研究生R語(yǔ)言考題

1、暨 南 大 學(xué) 考 試 試 卷教師填寫2010 - 2011_ 學(xué)年度第_2_學(xué)期課程名稱： 數(shù)據(jù)分析與r語(yǔ)言應(yīng)用 授課教師姓名：_王斌會(huì)_ 考試時(shí)間:_2011_年_11_月_8_日課程類別必修 選修 Ö 考試方式開(kāi)卷 Ö 閉卷 試卷類別(a、b) a 共 4 頁(yè)考生填寫 經(jīng)濟(jì)學(xué)院 學(xué)院(校) 數(shù)量經(jīng)濟(jì)學(xué) 專業(yè) 班(級(jí))姓名 劉偉 學(xué)號(hào) 1130111008 內(nèi)招 外招 題 號(hào)一二三四五六七八九十總 分得 分得分評(píng)閱人一、統(tǒng)計(jì)圖表（共1小題，共20分）1應(yīng)用r圖表對(duì)各類產(chǎn)品供貨走勢(shì)圖分析類別月份123456789101112彩電a1冰箱a2空調(diào)a3洗衣機(jī)a4（1） 要求：

2、數(shù)據(jù)由r隨機(jī)數(shù)函數(shù)生成，產(chǎn)生20,50間的均勻隨機(jī)數(shù)。解：首先對(duì)r進(jìn)行初始化，設(shè)定參數(shù)，再生成隨機(jī)數(shù)，代碼如下：rm(list=ls()options(digits=4)par(mar=c(4,4,2,1)+0.1,cex=0.75)a1=runif(12,20,50);a1a2=runif(12,20,50);a2a3=runif(12,20,50);a3a4=runif(12,20,50);a4（2）分析（圖形要進(jìn)行一定修飾）：1）繪制各類產(chǎn)品的月份趨勢(shì)線圖。解：趨勢(shì)線圖如下代碼如下：par(mfrow=c(2,2)plot(a1,type="l",ylab="

3、;銷售量",xlab="月份",main="彩電(a1)",xlim=c(1,12),ylim=c(0,50)plot(a2,type="l",ylab="銷售量",xlab="月份",main="冰箱(a2)",xlim=c(1,12),ylim=c(0,50)plot(a3,type="l",ylab="銷售量",xlab="月份",main="空調(diào)(a3)",xlim=c(1,12

4、),ylim=c(0,50)plot(a4,type="l",ylab="銷售量",xlab="月份",main="洗衣機(jī)(a4)",xlim=c(1,12),ylim=c(0,50)2）繪制各類產(chǎn)品的季度的柱形圖。解：首先對(duì)數(shù)據(jù)進(jìn)行整理，得出各自的季度數(shù)據(jù)。柱狀圖如下代碼如下：dat=data.frame(a1,a2,a3,a4)q1=c(dat1,1+dat2,1+dat3,1,dat4,1+dat5,1+dat6,1,dat7,1+dat8,1+dat9,1,dat10,1+dat11,1+dat12,1)q

5、2=c(dat1,2+dat2,2+dat3,2,dat4,2+dat5,2+dat6,2,dat7,2+dat8,2+dat9,2,dat10,2+dat11,2+dat12,2)q3=c(dat1,3+dat2,1+dat3,3,dat4,3+dat5,1+dat6,3,dat7,3+dat8,3+dat9,3,dat10,3+dat11,3+dat12,3)q4=c(dat1,4+dat2,4+dat3,4,dat4,4+dat5,4+dat6,4,dat7,4+dat8,4+dat9,4,dat10,4+dat11,4+dat12,4)dat1=data.frame(q1,q2,q3,

6、q4);dat1par(mfrow=c(2,2)barplot(dat1,1,xlab="季度",ylab="銷售量",main="彩電(a1)",ylim=c(0,150)barplot(dat1,2,xlab="季度",ylab="銷售量",main="冰箱(a2)",ylim=c(0,150)barplot(dat1,3,xlab="季度",ylab="銷售量",main="空調(diào)(a3)",ylim=c(0,1

