r語(yǔ)言與統(tǒng)計(jì)分析課后答案

1、第五章5.1 設(shè)總體x是用無(wú)線(xiàn)電測(cè)距儀測(cè)量距離的誤差，它服從( 出 (3)上的均勻分布，在200次測(cè)量中，誤差為xi的次數(shù)有ni次：Xi : 3 5 7 9 11 13 15 17 19 21Ni : 21 16 15 26 22 14 21 22 18 25求, B的矩法估計(jì)值a=U- 3s片u+ 3s程序代碼：x=seq(3,21,by=2)y=c(21,16,15,26,22,14,21,22,18,25)u=rep(x,y)u1=mean(u)s=var(u)s1=sqrt(s)a=u1-sqrt(3)*s1b=u1+sqrt(3)*s1b=u1+sqrt(3)*s1得出結(jié)果：a= 2

2、.217379b= 22.402625.2 為檢驗(yàn)?zāi)匙詠?lái)水消毒設(shè)備的效果，現(xiàn)從消毒后的水中隨機(jī)抽取50L,化驗(yàn)每升水中大腸桿菌的個(gè)數(shù)(假設(shè)1L水中大腸桿菌的個(gè)數(shù)服從泊松分布)，其化驗(yàn)結(jié)果如下表所示：試問(wèn)平均每升水中大腸桿菌個(gè)數(shù)為多少時(shí)，才能使上述情況的概率達(dá)到最大大腸桿菌數(shù)/L : 0 1 2 3 4 5 6水的升數(shù)：17 20 10 2 1 0 0丫 =u是最大似然估計(jì)程序代碼：a=seq(0,6,by=1)b=c(17,20,10,2,1,0,0)c=a*bd=mean(c)得出結(jié)果：d= 7.1428575.3 已知某種木材的橫紋抗壓力服從正態(tài)分布，現(xiàn)對(duì)十個(gè)試件做橫紋抗壓力試驗(yàn)，得數(shù)據(jù)如

3、下：482 493 457 471 510 446 435 418 394 469(1)求u的置信水平為0.95的置信區(qū)間程序代碼：x=c(482 493 457 471 510 446 435 418 394 469 )t.test(x)得出結(jié)果：data: xt = 6.2668, df = 9, p-value = 0.0001467 alternative hypothesis: true mean is not equal to 095 percent confidence interval:7.668299 16.331701sample estimates:mean of x12

4、由答案可得：u的置信水平為0.95的置信區(qū)間7.668299 16.331701(2)求(T的置信水平為0.90的置信區(qū)間程序代碼：chisq.var.test<-function(x,var,alpha,alternative="two.sided")options(digits=4)result<-list()n<-length(x)v<-var(x)result$var<-vchi2<-(n-1)*v/varresult$chi2<-chi2p<-pchisq(chi2,n-1)result$p.value<-pi

5、f(alternative="less")result$p.value<-pchaisq(chi2,n-1,loer.tail=F) else if(alternative="two.sider")result$p.value<-2*min(pchaisq(chi2,n-1),pchaisq(chi2,n-1,lower.tail=F)result$<-c(n-1)*v/qchisq(alpha/2,df=n-1,lower.tail=F),(n-1)*v/qchisq(alpha/2,df=n-1,lower.tail

6、=T)result)x<-c(482,493,457,471,510,446,435,418,394,469)y=var(x)chisq.var.test(x,0.048八2,0.10,alternative="two.side")得出結(jié)果：$ : 659,8 3357.0由答案可得：(T的置信水平為0.90的置信區(qū)間659,8 3357,05.4某卷煙廠生產(chǎn)兩種卷煙A和B現(xiàn)分別對(duì)兩種香煙的尼古丁含量進(jìn)行6次試驗(yàn)，結(jié)果如下：A:25 28 23 26 29 22B:28 23 30 35 21 27若香煙的尼古丁含量服從正態(tài)分布(1)問(wèn)兩種卷煙中尼古丁

7、含量的方差是否相等(通過(guò)區(qū)間估計(jì)考察)(2)試求兩種香煙的尼古丁平均含量差的95婿信區(qū)間(1)程序代碼:X=c(25,28,23,26,29,22)Y=c(28,23,30,35,21,27)Var.test(x,y)得出結(jié)果：F test to compare two variancesdata: x and yF = 0.2992, num df = 5, denom df = 5, p-value = 0.2115alternative hypothesis: true ratio of variances is not equal to 195 percent confidence i

8、nterval:0.04187 2.13821sample estimates:ratio of variances0.2992由答案可得：其方差不相等，方差區(qū)間為0.04187 2.13821 (2)5.5比較兩個(gè)小麥品種的產(chǎn)量，選擇 24塊條件相似地實(shí)驗(yàn)條，采用相同的耕作方法做實(shí)驗(yàn)，結(jié)果播種甲品種的12塊實(shí)驗(yàn)田的單位面積產(chǎn)量和播種乙品種的12塊試驗(yàn)田的單位面積產(chǎn)量分別為：A: 628 583 510 554 612 523 530 615 573 603 334 564B: 535 433 398 470 567 480 498 560 503 426 338 547假定每個(gè)品種的單位面積

