Chapter = 6.9, and for red and brown is ( ) ( ) ( ) ( ) ( ) ( )

Transcription

1 Chpter 11 Section 11.1 Check Your Understnding, pge 681: 1. 0 : pblue = 0.3, pornge = 0.3, pgreen = 0.15, pyellow = 0.15, pred = 0.1, pbrown = 0.1 : At lest one of the p i's is incorrect.. There were = 46 cndies in the bg. The expected count of both blue nd ornge , = 6.9, nd for red nd brown is cndies is ( ) = for green nd yellow is ( ) 46( 0.1) = ( ) ( ) ( ) ( ) ( ) ( ) χ = = = Check Your Understnding, pge 684: 1. The expected counts clculted in Exercise of the previous Check Your Understnding were ll t lest 5. We should use the chi-squre distribution with 6 1= 5df.. 3. From Tble C 0.05 < P-vlue < From the clcultor the P-vlue is Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht t lest one of the proportions of M&M s Penut Chocolte Cndies of prticulr color is different thn wht the compny reported. Check Your Understnding, pge 689: 1. Stte: We wnt to perform test t the α = 0.01significnce level of : pred-stright =, pred-curly =, pwhite-stright =, pwhite-curly = versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: The expected counts 9 3 in ech ctegory re red-stright: 00 = 11.5, red-curly nd white stright: 00 = 37.5, nd The Prctice of Sttistics for AP*, 4/e

2 1 white-curly: 00 = 1.5, 16 ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 000 pirs of fruit flies. The conditions re met. Do: The test sttistic is ( ) ( ) ( ) ( ) χ = = nd the distribution hs = 3df. The P-vlue is Conclude: Since the P-vlue is greter thn 0.01, we fil to reject the null hypothesis. We do not hve enough evidence to reject the biologists predicted distribution of trits in offspring. Exercises, pge 69: 11.1 () 0 : pcshews = 0.5, plmonds = 0.7, pmcdmi = 0.13, pbrzil = 0.08 versus : At lest one of the p i's is incorrect. (b) There ws rndom smple of 150 nuts. The expected count for cshews is 150( 0.5) = 78, for lmonds is 150( 0.7) = 40.5, for mcdmi nuts is = ( 0.13) 19.5, = nd for brzil nuts is ( ) 11. () 0 : p red =, pblck =, pgreen = versus : At lest one of the p i's is incorrect. (b) There ws rndom smple of 00 spins. The expected count for red is 00 = 94.74, for blck is = 94.74, 38 nd for green is 00 = ( ) ( ) ( ) ( ) χ = = = ( ) ( ) ( ) χ = + + = = () The expected counts clculted in Exercise 1 re ll t lest 5. Since there re 4 ctegories, use chi-squre distribution with 4 1= 3df. (b) Chpter 11: Inference for Distributions of Ctegoricl Dt 53

3 (c) Using Tble C, 0.05 < P-vlue < Using the clcultor the P-vlue is (d) Since the P- vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to reject the compny s clim bout the distribution of nuts () The expected counts clculted in Exercise re ll t lest 5. Since there re 3 ctegories, use chi-squre distribution with 3 1= df. (b) (c) Using Tble C, 0.10 < P-vlue < Using the clcultor, the P-vlue is (d) Since the P- vlue is greter thn 0.05, we fil to reject the null hypothesis nd find tht we do not hve enough evidence to sy tht the roulette wheel probbilities re not wht they should be Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : pfirs = 0.54, ppines = 0.40, pother = 0.06 versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: The expected counts in ech ctegory re firs: 156( 0.54) = 84.4, pines: 156( 0.40) = 6.4, nd other: 156( 0.06) = 9.36, ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 1560 red-brested nuthtches. The conditions re met. Do: ( ) ( ) ( ) The test sttistic is χ = + + = nd the distribution hs = df. The P-vlue is Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the nuthtches prefer prticulr types of trees when they re serching for seeds nd insects Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : psnd = 0.56, pmud = 0.9, procks = 0.15 versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: This ws rndomized experiment. Lrge smple size: The expected counts in ech ctegory re snd: 00( 0.56) = 11, mud: 00( 0.9) = 58, nd rocks: 00( 0.15) = 30, ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 000 segulls. The conditions re met. ( ) ( ) ( ) Do: The test sttistic is χ = + + = nd the distribution hs = df. The P-vlue is Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the segulls hve preference for the loction where they stnd. 54 The Prctice of Sttistics for AP*, 4/e