7、50)barplot(dat1,4,xlab="季度",ylab="銷售量",main="洗衣機(jī)(a4)",ylim=c(0,150)3）繪制各類產(chǎn)品的年度的餅圖。解：餅圖如下代碼如下：par(mfrow=c(1,1)y1=dat11,1+dat12,1+dat13,1+dat14,1y2=dat11,2+dat12,2+dat13,2+dat14,2y3=dat11,3+dat12,3+dat13,3+dat14,3y4=dat11,4+dat12,4+dat13,4+dat14,4x=c(y1,y2,y3,y4)pie(x,lab

8、els=c("彩電(a1)","冰箱(a2)","空調(diào)(a3)","洗衣機(jī)(a4)"),col=c("red","green","purple","blue")得分評(píng)閱人二、統(tǒng)計(jì)檢驗(yàn)（共2小題，每題10分，共20分）1. 兩臺(tái)銑床生產(chǎn)同一種型號(hào)的套管，平日兩臺(tái)銑床加工的套管內(nèi)槽深度都服從正態(tài)分布n(10，0.32)和n(8，0.22)，從這兩臺(tái)銑床的產(chǎn)品中分別抽出13個(gè)和15個(gè)，請(qǐng)分別按方差已知和未知檢驗(yàn)兩臺(tái)產(chǎn)品的深度是否不同(=0.0

9、5)？（1）兩臺(tái)銑床的產(chǎn)品內(nèi)槽精度(方差)有無(wú)顯著差別?解： x=rnorm(13,10,0.3)y =rnorm(15,8,0.2)var.test(x,y) f test to compare two variancesdata: x and y f = 3.899, num df = 12, denom df = 14, p-value =0.01785alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 1.278 12.502 sample e

10、stimates:ratio of variances 3.899由于 p-value =0.01785<0.05,故兩臺(tái)銑床的產(chǎn)品內(nèi)槽精度(方差)有顯著差別。（2） 兩臺(tái)產(chǎn)品的的深度是否不同? 解：1、方差未知時(shí) t.test(x,y)welch two sample t-testdata: x and y t = 17.61, df = 17.2, p-value = 1.934e-12alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 1

11、.605 2.042 sample estimates:mean of x mean of y 9.982 8.159 由于p-value = 1.934e-12<0.05,故兩臺(tái)產(chǎn)品的的深度是不同的。 2、方差已知時(shí) u.test=function(x,y,sigmax,sigmay) nx=length(x) ny=length(y) xbar=mean(x) ybar=mean(y) u=(xbar-ybar)/sqrt(sigmax2/nx+sigmay2/ny) p=pnorm(u,lower.tail=f) c(u=u,p=p) u.test(x,y,0.3,.02) u p

12、 2.187e+01 2.332e-106由于p=2.332e-106<0.05, 故兩臺(tái)產(chǎn)品的的深度是不同的。2. 如果還有一臺(tái)銑床生產(chǎn)同一種型號(hào)的套管，其加工的套管內(nèi)槽深度都服從正態(tài)分布n(12，0.42)，從這臺(tái)銑床的產(chǎn)品中抽出18個(gè)，請(qǐng)分別按方差已知和未知檢驗(yàn)三臺(tái)產(chǎn)品的深度是否不同(=0.05)？(1)、方差已知的情況 x1=rnorm(13,10,0.3)x2 =rnorm(15,8,0.2)x3=sample(rnorm(1000,12,0.4),18)n1=length(x1)n2=length(x2)n3=length(x3)se1=sqrt(0.32/n1+0.22/n

