is 3(since there are 10/40 large facilities in all, 25% of thesample could be large facilities, for 25% of 12 = 3), and the sample size from Stratum II is 12–3 or9.(b)Using the calculator, the stratum I selected facilities and responsesare:7 (Yes),10 (Yes),8 (No), and thestratum II selected facilities and responsesare:7 (No), 17 (No), 9 (Yes), 24 (No), 3 (No), 21 (No), 8 (Yes),12 (No),and1 (Yes),Using the random number tablewe will label the stratum Ifacilitiesas:1 has label 1, Facility 2 has label 2, ... ,Facility 10 has label 0.Then the selected facilities and responsesare:2 (Yes), 6 (Yes), 4 (Yes). We will labelthe stratum IIfacilities as:1 has label 01; Facility 2 has label 02;up toFacility 30 has label 30.Then theselected facilities and responsesare:05 (No), 36 (skip), 60 (skip), 42 (skip), 13 (Yes), 25 (Yes), 66 (skip), 92(skip), 64 (skip), 22 (Yes), ..., 04 (Yes), 06 (No), 20 (No), 12 (No), 18 (Yes).(c)Using the calculator we have:==estimateˆp416.0934030324010=+.Using the table we have:==estimateˆp667.0954030334010=+.(d)The true population proportion isp= 20/40 = 0.50.(e)In general, estimates may not be exactly equal top.Thecalculatorestimate was slightly too small, while therandom number table estimate was too large.2.75(a)Answers will vary.See the web site.(b)Summaries will vary.

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298Chapter 2

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Solutions to All Exercises299Interactive Statistics3rdEdition: Chapter 3FullSolutions3.1(a)Explanatory variable:Amount of coffee consumed.Response variable:Exam performance.(b)Explanatory variable:Hours of counseling per week.Response variable:Grade point average.(c)Explanatory variable:Number of roommates.Response variable:Physical health at the end of the semester.(d)Explanatory variable:Type of driver (levels are good or bad).Response variable:Reaction time on a driving test.3.2(a)A possible confounding variable is type of school (public versus parochial).(b)A possible confounding variable istype of sport of the subject (track–which are used to running, andwresting).(c)A possible confounding variable is type of job (modeling–which may have more fun, and temporary staff).3.3(a)Experiment.(b)The response variable is improvement in blood flow.The explanatory variable is the amount of flavonoid(High versus Placebo).(c)p-value ≤ 0.05(d)The 178.8 is a statistic and a sample mean.3.4(a)This was an observational study. The explanatory variable of hormone estrogen level was not actively imposedon the subjects, but rather simply recorded, along with the various qualityof sleep measurements.(b)The explanatory variable is hormone estrogen level. The general response variable is ‘sleep quality’ which wasmeasured and compared using different criteria---total sleep time, time spent awake during the night, andefficiency of sleep time (based on sleep stages and REM sleep).(c)The null hypothesiscouldnot be refuted, so thep-value must have been larger than 0.05. Two possible valuescould be 0.12 and 0.26.(d)This design technique is calledmatching (orpairing)and it was done to help reduce the effect of confoundingdue to age.(e)Some are: oxygen saturation level, muscle movement, hot flash patterns.3.5(a)Observational study.(b)Blood cholesterol level.(c)Frequency of egg consumption.(d)Diet, age, amount of exercise, and history of high cholesterol are a few possible confounding variables.3.6(a)Observational study.(b)Heart attack status.(c)Depression status.(d)Statistic.(e)Confounding variable.

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300Chapter 33.7(a)Observational study.(b)Homelessness.(c)Some are: separation status (whether or not they were separated from their parents), poverty Status (whether ornot they were raised in poverty), family problem status (whether or not they have family problems), abuse status(whether or not they were sexually or physically abused).(d)All homeless people in Los Angeles.(e)The homeless people in Los Angeles that were surveyed.(f)The 2.5% figure is a statistic since it is based on a sample of homeless people in Los Angeles.3.8(a)The response variable is payment status.(b)The explanatory variable is source of payment.(c)The levels of the explanatory variable are: private insurance, government, and other.(d)This is an observational study because the data were obtained from records.(e)(i)A statistic since it was computed from a sample of 120 private insurance records.(ii) Stratified random sampling.(iii) With the TI calculator we have: 377, 1138, 559, 850,and 1010.With the random number table we have: 705, 488, 448, 168,and 253.(f)The population is all hospital discharges in the U.S. or perhaps for hospitals in a certain state.(g)The sample consists of the 2000 records (120 private insurance records, 60 government records, and 20 “other”records) from the past two months (for the studied hospital.)3.9(a)This is a retrospective observational study.(b)The response variable is car crash status. The explanatory variable is gender.(c)We should take into account that men typically drive more than women. In this case, "miles driven" is aconfounding variable that can skew your results.Another confounding variable could be weather conditions atthe time of the crash.3.10This may not be valid. It is similar to trying to replace a "nonresponder" in a survey with a "responder". There canbe differences in these two groups, those who survive their suicide attempts perhaps had some uncertainty abouttheir interest, they may have been trying to get help.3.11(a)H1.(b)p-value was less or equal to 0.05.(c)(iii)Observational study, Prospective.3.12(a)This is an observational study and it is of the prospective type. A group of nurses was identified and followedup over time to ascertain whether they developed heart disease or not.(b)The response variable is heart disease status (whether or not the subject gets heart disease).The explanatoryvariable (risk factor) is the amount of stick margarine in their diet.3.13(a)Observational study (retrospective).(b)Response variable is kidney stones status (whether or not the subject develops kidney stones); Explanatoryvariable is amount of calcium in diet.(c)A diet high in calcium lowers the risk of developing kidney stones in men and women.