4 11.9 A chi-squre goodness-of-fit test would not be pproprite here becuse the dt for ech dy were not counts but rther mens, nd becuse the sme 50 students were used ech dy (so the observtions were not independent of ech other) A chi-squre goodness-of-fit test would not be pproprite here becuse mny of these students likely hd sme techers (so the observtions were not independent of ech other) () Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : p1 = 0.301, p = 0.176, p3 = 0.15, p4 = 0.097, p5 = 0.079, p6 = 0.067, p7 = 0.058, p8 = 0.051, p9 = versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: The expected counts in ech ctegory re first digit 1: 50( 0.301) = 75.5, first digit : 50( 0.176) = 44, first digit 3: 50( 0.15) = 31.5, first digit 4: 50( 0.097) = 4.5, first digit 5: 50( 0.079) = 19.75, first digit 6: , = 1.75, nd first digit 9: 50( 0.067) = 16.75, first digit 7: ( ) = first digit 8: ( ) 50( 0.046) = 11.5, ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 500 invoices t this compny. The conditions re met. Do: The test sttistic is ( ) ( 50 44) ( ) ( ) ( ) ( ) χ = ( ) ( ) ( ) + + = nd the distribution hs 9 1= 8df. The P-vlue is Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the invoices re inconsistent with Benford s lw. Follow-up nlysis: The brekdown of the chi-squre sttistic is: χ = From this we see tht the lrgest contributors to the sttistic re mounts with first digit 3, 4 nd 7. There re too mny mounts tht strt with 3 or 4 nd not enough tht strt with 7. Similrly there re too few tht strt with 1 or with 9. (b) A Type I error would be to sy tht the compny s invoices did not follow Benford s lw (suggesting frud) when in fct they were consistent with Benford s lw. A Type II error would be to sy tht the invoices were consistent with Benford s lw when in fct they were not. A Type I error would be more serious here lleging tht the compny hd committed frud when it hd not Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : pispnic = 0.8, pblck = 0.4, pwhite = 0.35, pasin = 0.1, pothers = 0.01 versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: The expected counts in ech ctegory re ispnic: 800( 0.8) = 4, Blck: 800( 0.4) = 19, White: 800( 0.35) = 80, Asin: 800( 0.1) = 96, nd Others: 800( 0.01) = 8, ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 8000 residents of lrge housing complex. The conditions re met. Do: The test sttistic is ( ) ( ) ( ) ( ) ( ) χ = = nd the distribution hs 5 1= 4df. The P-vlue is Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the residents of this housing complex do not follow the distribution by ethnic bckground of New York City s whole. Follow-up nlysis: The brekdown of the chi-squre Chpter 11: Inference for Distributions of Ctegoricl Dt 55

5 sttistic is: χ = From this we see tht the lrgest contributor to the sttistic, by fr, is the lst mount. There were mny more people in the Other ctegory thn we would hve predicted () 0 : plemon = 0., plime = 0., pornge = 0., pstrwberry = 0., pgrpe = 0. versus : At lest one of the p i's is incorrect. (b) Since ll 5 flvors hve the sme proportion, they ll hve the sme expected count: 60( 0.) = 1. (c) There re 5 1= 4df for this chi-squre sttistic. Using Tble C, the vlue for α = 0.05 is 9.49 nd for α = 0.01 is (d) Answers will vry. One possibility consists of 6 lemon, 6 lime, 16 ornge, 16 strwberry nd 16 grpe. The chi-squre sttistic is ( 6 1) ( 6 1) ( 16 1) ( 16 1) ( 16 1) χ = = 10 with P-vlue of () 0 : p1 = 0.1, p = 0.1, p3 = 0.1, p4 = 0.1, p5 = 0.1, p6 = 0.1, p7 = 0.1, p8 = 0.1, p9 = 0.1, p0 = 0.1 versus : At lest one of the p i's is incorrect. (b) Answers will vry. We perform the test t the α = 0.05 significnce level. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: The expected count in ech ctegory is: 00( 0.1) = 0, which is t lest 5. Independent: The clcultor creted these in n independent wy. The conditions re met. Do: In one test of the clcultor we got 18 0 s, 1 s, 3 s, 1 3 s, 1 4 s, 1 5 s, 17 6 s, 14 7 s, 1 8 s, nd 9 s. The test sttistic is ( 18 0) ( 0) ( 3 0) ( 1 0) ( 1 0) ( 1 0) ( 17 0) ( 14 0) χ = ( 1 0) ( 0) + + = nd the distribution hs 10 1 = 9 df. The P-vlue is Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We don t hve enough evidence to sy tht the clcultor s RndInt function is not working properly Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : p i = = for ll 1 strologicl signs versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple 1 size: Since the totl smple size is 4344, the expected count in ech ctegory is 4344 = 36 1 which is t lest 5. Independent: There re more thn 43,440 residents in the U.S. The conditions re met. Do: ( ) ( ) ( ) The test sttistic is χ = = nd the distribution hs = 11df. The P-vlue is Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the 1 signs re not eqully likely. Follow-up nlysis: The brekdown of the chi-squre sttistic is: χ = From this we see tht the lrgest contributors to the sttistic re Aries nd Virgo. There re fewer Aries nd more Virgos thn we would expect. There re lso more Librs nd fewer Scorpios nd Sgittrius thn would be predicted. This suggests tht there re more people born in the lte summer nd erly fll months thn we would predict if ll time periods were eqully likely. 56 The Prctice of Sttistics for AP*, 4/e

6 11.16 Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : p i = for ll flvors versus : At lest one of the p i's is incorrect. Pln: We should use chisqure goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge 1 smple size: Since the totl smple size is 10, the expected count in ech ctegory is10 = 0 6 which is t lest 5. Independent: There re more thn 100 Froot Loops. The conditions re met. Do: ( ) ( ) ( ) ( ) ( ) ( ) The test sttistic is χ = = nd the distribution hs 6 1= 5df. The P-vlue is Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to sy tht the six flvors re not eqully likely. Follow-up nlysis: Not necessry Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : psmooth = 0.75, pwrinkled = 0.5 versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used = 417, nd Lrge smple size: e observed 556 pes so the expected counts re smooth: ( ) wrinkled: 556( 0.5) = 139, both of which re t lest 5. Independent: It seems resonble tht there would be more thn 5560 pes. The conditions re met. Do: The test sttistic is ( ) ( ) χ = + = nd the distribution hs 1= 1df. The P-vlue is Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to dispute Mendel s beliefs. Follow-up nlysis: Not necessry Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : ptll-cut = 0.565, ptll-potto = , pdwrf-cut = , pdwrf-potto = versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: A totl of 1611 plnts were observed. The expected counts in ech ctegory re tll-cut: 1611( 0.565) = , tll = , nd dwrf- potto: 1611( ) , = dwrf-cut: ( ) potto: 1611( 0.065) = , ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 16,110 tomto plnts. The conditions re met. Do: The test sttistic is ( ) ( ) ( ) ( ) χ = = nd the distribution hs 4 1= 3df. The P-vlue is Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to dispute the genetic lws. Followup nlysis: Not necessry d c 11. d Chpter 11: Inference for Distributions of Ctegoricl Dt 57