13、2)se2=sqrt(0.32/n1+0.42/n3)se3=sqrt(0.22/n2+0.42/n3)x1bar=mean(x1) x2bar=mean(x2)x3bar=mean(x3) u1=(x1bar-x2bar)/se1 u2=(x1bar-x3bar)/se3 u3=(x2bar-x3bar)/se3chi=u12+u22+u32 p=2*pchisq(chi,3,lower.tail = f);p p< a=0.05，所以拒絕三臺(tái)產(chǎn)品的的深度相等的假設(shè)，三臺(tái)臺(tái)產(chǎn)品的深度不等。 (2)、方差未知的情況y=c(x1,x2,x3) group=c(rep(1,13),rep(2

14、,15),rep(3,18) oneway.test(ygroup) one-way analysis of means (not assuming equal variances)data: y and group f = 564.7, num df = 2.00, denom df = 28.28, p-value < 2.2e-16 p< a=0.05，所以拒絕三臺(tái)產(chǎn)品的的深度相等的假設(shè)，三臺(tái)臺(tái)產(chǎn)品的深度不等。得分評(píng)閱人三、統(tǒng)計(jì)建模（共1小題，共20分）為了解百貨商店銷售額x與流通費(fèi)率y（這是反映商業(yè)活動(dòng)的一個(gè)質(zhì)量指標(biāo)，指每元商品流轉(zhuǎn)額所分?jǐn)偟牧魍ㄙM(fèi)用）之間的關(guān)系，收集25個(gè)

15、商店的有關(guān)數(shù)據(jù)，這里假定x來(lái)自10,30上的均勻隨機(jī)數(shù)，en(0，0.32)的正態(tài)隨機(jī)數(shù)。（1） 設(shè)y=2+3*x+e，試用r擬合y=a+bx的線性回歸模型解： x=runif(25,10,30) e=rnorm(25,0,0.3) y=c(2+3*x+e)fm=lm(yx) fmcall:lm(formula = y x)coefficients:(intercept) x 2.06 3.00 故y=2.06+3.00x.（2） 設(shè)y=2+3*ln(x)+e，試用r擬合y=a+bln(x)的對(duì)數(shù)回歸模型解： x=runif(25,10,30) e=rnorm(25,0,0.3)y=c(2+3

16、*log(x)+e) fm=lm(yx) fmcall:lm(formula = y log(x)coefficients:(intercept) log(x) 2.03 3.02 故y=2.03+3.02ln(x).（3） 設(shè)y=2+3*x2+0.5*x3+e，試用r擬合y=a+b*x2+c*x3+e的回歸模型解： x=runif(25,10,30) e=rnorm(25,0,0.3) y=c(2+3*x2+0.5*x3+e) fm=lm(yi(x2)+i(x3) fmcall:lm(formula = y i(x2) + i(x3)coefficients:(intercept) i(x2

17、) i(x3) 2.296 2.999 0.500 故y=2.296+2.999*x2+0.500*x3 . 得分評(píng)閱人四、數(shù)據(jù)分析（共2小題，每題20分，共40分）1. let x be a matrix of random normal values (mean =0; sd=1) having 10 columns and n=100 rows. reset the values in the first row in the matrix to (1,1.5,1.4,3,1.9,4,4.9,2.6,3.2,2.4). assume that the first 5 columns of

18、 data for each row correspond to a group a, while the remaining 5 to another group b.1.) for each row of the matrix x, compute:a) the t-statistic comparing the groups a and b assuming equal variance and the p-value b) compute the probability to observe such a t-statistics only by chance, using a per

19、mutation analysis. the following strategy will be used: the columns will be randomly permuted nk=1000 times, and at each iteration the t-statistic will be computed again and recorded in a vector. at the end, compute the p-value as the number of times out of nk when the t-statistic with the permuted

20、data was at least as or more extreme than the t-statistics obtained with the real (non-permuted data). present the result as a data.frame with 4 columns: id= row number, t= t-score, p_theoretical=p-value assuming the asymptotic distribution; p_permutations=p-value from permutations;sort the data.fra

