Sample Examination Three-Answers Section I l. (E) The sum of the probabiiities must be one' 0.1 + 0.2+ 0.3+0'4+ 0 5 = 1'5 z. (c) ln the case of (E) the sttm is as for each increase in 1 second 0.8 minutes. The slope of 0. I 8 can be interpreted 0.18 rninutes interval ot-time in duratior-r of an eruption. there rviil be an increase of in duration would mean an the next eruptron. Therefbre, an increase of 60 seconds 1 until estimated increase of a -') . (c) 60(0'18)= 10'8 minutes' 62%.',rhe percent Variation in .r, explained by the least square regression of ,v on x isr2. Thevalueof rrcanbefoundbyusingtheformula'.b--rfi's"tuittgforrweget: ,=b r,\ s,=0.307 .sr=0. 73andbrepresentstheslopeofttreregressionlinewhichis given as: 0.90. Pluggirrr into tlte fortnula rve get: r'= we square 0.79. r U 79 tr = 6lo 4. (C) 0.4402This / n+g\ 0'96\i.r,r, = O'79To obtain r2' o observations anci each represents a binomial setting with independent observation having the sanre probability of success. rhe Tl-83 calculator. plug in I - N= 4O,p= 0.6 and k> 25, Using binomialcdf (40, 0.6.24')which is 1 - Probabiiity of 24 or less will prefer the brand ilame' 5'(B)Thesamplemeanrisanunbiasedestimatorofthepopulationmeanp. 6. (D) 0.783. This a binomial distribution because observations are independent and the observation' N= 15' P= 0'4 and probability of ou'ning a cell phone is 40% for each k=5.6.7,.."15.UsingtheTi-83:i-binomialcdf(15'0'4'4)=0'Zlg: '7. "\1 | fn. Et f# llo probability of at least probability of having no girls,tt (+) = one girl" is 1-(probabilityof nogirls)' The fr' so' t -#=ffi OR TI-83: 1 - binomialcdf (7' 0.5' 0) up to one, P(y= 1) must be 0'5 0.1. Since a probability distribution must add at'rd y= 1)=P(r= 2)xP(l= 1)since because0.l +0.5+ 0.4= 1. FurlhermoreP(r =2 s. (A) .r and ) are independent' t'et the P(x=2)=xz'P(y=1)=O'S and the P(x=2,!='s')=0'2.Then(.x:X0.5)=0.2'SolvingforX2'wegetxz=0.4.Thesumof 32 6;uthorized coPYing or reuslng any part of this Page is illegal &- Section Sample Examination Three-Answers I 33 probabilities in probability distribution must be l. To find xr or P (r =.- 2), add all the the distribution ofx and subtract from 1. 0.1 + 0.3 + 0. I + 0.4 = 0.9. Then P (x = -2)=1-0.9=.1 a 9. (D)83'9.Theexpectedcellcountcanbecalculated,,ineffi. instrument is Rorv 1' For the cell representing right-handed girls who play a musical Column l. the formula becornes {?$9 = 83'9 10. (B) 75. The formula which provides us with the smallest sample maximurn error, and { n =l a is: '=(##) confidence level required given a ,size ' This gives us (i.qo \(2.2)\' \i-]}.1/ | =1+.Zl.We round this up to 75. [''') 11. (C) 14. Creating a sample space for the 36 possible sums: n 0 0 6 6 o 6 12 12 11 IL 12 l8 18 6 6 t2 la LL 6 ta IL t2 t8 l8 12 6 6 lz ta l/- 6 6 la IL 1") 12 ta IL t8 18 t2 1a 18 r8 18 IL 18 18 4i t8 24 z+ aA 24 We find the probability distribution for the sum to be: Sum -Probability 1 0 -Jt) 6 6 -JO 1a t2 ;.10 l8 ;JO 12 .A LA 4 36 the multiplication rule for (These probabilities could have also been obtained by using events") The expected independent events and the addition rule for mutually exclusive varue is zx r(x)= o(+).r(*)+ r2(*)-' 't(*)+z+(*)='n or any part of this Page is illegal. 34 Sample Examination Three-Answers Section I t2. (E) As one increases speed, the time required to travel a given distance decreases. This indicates a negative correlation. ta lJ. (B) 0.07. This isageometricdistribution,where p=0.47 andl - p=0.53 andx=4. The probability is calculated by multiplying: 0.533 0.47 = 0.07. This is the probability of getting three girls and then one boy in that order. 14. (C) H,,: 15. H= 36. The null hypothesis is: Hu: F= hypotheticalmean. (B) 0.29. Standard deviation for difference between two means, standard deviations nr-= F;i- ,;lf * =V -l3O- * :16f unknown is: u.fr i6. (B) Since fi P(A) P(B)= = 0.29 (O.S+)(0.20)= 0.108 = P(AandB);the events are independent. 17. (B) Because of differences in reading levels between urban, suburban and rural schools, a stratified sampiing metliod is most appropriate in the proportion which exists in the population. 