7 11.3 The English grdes for the hevy reders re centered round pproximtely 3.7 with left skewed distribution. The IQR is pproximtely 0.4 nd there is one low outlier t pproximtely.9. The English grdes for light reders hve lower center t pproximtely These grdes re more symmetriclly distributed with no outliers nd the distribution hs n IQR of pproximtely 0.7. In generl, the grde distribution for light reders is hs lower nd is more spred out thn the grde distribution for hevy reders () The conditions required re Rndom: The smple ws rndomly selected. Norml: Both smple sizes were t lest 30 nd the only outlier ws not lrge outlier. Independent. Since the school ws described s being lrge high school, it seems resonble to believe tht there re more thn 790 students in the school. All conditions re met. (b) Stte: Our prmeters of interest re µ 1 = the men English grde of hevy reders nd µ = the men English grde of light reders. We wnt to estimte the difference µ 1 µ t 95% confidence level. Pln: We should use two-smple t intervl for µ 1 µ if the conditions re stisfied. We checked the conditions in prt (). The conditions re met. Do: From the dt, n 1 = 47, x 1 = 3.64, s x = 0.34, n 1 = 3, x = 3.356, nd s x = We will use the conservtive degrees of freedom which is 31 in this cse. So our 95% confidence intervl is ( 0.34) ( 0.38) ( ) ± = 0.84 ± = , Conclude: We re 95% 47 3 confident tht the intervl from to cptures the true difference in the men English grde of hevy nd light reders. This suggests tht the men English grde is between nd points higher for hevy reders thn for light reders. (c) No. This ws n observtionl study so no conclusion of cuse nd effect cn be mde () The y-intercept is 3.4. This mens tht we would predict n English grde of 3.4 for student who hd red no books. (b) The predicted English GPA for this student is y ˆ = ( 17) = This mens tht the residul is y yˆ = = (c) The reltionship between English grdes nd number of books red is not very strong. The vlue of r is only which mens tht only 8.3% of the vrition in English grdes is ccounted for by the liner reltionship with the number of books red () Wht Luis hs clculted is the probbility of getting 5 of one prticulr number. Wht he hs not tken into ccount is tht there re 6 different possible wys of getting Yhtzee one for ech number on the die. (b) No, Nssir should not be surprised. The probbility of getting Yhtzee is = The probbility of getting no Yhtzee is = Finlly, the probbility of getting 5 rolls with no Yhtzee is ( ) 5 = Section 11. Check Your Understnding, pge 698: 1. For the min cmpus the conditionl distribution is: 6.0% use Fcebook severl times month or less, 3.6% use it t lest once week, nd 70.3% use it t lest once dy. For the commonwelth cmpuses: 1.1% use it severl times month or less, 5.0% use it t lest once week, nd 6.8% use it t lest once dy. 58 The Prctice of Sttistics for AP*, 4/e

8 . It is importnt to compre proportions rther thn counts in Question 1 becuse there ws such big difference in the smple size from the two different types of cmpuses. 3. The biggest difference between the two types of cmpuses is tht students on the min cmpus re more likely to be everydy users of Fcebook thn students on the commonwelth cmpuses nd those on the commonwelth cmpuses re more likely to use Fcebook severl times month or less thn those students on the min cmpus. Check Your Understnding, pge 703: 1. 0 : There is no difference in the distributions of Fcebook use between students t Penn Stte's min cmpus nd its commonwelth cmpuses : There is difference in the distributions of Fcebook use between students t Penn Stte's min cmpus nd its commonwelth cmpuses.. First note tht there re = 1537 students in the smple. Also, there re = 131 totl students who use Fcebook severl times month or less. So, the expected count for the number of students t the min cmpus who use Fcebook severl times month or less is computed s 131 ( 910 ) = The other expected counts re computed in similr fshion nd re given in the 1537 tble below. Use Fcebook Min cmpus Commonwelth cmpus Severl times month Once week Once dy Chpter 11: Inference for Distributions of Ctegoricl Dt 59

9 3. ( ) ( ) ( ) ( ) ( ) ( ) χ = = Check Your Understnding, pge 705: 1. Using Tble C with df = ( 3 1)( 1) = P-vlue < Using the clcultor, the P-vlue is Assuming tht there is no difference in the distributions of Fcebook use between students on Penn Stte s min cmpus nd students t Penn Stte s commonwelth cmpuses, the probbility of observing smple tht shows difference in the distributions of Fcebook use mong students t the min cmpus nd the commonwelth cmpuses s lrge or lrger thn the one found in this study is bout 6 in 100, Since the P-vlue ws so smll, reject the null hypothesis. It ppers tht the distribution of Fcebook use is different mong students t Penn Stte s min cmpus nd students t Penn Stte s commonwelth cmpuses. Check Your Understnding, pge 708: 1.. Stte: We wnt to perform test of 0 : There is no difference in the distribution of qulity of life in Cnd nd the US : There is difference in the distribution of qulity of life in Cnd nd the US t the α = 0.01level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from seprte rndom smples. Lrge smple size: We used technology to get the following expected counts: Qulity of life Cnd US Much better Somewht better About the sme Somewht worse Much worse The Prctice of Sttistics for AP*, 4/e