21、me in descending order of p-values. 解：x1=matrix(rnorm(1000,0,1),100,10) x2=ma12:100,1:10v=c(1,1.5,1.4,3,1.9,4,4.9,2.6,3.2,2.4)ab=rbind(v,x2) #構(gòu)造如題中所述的矩陣ab，其前5列為a矩陣，后5列為b矩陣p_theoretical =apply(ab,1,function(x) t.test(x1:5,x6:10,var.equal=t)$p.value ) #計(jì)算ab矩陣每行兩個(gè)樣本t檢驗(yàn)的p值t=apply(ab,1,function(x) t.test

22、(x1:5,x6:10,var.equal=t)$statistic ) #計(jì)算ab矩陣每行兩個(gè)樣本t檢驗(yàn)的t值t1=abs(t) # 取t值的絕對(duì)值m=list() tt=list() # 構(gòu)造兩個(gè)列表m和tt用于下面的置換循環(huán)for(i in c(1:1000)mi=ab,sample(ncol(ab) #將ab的10列隨機(jī)排列，得到1000個(gè)新矩陣tti=abs(apply(mi,1,function(x) t.test(x1:5,x6:10,var.equal=t)$statistic ) #計(jì)算每個(gè)矩陣的每行兩個(gè)樣本t檢驗(yàn)的t值的絕對(duì)值，上述過(guò)程需要15分鐘左右時(shí)間u=do.call

23、(cbind,tt) #將列表tt轉(zhuǎn)換為矩陣up_permutations =array() #構(gòu)造數(shù)組p_permutationsfor(j in c(1:100) p_permutations j=sum(uj, >=t1j)/1000 #計(jì)算矩陣u每行中的數(shù)據(jù)大于t1中對(duì)應(yīng)t值絕對(duì)值的概率，并將值賦予p_permutations數(shù)組id=c(1:100) # 構(gòu)造row number result=data.frame(id,t, p_theoretical, p_permutations) #輸出結(jié)果 id t p_theoretical p_permutations1 1 -2

24、.8853207 0.02034 0.0152 2 -1.3143173 0.22517 0.2273 3 -2.6940739 0.02732 0.0064 4 -1.3802458 0.20485 0.2285 5 0.0473332 0.96341 0.9776 6 0.9959910 0.34842 0.2917 7 -0.4321985 0.67701 0.7228 8 -0.4268158 0.68077 0.6789 9 -0.1412908 0.89113 0.89910 10 1.3169132 0.22434 0.25411 11 -0.0489858 0.96213 0.

25、98812 12 1.4028308 0.19826 0.16913 13 1.1741980 0.27408 0.25614 14 -1.0676853 0.31682 0.31715 15 2.6462644 0.02943 0.01216 16 -1.9388080 0.08851 0.06117 17 -0.6731468 0.51982 0.53618 18 -0.3519451 0.73397 0.75519 19 0.5819389 0.57663 0.57720 20 0.6656648 0.52435 0.57721 21 -0.1460266 0.88751 0.74822

26、 22 -2.7035412 0.02693 0.02323 23 -0.7900807 0.45226 0.47224 24 -0.4385279 0.67260 0.62925 25 0.3676752 0.72265 0.74226 26 2.2177703 0.05738 0.03527 27 -0.1415220 0.89096 0.91828 28 -0.0004833 0.99963 1.00029 29 -0.4102665 0.69238 0.68730 30 0.0448795 0.96530 0.94431 31 1.4521009 0.18454 0.19332 32

27、0.2854683 0.78254 0.79333 33 0.1298656 0.89988 0.89634 34 -0.9887282 0.35175 0.34635 35 -1.4024164 0.19838 0.15836 36 0.1285964 0.90085 0.85037 37 3.0086702 0.01685 0.03238 38 0.6787512 0.51645 0.50339 39 1.0937833 0.30589 0.33140 40 1.9842917 0.08250 0.10241 41 -0.0968299 0.92524 0.96942 42 -3.1236