18. 19. (D) I -p by definitron of Pouer. Pon'er is one minus the probability of a Type II error. or r = 577.2. for .r. *e set: 84 = (B) 580 Solve the equation: : = + To get 0.84 for z, we Rounded to the nearest multrple of 10 rre get 580 *'hich is B.=# to the 1eft of the cune under z. The closest value for the area looked up 1- 0.2 or 0.8 using the table is 0.7995 *hich gires 1'ou a:-score of 0.84. This may also be solved using your graphing calculator b1 using 2nd DIST 3: invNorm (area to left of z, mean, standard deviation). Forourproblem rve rvould do invNonn(O.8,510,80). The answer from the calculator will vary slightll' frorn the table answer. Either way rounding to the nearest multiple of ten gives you the same answer. ^I where I'20. (A) (35.5, 44.5) The formula to use is: p + ,iry. for a95o/o confidence interval, for this example it would be: n, Since z = 1.96 04T051 /a5O\'-?5oi -o4+ 1.96\ -i53r=(355..1+5) --Zs3-v.*r4s0-..,-\ t80 +, o^ each population are independent. Two sample /-tests are used when the population standard deviations are not known. Two sample l-tests are used if samples are independent. If samples are dependent, then a matched pair procedure should be used. 21. (B) The samples from copying or reusrng any part of this page is illegal. Section I Sample Examination Three-Answers 35 and III only. The third quartile is 59.8 so try definition of the third quar1ile,25Yo of the values lie above 59.8. 31.2 and 59.8 are the first and third quarliles and again by definition, 50o/o of the values lie in this region. Selection I says the mean is less than 44.5 which is the median. This would be true if the graph was lefl skewed. In drawing the box plot there is no evidence to believe this. 22. (D) II t0 60 c 50 40 30 20 L). (E)1, Ilandlli.Oneproperll ofresidualsislre.siductls=0.Sincethemeanresiduai is Lresiduols . this ri,ould har,'e to be equal to 0. A curved pattern in a residual plot does , show that the relationship betr.veen the two variables can best be modeled by a nonlinear model. The residual plot can be a plot of residuals against the x values, as well as the residuals against the fttted y values. /-+. (A) I only. The normal curve is symmetric about the mean. If a normal curve does not have a mean of 0, it is not symmetric about 0. The empirical rule states that approximately 680/o (not 50%) of the values lie within 1 standard deviation of the mean. 25. (A) A continuous variable takes all values on the number line, Shoe sizes may come in 6 or 6.5 but cannot take on a value such as 6.23384 and therefore are not continuous. 26. (D) 1.699. To find the value of / we us the area under one t-value of |.6997 27. L+9q ,ur"*[email protected]= tail. Looking up 0.05 for 30 - 1 = 0.05. This gives degrees of freedom, one finds a . (B) A type I error is reject Hu. but Hu is true, a Type II error is when we fail to reject Hu, but Hu is not true. Since we are failing to reject the null hypothesis, this is an example of a type II error 78. (C) The interquartile range (lQR) is obtained by subtracting Qr from p:, therefore IQR= (49 - 27) = 22. To determine the existence of an outlier, compute 1.5 x IQR add this value to Q: and subtract it from Qt lf any piece of data falls outside the range of these two values, it is an outlier. (0,) + 1.5(22) = 49 + 33 = 82 (0,1 - t.s (22) = 27 - 33 = - 6 Since all the values in (C) fall above 82, they are outliers. zed copying or reusing any part of this page is illegal. 36 Sample Examination Three-Answers Section I 29' (B) The distribution of a sample mean when o is not known can be represented by a /-distribution, provided the sample is a simple random sample. and the population from which the sample is taken is normal. The level of significance does not play a factor in this decision at ail. nor does the value of the sample-mean. 30. (D) ijsing IJ.:trt = 15 and H^:,u< 15,r =t!?,-^t) =- 1.40, which results .43t in ap_value /36 of betrveen 0.05 and 0.10. usingthe r-score char1. (r-teston-fl-g3 yierds p= 0.0g59) at Jt. (E) i. 2' 3' -fhe basic principles of statistical design of experiments are: response, most easily by comparing several treatments. control of the effects of rurking variables onthe Randomizatio,n, the use of impersonal chance to assign subjects to treatments. Replication oithe experiment on many subjects to reJuce chance variation in the results. -12. (B) A census is a complete count of an entire population. The nervspaper is polling the entire population of their readers. Although the sampling design may not be random, this sample survei- is intended to make inferences about all the paper,s readers. 