10 All of these counts re t lest 5. Independent: We clerly hve less thn 10% of ll hert ttck victims in the U.S. nd in Cnd. All conditions hve been met. Do: The test sttistic is ( ) ( ) χ = = We use chi-squre distribution with = 4df nd find P-vlue of Conclude: Since the P-vlue is greter thn 0.01, we ( )( ) fil to reject the null hypothesis. There is not enough evidence to conclude tht there is difference in the distribution of qulity of life in Cnd nd the United Sttes. Check Your Understnding, pge 713: 1. This ws n experiment. Ech individul ws exposed to tretment tht involved how they were contcted.. Contct method Answered yes Answered no Phone Personl interview Written response Stte: We wnt to perform test of 0 : There is no difference in the proportion who nswer yes bsed on contct method : There is difference in the proportion who nswer yes bsed on contct method t the α = 0.05 level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from rndomized experiment. Lrge smple size: We used technology to get the following expected counts: Contct method Answered yes Answered no Phone Personl interview Written response All of these counts re t lest 5. Independent: Due to the rndom ssignment, these three groups of people cn be viewed s independent. Individul observtions in ech group should lso be independent: knowing one person s response gives no informtion bout nother person s response. Do: The test ( ) ( ) sttistic is χ = = We use chi-squre distribution with = df nd find P-vlue of Conclude: Since the P-vlue is less thn 0.05, we reject ( 3 1)( 1) the null hypothesis nd conclude tht there is convincing evidence of difference in the proportion of people who nswer yes bsed on how they re contcted. Check Your Understnding, pge 718: 1. Stte: We wnt to perform test of 0 : There is no ssocition between territory type nd success or not in the popultion of frnchises : There is n ssocition between territory type nd success or not in the popultion of frnchises t the α = 0.01level. Pln: We should use chi-squre test of independence/ssocition if the conditions re stisfied. Rndom: This ws rndom smple of frnchises. Lrge smple size: We used technology to get the following expected counts: Exclusive territory Success Filure Yes No Chpter 11: Inference for Distributions of Ctegoricl Dt 61

11 All of these counts re t lest 5. Independent: We smpled 170 frnchises. There re more thn 1700 frnchises in the US. The conditions re met. Do: The test sttistic is ( ) ( ) χ = = We use chi-squre distribution with = 1df nd find P-vlue of Conclude: Since the P-vlue is greter thn 0.01, we ( )( ) fil to reject the null hypothesis. We do not hve enough evidence to conclude tht there is n ssocition between whether frnchises hve n exclusive territory or not nd whether they re successful or not. Exercises, pge 74: 11.7 () There were 67 women nd 67 men in the two smples so the conditionl distributions for femles nd mles re: Gol Femle Mle SC-M = = SC-LM LSC-M LSC-LM (b) (c) In generl it ppers tht femles were clssified mostly s low socil comprison wheres mles were clssified mostly s high socil comprison. Mles were most likely to be in the high socil comprison/high mstery group wheres femles were most likely to be in the low socil comprison/low socil mstery group. 6 The Prctice of Sttistics for AP*, 4/e

12 11.8 () There were 0 in ech of the blck prents nd ispnic prents groups nd 01 in the white prents group so the conditionl distributions re: Survey result Blck prents ispnic prents White prents Excellent = = = Good Fir Poor Don t know (b) (c) In generl, ispnic prents re more likely to think tht the high schools re doing n excellent job. Whites re more likely thn either of the other two groups to think tht the high schools re doing either n excellent or good job. Blck prents re more likely to think tht schools re only doing fir job () 0 : There is no difference in the distribution of sports gols for femle nd mle undergrdutes : There is difference in the distribution of sports gols for femle nd mle undergrdutes (b) First note tht there re = 134 students in the smple. Also, there re = 45 totl students who re clssified s SC-M. So, the expected count for the number of femle students 45 clssified s SC-M is computed s ( 67 ) =.5. The other expected counts re computed in 134 similr fshion nd re given in the tble below. Gol Femles Mles SC-M.5.5 SC-LM LSC-M LSC-LM (c) ( ) ( ) χ = = Chpter 11: Inference for Distributions of Ctegoricl Dt 63

13 11.30 () : There is no difference in the distribution of opinions bout high school mong blck, ispnic 0 nd white prents : There is difference in the distribution of opinions bout high school mong blck, ispnic nd white prents (b) First note tht there re = 605 prents in the smple. Also, there re = 68 totl prents who thought the high schools re excellent. So, the expected count for the 68 number of blck prents who think the high schools re excellent is computed s ( 0 ) =.70. The 605 other expected counts re computed in similr fshion nd re given in the tble below. Survey result Blck prents ispnic prents White prents Excellent Good Fir Poor Don t know (c) ( ) ( ) χ = = () Rndom: The dt cme from rndom smples. Lrge smple size: The expected counts computed in exercise 9 were ll t lest 5. Independent: Clerly we hve less thn 10% of ll mle or femle students possible. (b) Using Tble C with df = 3,we get P-vlue < Using the clcultor, the P-vlue is (c) Assuming tht there is no difference in the distributions of gols for plying sports mong mles nd femles, the probbility of observing smple tht shows difference in the distributions of gols in plying sports mong mles nd femles s lrge or lrger thn the one found in this study is bout in 100,000. (d) Since the P-vlue is so smll, we reject the null hypothesis. There is convincing evidence of difference in the distributions of gols for plying sports mong femle nd mle undergrdutes () Rndom: The dt cme from rndom smples. Lrge smple size: The expected counts computed in exercise 30 were ll t lest 5. Independent: Clerly we hve less thn 10% of ll prents possible in ech group. (b) Using Tble C we get < P-vlue < Using the clcultor, the P- vlue is (c) Assuming tht there is no difference in the distributions of opinions bout high school mong blck, ispnic, nd white prents, the probbility of observing smple tht shows difference in the distributions of opinions bout high school mong blck, ispnic, nd white prents s lrge or lrger thn the one found in this study is bout 4 in (d) Since the P-vlue is so smll, we reject the null hypothesis. There is convincing evidence of difference in the distributions of opinions bout high school mong blck, ispnic, nd white prents () The two-wy tble of the dt is given below: Environment tched Did not htch Totl Cold Neutrl ot Totl The Prctice of Sttistics for AP*, 4/e