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Solutions to All Exercises3013.14(a)This is an observational study and it is of the retrospective type. Women with breast disease were identified andthen levels of PCBs measured.(b)These are called confounding variables.(c)Older women having been exposed for a longer period of time to PCBs might have a higher level of PCBs intheir breast tissue. It would be hard to tell whether it is age or level of PCBs that is increasing the incidence ofmalignant breast cancer.3.15(a)Observational study (retrospective).(b)Response variable is Alzheimer’s disease status, explanatory variable is linguistic ability.(c)The chance of low linguistic ability given a person has Alzheimer’s.3.16(a)This study is an observational study---observations were collected from the U.S. Census. The sample size wasvery large, approximately 3 million.(b)The response variable is divorce status (whether or not the couple hasa divorce) and the explanatory variable isgender of the child (or the number of girl children versus number of boy children).(c)With such a large sample size, we would expect the various groups being compared to be somewhat balancedwith respect to any other confounding or lurking variables. With such a large sample size, we are not surprisedto see the difference of 5% being statistically significant.Still, a statistically significant difference does notimply the explanatory variable caused the response.And in fact, Steven’s presents a possible third factorinvolving the concept of importance of inherited wealth that fits with the statistics and doesn’t rely on thepreference for boys as the causal variable.3.17(a)It is an experiment because the various treatment combinations were actively imposed on the experimentalunits.(b)The experimental units are the metal clutches.(c)The response variable is the lifetime of the clutch.(d)The factors or explanatory variables aretype of oven and temperature. We have4 types of oven and 3 levels oftemperature.(e)There are 12 treatments.(f)We would need24 clutches, 2 at each of the 12 treatments.(g)A design layout table for this experiment is given as:Factor 2:TemperatureFactor 1: Type of Oven2 clutchesgears2 clutches2 clutches2 clutches2 clutches2 clutchesType 1Type 2Type 3Temp 1Temp 2Type 42 clutches2 clutchesTemp 32 clutches2 clutches2 clutches2 clutches3.18Six treatment combinations, so(6)(4) = 24experimental units or pizzas.3.19(a)Durability.(b)Dye color and type of cloth.(c)20(d)20 x 6 = 120.

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302Chapter 33.20(a)Progress status.(b)Teacher and Method of teaching.(c)(Grade)(Teacher)(# students) = (5)(4)(6) = 120students.3.21(a)Batches of feed stock.(b)Yield of the process.(c)Temperature (2 levels) and Stirring Rate (3 levels).(d)6 treatments.(e)24 units.(f)Design Layout Table:3.22(a)The experimental units are the subjects who responded to a local questionnaire and who own other advancedelectronic equipment at home.(b)The response variables are: subject's recall of the ad, attitude toward the camera, and intention to purchase it.(c)Factor 1 is length (2 levels) and Factor 2 is frequency (3 levels).(d)The number of treatments is 6, obtained by multiplying the 2 levels for Factor 1 by the 3 levels for Factor 2.(e)One way to construct the design layout table is given below.Factor 1: Length30 sec90 secFactor 2:Frequency1353.23(a)Weight.(b)Fiber content and carbohydrate content.(c)Fiber levels (3) Low, Medium and High. Carbohydrate levels (2) Low and Medium.(d)3 x 2 x 20 = 120.3.24(a)Number of apple per tree.(b)Type of Fertilizer and Soil Moisture status.(c)Type of fertilizer: 3 levels. Soil moisture: two levels.(d)3x2x3 = 18.3.25Answer is (d).3.26The correct answer is (a), treatments are assigned to subjects.3.27