33. (B) Since an outlier can stronsly' affect the ralLre of the mean (1t), it is non-resistant. since both the standard cleviation (o) and ccrrelarion coefficilnt (r) depend on the mean' they are also non-resistant. The meclian i-s not affected by an outlier, so it is resistant. a1 J{. (C) To select a stratifled random sample. first divide the population into groups of similar individuals, called strata. l-hen choose a separate SRS in each stratum and combine these SRS's to form the full sample. 35. (D) Increasing each grade by 5 points ',vill not affect the distance each grade is from the nlean. therefbre the standard deviation will not change. However, measures of center like the mean and rnedian rvill increase by 5. Jo. (A) Substituting the value x= 3 into the equation of the least-squares regression line results in i= i.3 +0.21(3)= 1.3 + 0.81=2.i1. The value of the residual is observed predicted -17. - : 2 - 2.1 1 =- 0.1 i. (B) fhe distribution is skewed to the reft, w'hich impries thar the mean is less than the median, since the mean is non-resistant. The mode must be 90 and the median is 85. so the mean is less than the mode as r.vell. 38' (D) The fact that a distribution has a larger mean does not have any bearing on how tightly the data is clustered around the mean. copylng or reusing any pari ot thls page is iltegal. Sample Examination Three-Answers Section I 39. (B) The pooled sample proportion for a 2-sample test of hypothesis is gi'ren p= #&rr=2#ffi= 40. (E) All of JI by: 0.3025 tl-re statements are true regarding the effects of lurking variables. trn fact, lurking variable can falsely suggest a strong relationship between the variables. or cari hide a relationshrp that really exists. copyrng oi" any pad of this page ls illegal 38 Sample Examination Three-Answers Section Section II II Question One l. Let P be the event an employee passes the dexterity test, then P' is the event an employee does not pass. Let W be the event that the employee is a \voman, then W' is the event the employee is a man. (a) The probability that the empioyee passed the test if it is known that the employee . P(PnW)-# -48-., pfw) -\ ' =i{=ii=0.64 185 --,-,--.\ isa r.roman is P(PiW)= (b) The probability that the employee is a woman. gii en that it is an employee who passed the test is denoted as (c) P(wlo)= r./ i$*a p(p) = #3 = #= 0.387 L?!-tz4-v.Je 185 Are the events "selecting an employee r,r'ho passed the test" and "selecting an emplovee rvho is a woman" independent events? Justify' your answer. Student must perform a chi-square test: H,,'. Gender and passing lhe test are inclepenclent Ho: Gender ancl passing the Iest ctre not inc/epertdenl Assumption that all expeced counts > 5 is true. Row Tolal x Coluryn Total Exoectecl _ ' Lrrl]no I OtJl e.g. Expected courrt fbr "men" uho "passed" ll0x 124 - -T85-_ 27=?8 JI z - 44 expecred -('^-2#) )t .,.-- /' =(observcd-expecred)2 ^ | 2A _1342f \r*--JTl ----ItT-- -tT - (nt-tfl) l{qq .2 - 37 2 s _91 \ "- 3l l \r'| 915 -iT =r,..5'-i d.f.