14 The proportions (conditionl distributions) re given in the tble below. Note tht the conditionl distributions go cross rows in this cse. Environment tched Did not htch Cold = Neutrl = 0.31 ot = 0.79 This dt does support the reserchers belief to certin extent. A smller proportion of eggs htched in the cold wter. But mjority of the eggs in cold wter did still htch. (b) Stte: We wnt to perform test of 0 : There is no difference in the proportion of eggs tht htch bsed on wter temperture : There is difference in the proportion of eggs tht htch bsed on wter temperture t the α = 0.05 level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from rndomized experiment. Lrge smple size: We used technology to get the following expected counts: Environment tched Did not htch Cold Neutrl ot All of these counts re t lest 5. Independent: Due to the rndom ssignment, these three groups of eggs cn be viewed s independent. Individul observtions in ech group should lso be independent: knowing whether one egg htches does not give informtion bout whether nother egg htches or not. ( ) ( 9 3.6) The conditions re met. Do: The test sttistic is χ = = We use = df nd find P-vlue of Conclude: Since the P- chi-squre distribution with ( 3 1)( 1) vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to sy tht different proportions of eggs htch in different wter tempertures () The two-wy tble of the dt is given below: Supplement Mle Femle Totl PBM NLCP PL-LCP TG-LCP Totl Chpter 11: Inference for Distributions of Ctegoricl Dt 65

15 The proportions (conditionl distributions) re given in the tble below. Note tht the conditionl distributions go cross rows in this cse. Supplement Mle Femle PBM = 0.55 NLCP = PL-LCP = TG-LCP = It does pper tht the groups re roughly blnced. (b) Stte: We wnt to perform test of 0 : There is no significnt difference in the proportions of femles in different supplement groups : There is significnt difference in the proportion of femles in different supplement groups t the α = 0.05 level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from rndomized experiment. Lrge smple size: We used technology to get the following expected counts: Supplement Mle Femle PBM NLCP PL-LCP TG-LCP All of these counts re t lest 5. Independent: Due to the rndom ssignment, these three groups of infnts cn be viewed s independent. Individul observtions in ech group should lso be independent: knowing whether one infnt is femle gives no informtion bout ny other infnt s gender. The ( ) ( ) conditions re met. Do: The test sttistic is χ = = We use chi = df nd find P-vlue of Conclude: Since the P-vlue is squre distribution with ( 4 1)( 1) 3 greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to sy tht the groups differ significntly bsed on gender We cnnot use chi-squre test with this dt becuse we do not hve the ctul counts of the trvelers in ech ctegory. We lso do not know if the smple ws tken rndomly, nor if the sme people could hve been counted more thn once (trveling both for leisure one time nd business nother) The observtions in this tble re not independent of ech other. Dt on ech womn occurs in ech row () The dt re in the tble below. Tretment Success Filure Totl Nicotine ptch Drug Ptch plus drug Plcebo Totl The Prctice of Sttistics for AP*, 4/e

16 (b) The best success rte is with both the ptch nd the drug. It lso ppers tht the ptch lone is not much better thn the plcebo. (c) The null hypothesis given sys tht ech of the four tretments leds to the sme probbility of success. (d) There re totl of 893 subjects in the smple. Also, there re 6 subjects who were successful. So, the expected count for the number of successes mong those who got 6 the nicotine ptch is computed s ( 44 ) = The other expected counts re computed in 893 similr fshion nd re given in the tble below. Tretment Success Filure Nicotine ptch Drug Ptch plus drug Plcebo () The dt re in the tble below. Tretment Stroke No stroke Totl Plcebo Aspirin Dipyridmole Both Totl Chpter 11: Inference for Distributions of Ctegoricl Dt 67

17 (b) It ppers tht ll 4 tretments hve bout the sme success rte. (c) The null hypothesis given sys tht ech of the four tretments leds to the sme probbility of success. (d) There re totl of 660 subjects in the smple. Also, there re 84 subjects who hd strokes. So, the expected count for the number of 84 who hd strokes mong those who got the plcebo is computed s ( 1649 ) = The other 660 expected counts re computed in similr fshion nd re given in the tble below. Tretment Stroke No stroke Plcebo Aspirin Dipyridmole Both Stte: We wnt to perform test of 0 : There is no difference in the smoking cesstion rtes of the different tretments : There is difference in the smoking cesstion rtes of the different tretments t the α = 0.05 level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from rndomized experiment. Lrge smple size: We computed the expected counts in Exercise All of them re t lest 5. Independent: Due to the rndom ssignment, these four groups of subjects cn be viewed s independent. Individul observtions in ech group should lso be independent: knowing whether one person quits smoking gives no informtion bout ny other subject s smoking sttus. The conditions re met. Do: The test sttistic ( ) ( ) is χ = = We use chi-squre distribution with ( 4 1)( 1) = 3df nd find P-vlue of pproximtely 0. Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the smoking cesstion rte is different for t lest one of the four tretments Stte: We wnt to perform test of 0 : There is no difference in the stroke rtes of the different tretments : There is difference in the stroke rtes of the different tretments t the α = 0.05 level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from rndomized experiment. Lrge smple size: We computed the 68 The Prctice of Sttistics for AP*, 4/e