28、870 0.01415 0.01643 43 0.6596542 0.52801 0.50144 44 1.2064446 0.26211 0.26745 45 -0.2720078 0.79250 0.81246 46 -0.1762814 0.86445 0.86047 47 -0.2106570 0.83842 0.87548 48 1.9866318 0.08220 0.07249 49 -0.0020090 0.99845 1.00050 50 -2.0646728 0.07283 0.03151 51 -0.2564763 0.80406 0.79752 52 2.8462192

29、0.02160 0.02853 53 -0.0657272 0.94921 0.96654 54 0.6318659 0.54510 0.58955 55 0.7555194 0.47159 0.42556 56 1.5005975 0.17185 0.17957 57 1.2630394 0.24214 0.24358 58 -0.6921399 0.50844 0.53959 59 -1.3837198 0.20382 0.19160 60 0.4838255 0.64148 0.63961 61 -0.1140807 0.91198 0.94462 62 1.1902273 0.2680

30、8 0.25863 63 0.2112508 0.83797 0.83664 64 0.8978211 0.39550 0.37365 65 -0.4111239 0.69177 0.71266 66 3.1668091 0.01326 0.01667 67 1.6002204 0.14822 0.14268 68 0.8488613 0.42063 0.39769 69 0.7102140 0.49775 0.49370 70 0.0780357 0.93972 0.93071 71 1.2471030 0.24763 0.25972 72 0.9615742 0.36442 0.33073

31、 73 1.2386493 0.25059 0.27274 74 -0.6696054 0.52196 0.53775 75 1.2152901 0.25890 0.25476 76 -0.1259623 0.90287 0.92477 77 -0.0791718 0.93884 0.96478 78 -3.6579469 0.00642 0.01679 79 0.2499050 0.80896 0.78080 80 -2.3661564 0.04552 0.05281 81 0.0254582 0.98031 0.98882 82 -1.9755606 0.08362 0.08583 83

32、0.0015678 0.99879 1.00084 84 0.1471918 0.88662 0.87185 85 0.0295263 0.97717 0.98486 86 0.2628059 0.79934 0.80987 87 0.0846366 0.93463 0.92888 88 -0.5474582 0.59900 0.63589 89 -2.7182033 0.02632 0.01390 90 0.5733288 0.58218 0.67891 91 2.3551061 0.04631 0.04892 92 0.7709098 0.46292 0.43593 93 -0.41786

33、80 0.68703 0.63494 94 0.2916059 0.77801 0.73695 95 0.2728228 0.79190 0.76296 96 1.5964084 0.14906 0.15597 97 1.3246252 0.22188 0.26698 98 0.6856973 0.51228 0.51199 99 -1.2272690 0.25461 0.248100 100 0.5488290 0.59810 0.5912.) plot the distribution (see hist) of the resulting vector of t-scores obtai

34、ned at step 1a) after excluding the first element (corresponding to the first row) and on the same graph show a vertical line for the t-value of the first row. 解：t2=t2:100hist(t2,main="t值分布的直方圖")abline(v=t1)2. olympic medalsduring both summer and winter olympic games the medal table is oft

35、en of interest to spectators and the media. the medal table is a tally of the number of medals which have been won by each participating country during the games. a good performance on the medal table is often a source of pride for a country. however, it is to be expected that large countries will w

36、in more medals than smaller countries, due to the fact that they have a larger pool from which to recruit athletes. thus smaller countries often argue that a better measure of performance would be medals per capita. however, it is possible that medal tally shouldn't be expected to increase in di

37、rect proportion to population. further, it is reasonable to think that the medal tally will also depend on the resources available to athletes in a country, or on the climate (for example, access to snow).the objective of this analysis is to explore the relationship between a country's medal tal

38、ly, population size, wealth (measured by gdp) and climate (approximated by latitude). further, it is proposed that in future a standardised measure of a country's medal tally should be developed which corrects for population size, climate and wealth. your should investigate the feasibility of th