=(row total- 1)x(coiumn total- 1)= (2-l)(Z-t1=1 p-value =0.4696. We rvould reject the null hypothesis if xr> 3.84 for a 0.05 level of significance. Due to large p-value. rve do not reject H". Conclusion: Insuffrcient evidence to reject the null hypothesis. Our data do not reject the claim that "selecting an employee rvho passed the test" and "selecting an employee who is a woman" are independent. Note: Students may try to use the multiplication rule for determining independence. This is only partiallv correct because variability is not being taken into consideration. or any part of this page is illegal Section Il Sample Examination Three-Answers Question Two Correct answer #I Two hundred patients with this type of cancer are randomly selected to get one of the A, B, and C. Randomizarion could be accomplished by assigning each of the 200 patients a number between I and 200. Select the first 66 of the patients using a random number table to get treatment A, the next 66 patients to receive treatment B and the remaining 67 Io geI treatment C. three procedures After a select period of time, results are collected for each group and compared. Randomly select ,/ 2oo Patientsto each type 4t of \ Procedure \ Proceclure '4 Prttcedure B Comoare results Procedure C Correct answer #2: If there is reason to believe that t.omen react differently than men then blocking by gender is appropriate. The 60 rvomen uould be randomly selected to each of the three procedures, Number 60 women 1-60. and using a random number table assign the first randomly selected 20 women to receive procedure A, the next 20 to receive procedure B and the remaining 20 to receive procedure C. Follow the same procedure for the 140 men. (i.e. the first 46 randomly selected get procedure A, the next47 to get procedure B and the last 47 to get procedure C). Compare results within the blocks. Procedure A Randomly assign the 60 \vomcn to each procedure ,/ <\ Procedure B Compare results Procedure C Procedure A Ranclomly the 140 assign men + toeachprocedure ,/ Procedure B \ Comoare results Procedure C Note: In both answers students could add a fourth group as a control group. This group would receive an old treatment or a placebo. However, a complete answer would address why or why not a control group would be used in these circumstances mindful of ethics, severify of disease, etc. Ideally, all experiments have a control group. copying or reusing any part of this page ls lllegal 40 Section Sample Examination Three-Answers II Question Three Tlie test required fbr this problem is a two sample difference of means I test since the groups are independent and since sigma is not known. This test requires the two samples to be independently selected simpie randoni samples and for the two populations to be normally distributed. It is not verified how these classes are selected, so \\'e will state our assumption that rve have tu,o independent simple random samples. Checking for normalcy of the population we ivill use normal quantile plots. Normal Quantile Plol for \\'ith \'lanipuletrves data Normal Quantile Plot for Without Manipulatives Data tr tr s d Since the plots do not shou anv severe skew or outliers, we continue under the assuilption that both popr-tlatior-is are approximately normal. i +Fl | tll a6) 15 -t-- ll with nanipulatives without manipulatives The boxplots indicates there is no evidence of outliers, an indication that the condition of normality 'nvould be satisfied. . 2. I Does manipulative increase test scores? LL,'. /Jt= /)z Il,.'. lJL> Pz 3. Where pr is the mean test score of those using manipulatives and test score of those not using manipulatives. a = 0.05 Unauthorized copying or reusing any part of this page is illegal pz is the mean Section II Sample Examination Three-Answers A ,_(x,-rr)-(ttta.r-T ,'l ru- n ltz) 6. Approximately / distribution with d.f. equal to the smaller of Reiect H" if r> 1.833 7. test statistic' r = 5. 4l -s3.37', , 5.4', / -To-- - 10 = 1.99 rir- | or nz- 1. p-value:0.33 1 TVro sample T for With Manipulatives vs Without Manrpuiatives N Mean StDev 5.40 SE Mean with manipulatives 1C 84.00 r.1 wi f horrf man'i nrrlatives 10 80.00 3 .37 1.1 T-Test mu with m=mu without (vs >) : T= L.99 P= 0.033 DF= 15 8. Conclusion: At the 0.05 level of significance there is sufficient evidertce to conclude that test scores of students using manipulatives are higher than those of students not using manipulatil'es. Note: a student using the TI-83 calculator's 2-sample t-test function u'ill obtain p-value of 0.031 with 18 degress of freedom using the pooled option, or a p-value of 0.033 rvith 15.085 degrees of freedom using the unpooled option. a Because there was no random assignment of