18 expected counts in Exercise All of them re t lest 5. Independent: Due to the rndom ssignment, these four groups of subjects cn be viewed s independent. Individul observtions in ech group should lso be independent: knowing whether one person hs stroke gives no informtion bout whether ny other subject hd stroke or not. The conditions re met. Do: The test sttistic is ( ) ( ) χ = = We use chi-squre distribution with ( 4 1)( 1) = 3df nd find P-vlue of pproximtely 0. Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the stroke rte is different for t lest one of the four tretments The sttistic breks down into its individul components s follows: χ = = The lrgest component of this eqution comes from those who hd success using both the ptch nd the drug. Fr more people fell in this group thn would hve been expected. The next lrgest component comes from those who hd success using just the ptch. Fr fewer were in this group thn would hve been expected. This mens tht the ptch lone relly is not ll tht helpful, but the ptch in combintion with the drug is good tretment The sttistic breks down into its individul components s follows: χ = = The lrgest component of this eqution comes from those who hd strokes on both spirin nd Dipyridmole. Fr fewer were in this group thn would hve been expected. The next lrgest component comes from those who hd strokes while on the plcebo. Fr more people were in this group thn would hve been expected. This mens tht the two drugs together fr outperform the plcebo nd should probbly be used if the side effects re not too gret () The hypotheses re s follows: 0 : There is no difference in the improvement rtes for the two tretments : There is difference in the improvement rtes for the two tretments Assuming tht there is no difference in the improvement rtes between gstric freezing nd the plcebo, the probbility of observing smple tht shows difference in the improvement rtes between gstric freezing nd the plcebo s lrge or lrger thn the one found in this study is bout 57 in 100. These dt do not provide convincing evidence of difference in improvement rtes for gstric freezing nd plcebo (b) The P-vlue for this test is identicl to the P-vlue for the test in prt (). And since the hypotheses for the two-smple z test re the sme (though usully stted in symbols rther thn words), the conclusions to this test re the sme s for the test in prt () () The hypotheses re s follows: 0 : There is no ssocition between support for the deth penlty nd eductionl level : There is n ssocition between support for the deth penlty nd eductionl level Assuming tht there is no ssocition between support for the deth penlty nd eduction level, the probbility of observing difference in support of the deth penlty mong those with high school degree nd those with Bchelor s degree s lrge or lrger thn the one found in this study is pproximtely 0. This suggests tht there is n ssocition between support for the deth penlty nd eductionl level in the popultion. (b) The P-vlue for this test is identicl to the P-vlue for the test in prt (). And since the hypotheses for the two-smple z test re the sme (though usully stted in symbols rther thn words), the conclusions to this test re the sme s for the test in prt (). Chpter 11: Inference for Distributions of Ctegoricl Dt 69

19 11.45 () Buyers re much more likely to think the qulity of recycled coffee filters is higher while nonbuyers re more likely to think the qulity is lower. (b) Stte: We wnt to perform test of 0 : Buying recycled products is independent of their perceived qulity rting : Buying recycled products is not independent of their perceived qulity rting t the α = 0.05 level. Pln: We should use chi-squre test for independence if the conditions re stisfied. Rndom: The dt cme from rndom smple. Lrge smple size: The computer printout shows the expected counts. All of them re t lest 5. Independent: The smple hd 133 people in it. There re more thn 1330 dults who could hve been prt of the smple. The conditions re met. Do: The test sttistic is given by the output s being We use chi-squre distribution with ( 3 1)( 1) = df nd find P-vlue of 0.0. Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht whether people buy recycled coffee filters or not is not independent of their opinion of the qulity of recycled coffee filters () 70 The Prctice of Sttistics for AP*, 4/e

20 The higher the degree erned, the less likely people re to think tht strology is scientific. (b) Stte: We wnt to perform test of 0 : Eduction level nd belief bout strology re independent : Eduction level nd belief bout strology re not independent t the α = 0.05 level. Pln: We should use chi-squre test for independence if the conditions re stisfied. Rndom: The dt cme from rndom smple. Lrge smple size: The computer printout shows the expected counts. All of them re t lest 5. Independent: The smple hd 687 people in it. There re more thn 6870 dults in the US who could hve been prt of the smple. The conditions re met. Do: The test sttistic is given by the output s being We use chi-squre distribution with ( 3 1)( 1) = df nd find P-vlue of Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht the eduction level nd belief bout the scientific nture of strology re not independent () Of those who spend less thn hours on extrcurriculr ctivities per week, the proportion who ern C or better is 11 = 0.55 nd the proportion who ern D or worse is Of those who spend 0 to 1 hours on extrcurriculr ctivities per week, the proportion who ern C or better is nd the proportion who ern D or worse is Of those who spend more thn 1 hours on extrcurriculr ctivities per week, the proportion who ern C or better is nd the proportion who ern D or worse is A br grph is given below. It ppers tht those who re modertely involved in extrcurriculr ctivities ( to 1 hours per week) ern higher grdes thn those who re very ctive nd those who re not ctive. (b) A chi-squre test should not be performed in this setting becuse there were only 8 students observed in the over 1 hours ctegory. The expected count for t lest one of these two cells will, necessrily, be less thn 5. We lso do not know how the students were chosen for this survey other thn the fct tht they were in required chemicl engineering course. We do not know if they were rndomly selected or not, for instnce. Chpter 11: Inference for Distributions of Ctegoricl Dt 71