39、is proposal, and discuss your finndings.the file medals.rdataload(medals.rdata) in r is an r data frame with one row for every country that has won at least one olympic medal in the previous four olympic games.the variable descriptions are as follows:country name of the competing country (only count

40、ries which have won at least one medal since 2004 are included).latitude latitude of the capital city.summer2004 total number of medals (gold, silver and bronze) won at the summer olympics in 2004.summer2008 total number of medals won at the summer olympics in 2008.winter2006 total number of medals

41、won at the winter olympics in 2006.winter2010 total number of medals won at the winter olympics in 2010.population2007 the population in 2007 (source: world fact book).gdp2009 gross domestic product in billions of us dollars (source: world fact book).一、引言眾所周知，歷屆奧運(yùn)會(huì)都以獲得的獎(jiǎng)牌的總數(shù)來(lái)衡量一個(gè)國(guó)家的體育發(fā)展水平，同時(shí)獲得更多的獎(jiǎng)牌也

42、成為一個(gè)國(guó)家的驕傲。然而，不同的國(guó)家所處氣候緯度、人口規(guī)模、gdp總量等因素是不同的，而僅僅靠獎(jiǎng)牌的總量來(lái)衡量一國(guó)的體育水平顯然不夠全面，也不太公正。本文旨在找出與影響獲得獎(jiǎng)牌數(shù)的一些因素，如：氣候（用緯度表示）、人口規(guī)模、gdp總量，通過(guò)相關(guān)分析和回歸分析研究它們的內(nèi)在關(guān)系，并在此基礎(chǔ)上提出更加全面的衡量一國(guó)體育發(fā)展水平的新指標(biāo)。二、相關(guān)性分析 為了分析各個(gè)國(guó)家的金牌數(shù)與其氣候（緯度表示）、人口規(guī)模和收入（gdp）的關(guān)系，首先對(duì)其進(jìn)行相關(guān)分析。以下就通過(guò)相關(guān)系數(shù)矩陣和散點(diǎn)圖考察它們之間的關(guān)系。1、2008年夏季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)及其影響因素相關(guān)分析fix(medals) #對(duì)原始數(shù)據(jù)進(jìn)行適當(dāng)編輯，

43、用負(fù)數(shù)表示南緯，朝鮮的gdp為280億美元，以此填充nall值x1=medals$summer2008;x1x2=medals$winter2010;x2x3=medals$winter2006;x3x4=medals$sumer2004;x4y1=medals$latitude;y1y2=medals$population;y2y3=medals$gdp;y3a=data.frame(x1,y1,y2,y3);acor(a) x1 y1 y2 y3x1 1.0000000 0.16387884 0.48614648 0.6624589y1 0.1638788 1.00000000 0.010

44、97671 0.1246518y2 0.4861465 0.01097671 1.00000000 0.1685161y3 0.6624589 0.12465175 0.16851611 1.0000000 par(mfrow=c(1,3)plot(y1,x1)plot(y2,x1)plot(y3,x1)由相關(guān)系數(shù)矩陣可知，2008年各個(gè)國(guó)家的金牌數(shù)與其所處的維度、人口規(guī)模和gdp存在正相關(guān)性，它們的相關(guān)系數(shù)分別為0.164、 0.486、 0.662，并且獎(jiǎng)牌數(shù)與人口規(guī)模和gdp中度相關(guān)，與維度輕度相關(guān)。從三點(diǎn)圖可以發(fā)現(xiàn)以下幾個(gè)現(xiàn)象。第一，獲得獎(jiǎng)牌的國(guó)家基本上分布于北緯0°到北緯6