9、產(chǎn)量服從正態(tài)分布，甲品種產(chǎn)量的方差為2140,乙品種產(chǎn)量的方差為3250,試求這兩個(gè)品種平均面積產(chǎn)量差的置信水平為0.95的置信上限和置信水平為0.90的置信下限。程序代碼：two.sample.ci=function(x,y,conf.level=0.95,sigma1.sigma2)options(digits=4)m=length(x); n=length(y)xbar=mean(x)-mean(y)alpha=1-conf.levelzstar=qnorm(1-alpha/2)*(sigma1/m+sigma2/n)八(1/2)xbar+c(-zstar, +zstar)x=c(628

10、,583,510,554,612,523,530,615,573,603,334,564)y=c(535,433,398,470,567,480,498,560,503,426,338,547)sigma1=2140sigma2=3250two.sample.ci(x,y,conf.level=0.95,sigma1.sigma2)得到結(jié)果：31.29 114.37程序代碼：two.sample.ci=function(x,y,conf.level=0.95,sigma1.sigma2)options(digits=4)m=length(x); n=length(y)xbar=mean(x)-

11、mean(y)alpha=1-conf.levelzstar=qnorm(1-alpha/2)*(sigma1/m+sigma2/n)八(1/2)xbar+c(-zstar, +zstar)x=c(628,583,510,554,612,523,530,615,573,603,334,564)y=c(535,433,398,470,567,480,498,560,503,426,338,547)sigma1=2140sigma2=3250two.sample.ci(x,y,conf.level=0.90,sigma1.sigma2)得到結(jié)果：37.97 107.695.6有兩臺(tái)機(jī)床生產(chǎn)同一型號(hào)

12、的滾珠，根據(jù)以往經(jīng)驗(yàn)知，這兩臺(tái)機(jī)床生產(chǎn)的滾珠直徑都服從正態(tài)分布，現(xiàn)分別從這兩臺(tái)機(jī)床生產(chǎn)的滾珠中隨機(jī)地抽取7個(gè)和9個(gè)，測(cè)得它們的直徑如下：機(jī)床甲:15.2 14.5 15.5 14.8 15.1 15.6 14.7機(jī)床乙：15.2 15.0 14.8 15.2 15 14.9 15.1 14.8 15.3試問(wèn)機(jī)床乙生產(chǎn)的滾珠的方差是否比機(jī)床甲生產(chǎn)的滾珠直徑的方差??？程序代碼：x=c(5.2,14.5,15.5,14.8,15.1,15.6,14.7)y=c(15.2,15.0,14.8,15.2,15,14.9,15.1,14.8,15.3)var.test(x,y)得出結(jié)果：F test to

13、 compare two variances data: x and yF = 430.1, num df = 6, denom df = 8, p-value = 2.723e-09 alternative hypothesis: true ratio of variances is not equa l to 195 percent confidence interval:92.47 2408.54sample estimates: ratio of variances430.1由結(jié)果可得：其甲機(jī)床的滾珠半徑遠(yuǎn)超出乙機(jī)床的滾珠半徑5.7某公司對(duì)本公司生產(chǎn)的兩種自行車(chē)型號(hào) A, B的銷(xiāo)售情況

14、進(jìn)行了 了解，隨機(jī)選取了 400人詢(xún)問(wèn)他們對(duì)A B的選擇，其中有224人喜歡 A,試求顧客中喜歡A的人數(shù)比例p的置信水平為0.99的區(qū)間估計(jì)。 方程代碼：Binom.test(224,400,conf.level=0.99)得出結(jié)果：Exact binomial testdata: 224 and 400number of successes = 224, number of trials = 400, p-value =0.01866alternative hypothesis: true probability of success is not equal to 0.599 percen

15、t confidence interval:0.4944077 0.6241356sample estimates:probability of success0.56由結(jié)果可得：顧客中喜歡a的人數(shù)比例p的置信水平為0.99的區(qū)間 估計(jì)：0.4944077 0.62413565.8 某公司生產(chǎn)了一批新產(chǎn)品，產(chǎn)品總體服從正態(tài)分布，現(xiàn)估計(jì)這批 產(chǎn)品的平均重量，最大允許誤差為1,樣本標(biāo)準(zhǔn)差s=10,試問(wèn)在0.95 的置信水平下至少要抽取多少個(gè)產(chǎn)品程序代碼：Size,norm2=function(s,alpha,d,m)t0=qt(alpha/2,m,lower.tail = FALSE)n0=(t0*s/d)A2t1=qt(alpha/2,n0,lower.tail = FALSE)n1=(t1*s/d)A2while(abs(n1-n0)>0.5)n0=(qt(alpha/2,n1,lower.tail = FALSE)*s/d2n1=(qt(alpha/2,n0,lower.tail = FALSE)*s/d2n1 )Size.norm2(10,0.01,2,100)得出結(jié)果：98.44268由結(jié)果可得，在0.95的置信