21 11.48 () The second row in the tble shown is just continution of the first row tht is, the sme people re still being mesured. Also, the tble does not show the people who did not get rid of the wrts, so this tble only shows the successes, not the filures. The tble we should use is s follows: Result Tretment Control Totl Wrts gone Wrts remin Totl (b) It is still not pproprite to use chi-squre test becuse we do not know how the experiment ws conducted. There is no mention of rndomiztion so we hve to wonder bout tht nd independence () The mjority of people in most eduction levels seem to oppose such lw. The one eduction level where there is n even split is the group tht hs less thn high school eduction. (b) Stte: We wnt to perform test of 0 : There is no ssocition between eduction level nd support of gun lw in the popultion of dults : There is n ssocition between eduction level nd support of gun lw in the popultion of dults t the α = 0.05 level. Pln: We should use chi-squre test for ssocition/independence if the conditions re stisfied. Rndom: The dt cme from rndom smple. Lrge smple size: We used technology to get the following expected counts: Eduction Yes No Less thn high school igh school grd Some college College grd Postgrdute degree All of these counts re t lest 5. Independent: Our smple includes 101 people. This is less thn 10% of the popultion of the US. The conditions re met. Do: The test sttistic is ( ) ( ) χ = = We use chi-squre distribution with = 4df nd find P-vlue of Conclude: Since the P-vlue is greter thn 0.05, we ( )( ) 7 The Prctice of Sttistics for AP*, 4/e

22 fil to reject the null hypothesis. We do not hve enough evidence to sy tht there is n ssocition between eductionl level nd support of gun lw in the popultion of dults () In soft wter, the new product is slightly preferred over the stndrd product nd there is not much difference between the wrm wter wsh nd the hot wter wsh. In hrd wter, the new product hs lrger mjority of the preference nd is slightly more preferred mong those using wrm wter wsh. (b) Stte: We wnt to perform test of 0 : There is no ssocition between type of wsh nd support for the new product mong people who don't use the estblished brnd : There is n ssocition between type of wsh nd support for the new product mong people who don't use the estblished brnd t the α = 0.05 level. Pln: We should use chi-squre test for ssocition/independence if the conditions re stisfied. Rndom: The dt cme from rndom smple. Lrge smple size: We used technology to get the following expected counts: Product preference Soft, wrm Soft, hot rd, wrm rd, hot Stndrd New All of these counts re t lest 5. Independent: Our smple includes 354 people. This is less thn 10% of the popultion of the US who don t currently use the estblished brnd. The conditions re met. Do: The test sttistic is ( ) ( ) χ = =.058. We use chi-squre distribution with ( 4 1)( 1) = 3df nd find P-vlue of Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to sy tht there is n ssocition between type of wsh nd support for the new product mong people who don t use the estblished brnd () Since this represents one smple nd the subjects were then clssified by their nswer nd their gender, this would be chi-squre test of ssocition/independence. (b) The hypotheses re Chpter 11: Inference for Distributions of Ctegoricl Dt 73

23 0 : Gender nd where subjects live re independent : Gender nd where subjects live re not independent (c) Rndom: This ws rndom smple. Lrge smple size: The expected counts re given in the output nd they re ll t lest 5. Independent: This smple of 4854 young dults is less thn 10% of ll young dults who could hve been smpled. (d) If gender nd plce of living re independent, then we hve 1.% chnce of finding smple with s much ssocition or more thn we found. This is firly unlikely (less thn 5% chnce) so we reject the null hypothesis nd conclude tht gender nd where young dults live re not independent () Since there were two seprte smples, one of Americn students nd one of Asin students, we should perform chi-squre test for homogeneity. (b) The hypotheses re 0 : There is no difference in the resons for shopping from ctlogs between the two groups of students : There is difference in the resons for shopping from ctlogs between the two groups of students (c) Rndom: This dt cme from rndom smples. Lrge smple size: The expected counts re given in the output nd they re ll t lest 5. Independent: There re t lest 950 Americn students nd 60 Asin students. (d) If the distribution of resons for shopping from ctlogs is the sme for both Asin nd Americn students, we hve 0.01% chnce of finding s big difference or bigger thn we found between the two smples. This is quite unlikely, so we reject the null hypothesis nd conclude tht there re differences between Americn nd Asin students in terms of why they shop from ctlogs e c d b b () One-smple t intervl for the men. (b) Two-smple z test for the difference between two proportions () Chi-squre test for ssocition/independence. (b) Two-smple t intervl for the difference between two mens This ws n experiment. The subjects were rndomly exposed to tretment (the type of response they were llowed to mke) A chi-squre goodness-of-fit test is only pproprite for testing whether single ctegoricl vrible hs specified distribution () The men for the 1-5 scle is x1 5= ( ( 1 ) + 3 ( ) + 1 ( 3 ) + 13 ( 4 ) + 3 ( 5 )) = Using technology, the stndrd devition is (b) The new men for the 0-4 scle is x( 0 4) + = = 4.1. Adding one to ll vlues does not chnge the stndrd devition so it remins (c) No, it would not be pproprite to compre the mens from () nd (b) using two-smple t 74 The Prctice of Sttistics for AP*, 4/e