45、0°之間，只有11個(gè)獲得獎(jiǎng)牌的國(guó)家位于南半球，且獲得獎(jiǎng)牌較多的國(guó)家基本上位于北緯30°到北緯50°，即亞熱帶和溫帶地區(qū)，這與現(xiàn)實(shí)所表現(xiàn)出來(lái)的是一致的，這是因?yàn)榉植荚谀习肭虻膮⑴c奧運(yùn)會(huì)的國(guó)家相對(duì)較少，自然獲得的金牌也少。第二，獲得獎(jiǎng)牌數(shù)與人口規(guī)模顯然正相關(guān)。人口規(guī)模越大，獲得獎(jiǎng)牌數(shù)越多，中國(guó)和美國(guó)是典型，但印度除外。第三，獲得獎(jiǎng)牌數(shù)與gdp也是正相關(guān)的，gdp規(guī)模越大，獲得獎(jiǎng)牌數(shù)越多。2、2010年冬季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)及其影響因素相關(guān)分析cor(b) x2 y1 y2 y3x2 1.0000000 0.31292367 0.17071234 0.6459346y1 0.

46、3129237 1.00000000 0.01097671 0.1246518y2 0.1707123 0.01097671 1.00000000 0.1685161y3 0.6459346 0.12465175 0.16851611 1.0000000plot(y1,x2)plot(y2,x2)plot(y3,x2)從相關(guān)系數(shù)矩陣來(lái)看，2010年冬季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)與緯度、人口規(guī)模和gdp有這正相關(guān)性，相關(guān)系數(shù)分別為0.3129、0.1707、0.6459。從散點(diǎn)圖來(lái)看，2010年冬季奧運(yùn)會(huì)獲得獎(jiǎng)牌的國(guó)家出澳大利亞外都位于北緯40°與北緯65°之間。這與2008年夏季運(yùn)動(dòng)會(huì)獲

47、獎(jiǎng)牌國(guó)家的緯度相比更加偏北。這是因?yàn)槎緤W運(yùn)會(huì)時(shí)，南半球國(guó)家正值夏季，很多冬季運(yùn)動(dòng)項(xiàng)目不宜開(kāi)展。就獎(jiǎng)牌數(shù)與人口規(guī)模和gdp的關(guān)系，與2008年夏季運(yùn)動(dòng)會(huì)相比變化不大。3、2006年冬季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)及其影響因素相關(guān)分析cor(c) x3 y1 y2 y3x3 1.0000000 0.33983286 0.15271725 0.4910002y1 0.3398329 1.00000000 0.01097671 0.1246518y2 0.1527172 0.01097671 1.00000000 0.1685161y3 0.4910002 0.12465175 0.16851611 1.00000

48、00 plot(y1,x3)plot(y2,x3)plot(y3,x3) 從相關(guān)系數(shù)矩陣和散點(diǎn)圖可以看到，2006年冬季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)與緯度、人口規(guī)模和gdp的關(guān)系與2010年冬季奧運(yùn)會(huì)相似。4、2004年夏季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)及其影響因素相關(guān)分析cor(d) x4 y1 y2 y3x4 1.00000000 0.05056714 0.29117180 0.5879380y1 0.05056714 1.00000000 0.01097671 0.1246518y2 0.29117180 0.01097671 1.00000000 0.1685161y3 0.58793800 0.12465175 0.16851611 1.0000000 plot(y1,x4)plot(y2,x4)plot(y3,x4) 由相關(guān)系數(shù)矩陣可以看到，2004年夏季奧運(yùn)會(huì)獎(jiǎng)牌數(shù)與緯度、人口規(guī)模和gdp的關(guān)系與2008年夏季奧運(yùn)會(huì)相似。 通過(guò)相關(guān)分析我們可以知道，一個(gè)國(guó)家在奧運(yùn)會(huì)中獲得獎(jiǎng)牌的數(shù)量和這個(gè)國(guó)家所處的緯度、人口規(guī)模、gdp總量都是有關(guān)系的，并且與人口規(guī)模和gdp總量的還有較為密切的關(guān)系。而冬奧會(huì)與夏奧會(huì)的獎(jiǎng)牌數(shù)分布于緯度（也就是氣候）有密切聯(lián)系，冬奧會(huì)獲得獎(jiǎng)牌的除個(gè)別的國(guó)家外都是處在北半球，而夏奧會(huì)