students to the two groups, this is an observational study and caution must be applied in reaching a conclusion. There seems to be an association between higher scores and use of manipulatives but higher scores may be due to other underiying conditions that have not been controlled. or any part of this page ls lllegal 42 SampleExaminationThree-Answers Section II Question Four The test required for this question is the chi-square test for Independence. We are looking for statistical evidence that proportion of subjects with low, average and elevated cholesteroi levels differs for walker and non-walkers. The assumptions for this test are that all expected counts are at least 5 and that the data are from a simple random sample. The first condition will be shown to be satisfied when the calculations are complete. The second condition is not clearly satisfied in the description of the experiment. W'ere subjects randomly selected to be walkers and not walkers or was the experiment more likely an observational study? We will proceed as if this condition was satisfied. 1. 2. Is there a relationship between aerobic walking and cholesterol levels? Hu : There is no relationship between walking and cholesterol level H" : There is a relationship between walking and cholesterol level 3. 4. 5. 0 6. 7. = 0.05 Test Statist ic: X2= t X2distribution with (rows - 1)(columns - 1) degrees of freedom. (3- 1X2-I)=2 df. Criterion for Rejection: Reject Hu for x2> 5.99 for 2 degrees of freedom Xz = 9,609 Note all expected cell counts above five. Cholesterol Level Walkers Lorl Average Elevated 51 86 31 lse2) (e6) f+ss) L-) 94 30 t"j 1.", Row Total 168 Non-Walkers 518 *t) 427\ 15 l t47 (84) Column Total '/4 180 61 315 Expected Counts: in parenthesis Expected counts can be calculated by rake,n. , -^^ ,2 , (r'-#) x- = ---sgT: * i5 t n '',, ,2 ,.^ aLt (so -s6\2 --96- + ffi (r'-ff)' (rr-#)' * 04-84)2 ---ZE8 * ---ilE15 15 --T-4- I \-" 15 / +-zn: l) p-value 8. : = 9'609 0.008 or using table for 2 d.f. 0.005 <p < 0.01 There is evidence to reject the null hypothesis and conclude that level of cholesterol is related to aerobic walking. copying or re any part of thls page is illegal. Section II Sample Examination Three-Answers 43 Quesion Five (a) Because we have two independent random samples from two distinct populations (men and rvomen) and rve are looking to see the difference between the proportion of men and women rvho favor legislation which restrict cell phone use while driving. the choice of interval u,ould be large sample confidence interval ri(r_it;-frr!_fr, forcomparingtwoproportions:formula:(i,,-i,')t,'l#+-' Assumplions inclttde: Trvo independent sirnple random samples fiom tu'o large populations , nr ftr , nz fu2, ir r (1 - fut). nz(l - irt) more than 10. Note: In probiems that deal rvith proportions, some texts use ntf)t,ttz pr,n'(l -ir'),n,(l -p,)are all greater than 10 when checking assumptions, r.r'hile other texts use the value of 5. Please be advised that as long as the student shou's that the assumptions har e been checked, either value is acceptable. bt= l?n Iuj 2i# )uu= 0.-16 tit.- = li] 500 = rl. jSU tt - 500 rz:= 5C)0 Clearly a/l o.f the conclition.y ctre .satis/ietl. \\'e also asslllre that there are at least 5000 rnen and 5000 \\omen in the countn For a 95%o confidence interval. the value of z = 1.96 The interval.for p,,,,- pnon*i.r (0.0109, 0.1331). (b) We are 95% confident that the difference in population proporlions is between 0.0109 and 0.1331. Because the interval does not contain 0, $'e are confident that men have a stronger preference for legislation that would restrict the use of cell phones while driving. (c) A99% confidence interval would change the value of z in the formula above from 1.96 to 2.576 thus making the interval wider. The new interval would become: (-0.0083, 0.15229). This wider interval would contain 0 and the conclusion reached concerning whether there u,as a difference in the proporlion of men and women who favored legislation would be different than in part (a), i.e. there r.vould appear to be no statistically significant difference between the two groups. zea or reusing any part of this page ls illegal. 