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Solutions to All Exercises303The rats are labeled 1 through 8. Use a calculator with seed= 1209 or a random number table with row=24,column=6 to select the 4 rats to receive the treatment. The other 4 rats will not receive the treatment. At the end ofthe week the effectiveness of the vaccine against the virus will be measured.3.28Assign a number to each dog. Rover=1, Spot=2, Jasmin=3, Chelsea=4, Klaus=5, Charlie=6, Maggie=7, Bo=8. Withthe calculator use the seed=27 to select the first 4 different numbers at random from 1-8. The dogs with thosenumbers will receive the treatment. The first four numbers are: 2, 1, 6,and5.With the random numbers table, thefour different numbers selected are: 8, 3, 4,and7.3.29(a)Experiment.(b)Cause of death.(c)Estrogen therapy with two levels: receive the treatment, do not receive the treatment.(d)Placebo.(e)H0: Women who receive Estrogen therapy do not have a lower risk of dying from a heart disease than womenwho do not receive the Estrogen therapyH1: Women who receive Estrogen therapy have a lower risk of dying from a heart disease than women who donot receive the Estrogen therapy.(f)Thep-value was less than or equal to 0.05.(g)Yes, Type I error.3.30(a)Here is one possibleallocation schemethat we will use: The first 10 randomlyselected subjects will beassigned to the aspirin group. The remaining 10, which donot have to be randomly selected because they are allthat are left, will be assignedto the placebo group.Another possible allocation scheme would be to alternate:The first randomly selected subject will be assigned to the aspirin group, the next randomly selected subject willbe assigned to the placebo group, and so on, until all subjects are assigned to one of the two groups. We will usethe first allocation scheme—the first 10 selected will be allocated to the treatment group.(b)Using thecalculator, the first 10 selected labels are:4,19,16,17,2,13,11,10,9,and5.So the subjects inGroup 1 (treatment group)are:Steve,James,Eric,Ralph,Kyle,Stan,Ed,Herb,Ryan,andPablo.Thus, theremaining subjects are assigned to the placebo group:Lee,Mark,Bill,Matt,Tim,Robb,Phil,Mike,Doug,andHenry.Using the table, the first 10 selected labels are:06,16,07,05,15,04,17,03,12,and19.So thesubjects in Group 1 (treatment group)are:Bill,Eric,Matt,Pablo,Mike,Steve,Ralph,Mark,Robb,andJames.Thus theremaining subjects are assigned to the placebo group:Lee,Kyle,Tim,Ryan,Herb,Ed,Stan,Phil,Doug,andHenry.3.31(a)With the calculator the selected rats were: 14, 13, 5, 16, 9, 3, 4, 2, 7,and 10. With the random number table theselected rats were: 4, 13, 8, 5, 12, 15, 2, 6, 9,and 20.(b)Have another researcher make and record the measurements.3.32(a)Completely randomized block design.(b)Paired design.(c)Randomized block design.3.33Blocking can help to reduce the bias due to possible confounding variables that you know of and thus build into thedesign of the experiment. Randomization can then be used to help reduce the bias due to other possible confoundingvariables that you do not know of or have not measured (sometimes these are referred to as lurking variables).

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304Chapter 33.34(a)A treatment was actively imposed.(b)With calculator: 2651, 4194, 4510, 6646, 754. With random numbers table: 3744, 6206, 6945, 4690, 405.(c)240.(d)Neither the doctor nor the patients know who receive each treatment.(e)(1)We don't know whether the result of the study would be published if the data were not favorable toCelebrex.(2)Maybe the other two drugs need more time to produce relieve.3.35Answers will vary.3.36Answers will vary.3.37Answers will vary.3.38Answers will vary.3.39Observational study. No active treatment was imposed.3.40For the TI calculator with seed 133, we have:Treatment 1: (2, 7); Treatment 2: (8, 4); Treatment 3: (3, 6); Treatment 4: (1, 5)For the random number table with row 18 and column 1, we have:Treatment 1: (1, 5); Treatment 2: (4, 2); Treatment 3: (3, 6); Treatment 4: (7, 8)3.41(a)Temperature (3 levels) and Baking Time (2 levels).(b)Taste.(c)6 treatments.(d)36 batches.3.42(a)There are 3 explanatory variables: time (3 levels), temperature (3 levels), and cheese(2 levels). So we have3x3x2 = 18 treatments.(b)With 5 batches for each treatment, we have 5x18 = 90 total units.3.43Lack of blinding of the subjects and of the experimenter(s) or evaluator(s).3.44(a)An experiment. A treatment was actively imposed.(b)Explanatory variable: Whether or not marigolds are planted withbroccoli.Response variable: nematode damage status.(c)Control for confounding variables.3.45(a)Experiment.(b)The response variable is skin rash status. The explanatory variable is skin rash treatment.(c)Using the calculator, the selected labels are: 10, 14, 23, 12, 11, 18, 3,25, 26, 22,27, 24, 13, 2,and4. Using therandom number table, the selected labels are: 09, 26, 06, 17, 13, 25, 18, 10, 16, 19, 24, 28, 14, 30, and 02.(d)The 20% value is a statistic.(e)Two possible values are 0.08 and 0.10. Any two values larger than 0.05 (but less than 1) would work.

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