24 test becuse the originl mesurements re ctegoricl not quntittive. Yes, they re recorded s numbers, but the distnce between 1 nd my not be the sme s the distnce between 3 nd 4 in people s minds It is certinly possible tht those who do not respond hve similr views bout the cfeteri food. For exmple, mybe people who relly do not like the food feel uncomfortble nswering this question. Chpter Review Exercises (pge 731) R11.1 () In this school the lrgest clss consists of the freshmen nd the clss sizes decrese ech yer. But in the smple, the lrgest groups were the sophomores nd the juniors. This suggests tht the smple my not be representtive of the school s whole. (b) Stte: We wnt to perform test t the α = 0.05 significnce level of 0 : pfresmn = 0.9, psophomore = 0.7, pjunior = 0.5, psenior = 0.19 versus : At lest one of the p i's is incorrect. Pln: We should use chi-squre goodness-of-fit test if the conditions re stisfied. Rndom: A rndom smple ws used. Lrge smple size: The expected counts , = 55.6, junior: in ech ctegory re freshmn: ( ) = sophomore: ( ) 06( 0.5) = 51.5, nd senior: 06( 0.19) 39.14, = ll of which re t lest 5. Independent: It seems resonble tht there would be more thn 060 students in lrge high school. The conditions re met. ( ) ( ) ( ) ( ) Do: The test sttistic is χ = = nd the distribution hs 4 1= 3df. The P-vlue, then, is Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to reject the ide tht the smple is representtive of the popultion of students. Chpter 11: Inference for Distributions of Ctegoricl Dt 75

25 R11. The expected counts re listed in the tble below. Techer type Knowledgeble Ignornt Note tht 6 of the cells hve expected counts tht re less thn 5. R11.3 () Tretment Suffered crdic event No crdic event Totl Stress mngement Exercise Usul cre Totl (b) The success rte for stress mngement is 30 = 0.909, for exercise nd for the usul cre (c) Stte: We wnt to perform test of 0 : There is no difference in the ctul proportion of crdic events for the three tretments : There is difference in the ctul proportion of crdic events for the three tretments t the α = 0.05 level. Pln: We should use chi-squre test for homogeneity if the conditions re stisfied. Rndom: The dt cme from rndomized experiment. Lrge smple size: We used technology to get the following expected counts: Tretment Crdic event No Crdic Stress mngement Exercise Usul cre All of these counts re t lest 5. Independent: Due to the rndom ssignment, these three groups of ptients cn be viewed s independent. Individul observtions in ech group should lso be independent: knowing whether person hs crdic event gives no informtion bout whether nother person hs crdic event or not. The conditions re met. Do: The test sttistic is ( ) ( ) χ = = We use chi-squre distribution with ( 3 1)( 1) = df nd find P-vlue of Conclude: Since the P-vlue is greter thn 0.05, we fil to reject the null hypothesis. We do not hve enough evidence to sy tht there is difference in the ctul proportion of crdic events for the three tretments. R11.4 () This should be chi-squre test for ssocition/independence becuse one rndom smple ws used nd people in the ds were clssified both by gender of the trget udience nd whether the d ws sexul or not. (b) The hypotheses re 0 : Gender of trget udience nd whether or not the d is sexul re independent : Gender of trget udience nd whether or not the d is sexul re not independent (c) The first number is the percentge of ds in women s mgzines which re found to be not sexy. This is computed s 351 = The second number is the expected count for the number of ds in women s mgzines found to be not sexy nd is computed s ( 576 ) = The third number is 1509 the contribution to the chi-squre sttistic from this sme cell nd is computed s ( ) = (The difference is due to rounding error.) (d) Since the P-vlue is so smll (essentilly 0), we would 76 The Prctice of Sttistics for AP*, 4/e

26 reject the null hypothesis nd conclude tht the gender of the trget udience nd whether mgzine ds re sexul re not independent. R11.5 () Both groups of children hve the lrgest percentge reporting grdes s the gol. But fter tht, boys were more likely to pick sports wheres girls were more likely to pick being populr. (b) Stte: We wnt to perform test of 0 : There is no ssocition between gender nd gol t school for 4th, 5th, nd 6th grde students : There is n ssocition between gender nd gol t school for 4th, 5th, nd 6th grde students t the α = 0.05 level. Pln: We should use chi-squre test for ssocition/independence if the conditions re stisfied. Rndom: The dt cme from rndom smple. Lrge smple size: We used technology to get the following expected counts: Gender Grdes Populr Sports Femle Mle All of these counts re lest 5. Independent: Our smple includes 478 children. This is less thn 10% of the popultion children in these grdes in the US. The conditions re met. Do: The test sttistic is ( ) ( ) χ = = We use chi-squre distribution with ( 3 1)( 1) = df nd find P-vlue of Conclude: Since the P-vlue is less thn 0.05, we reject the null hypothesis nd conclude tht there is n ssocition between gender nd gol t school for 4 th, 5 th, nd 6 th grde students. (c) The chi-squre sttistic, broken down, is χ = = The lrgest contributions re from the mle/sports cell nd the femle/sports cell. More mles thn expected chose sports nd fewer femles thn expected chose sports. R11.6 () The two-wy tble is Type of subject Answer yes Answer no Totl Student Non-student Totl Chpter 11: Inference for Distributions of Ctegoricl Dt 77