44 Sample Examination Three-Answers Section II Question Six (a) 0'60' using the + The probability of B or o blood is P(B) + P(O) = 0.15 0'45 = B or O blood probability of b.OO, \ve can assign six of the ten integers to be a is to solution possibie iype and assign four to be some other blood type. One the random urrlgn 0 throigh 5 as blood type B or O. We then need to use values three at number table and count how many digits it takes to arrive are needed to get three betu,een 0 and 5. This would give us how many donors donors with blood tYPes B or O' (b)Startingintheupperlefthandcornerw.euses,4,l,T'7and0.Westopat0 0 and 5 have found three values 4, 1, and 0 that are values between 'ne type. It took 6 donors to and represent three donors with type B or O blood again where we left off-"uve arrive at three uith the desired blood type" Beginning at I because we have founc use the number 6.7.5. 7.1.1,6, and 1. we stop donors before finding three integers bet$'een 0 and 5. l'his time it took eight we again start where we three with the desired blood type. For the last simulation these numbers are left ofr. The numbers begin rvith 3, 1, and 5. All three of donors with type B or O integers betu,een 0 and 5 and therefore u'e got three that it took first tl.rree numbers. Three trials of our simulation found blood in 'ur the blood type B or O' six, eight and three clonors io get three donors r'vith because (c) protrability of getting a Distribution B is the correct distributionBecause the probability of finding a donor donor *,ith tlpe B or o blood is greater tharr the donors with nith t.vpe A or AB blood. it $ou1d take fe$'er donors to find three mean and a lower maximum type ts or o blood t1,pe. Distribution B has a lorl'er as well. (d)UsingdistributionB,theexpectetlnumberisfoundusingtheformula: (.x) can be found by raking the frequencies for E (ri= x , /(r). The vah_ies of l of the frequencies which each:rvalue of the histogram and dividing by the sum isl00.Thevaluesofxaretlrevaluesonthehorizontalaxisofthehistogram. Thisgivesus:0.22(3)+0.27(4)+0.21(5)+0.15(6)+0.07(7)+0.05(8) +0.02(9) + 0 0i (i1) = 4'81 (e) of type o or B donors Blood bank A would be more likely to have a percentage donors each day as a to be more than seventy-five percent. lf r.r'e treat the o anci B donors for blood bank sample. the distribution of the percentage of type Awouldbeapproximatelynormallydistributedwithameanof0.60anda standard deviation "t m = 0.069. The distribution of the percentage typeoandBdonorsforbloodbankBhasameanof0.60andastandard deviation"r/W=oo35' Unauthorized coPYing or reus any part of this Page is illegal' of Section II Sample Examination Three-Answers 45 We can find the appropriate probabilites by using the z-score table and the ) l7 fbrmula for z-scores. The z-score for blood bank A would be 0'7i =-p='60 = 0.069 -L'tt' [email protected] The z-score for blook bank B '. curve to the right of 2.17 is 0.0150. = 4.29.The area under the normal T'he area under the curve to the right of 4.29 is less than 0.0002. OR Using the TI-83 calculator for blood bank A: Press: 2nd distribution Use 2: normalcdf as shorvn below: norma L sdf ( E. 75 ' I E16' 9.6' B.869) .8148557881 I Using the TI-83 calculator for blook bank B: Fress: 2nd distribuiion Use 2: normalcdf as shown belou,: rdf { B. 75 I 8.6' E.835) '-5 nBFr'ia l E16' T 9. 11484S649e The probability of having at ieast T5ohtype B or O donors is greater for blood bank A than for biood bank B. This also stands to reason since there are f-ewer donors at Blood Bank A and the Law of Large Numbers implies that the larger sample (Blood Bank B) is more likely to have a sample proportion that is closer to the true population proportion of 6A9'o. The higher variability in Blood Bank A's distribution is caused by the smaller sample size. With the higher variability', the chances of obtaining a sample percentage of 75o/, or more are higher for Blood Bank A. Unauthorized copying or reusing any part of thls page is illegal

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