畢業(yè)論文外文翻譯---實(shí)驗(yàn)誤差與數(shù)據(jù)分析

1、<p><b>　　附錄1 中文翻譯</b></p><p><b>　　實(shí)驗(yàn)誤差與數(shù)據(jù)分析</b></p><p>　　by J.C. de Paula</p><p>　　1、隨機(jī)誤差和系統(tǒng)誤差</p><p>　　每個實(shí)驗(yàn)的結(jié)果都是存在誤差的，我們可以嘗試用最小誤差但是卻不能完全消除

2、誤差，實(shí)驗(yàn)誤差可以分為兩大類：隨機(jī)誤差和系統(tǒng)誤差。</p><p>　　隨機(jī)誤差是由于被測物理量的自然屬性所引起的。例如，重復(fù)測量同一物理量會出現(xiàn)不同的結(jié)果，即使在同一個經(jīng)過校對正常運(yùn)行的儀器下進(jìn)行測量，這種變化決定著測量的精度。精度同時也涉及重復(fù)性。</p><p>　　系統(tǒng)誤差是由于測量過程中出現(xiàn)一些錯誤所引起的。例如，不正常運(yùn)行的儀器就會產(chǎn)生系統(tǒng)誤差，測量不準(zhǔn)確度，用壞的儀器重復(fù)測量

3、會出現(xiàn)重復(fù)性問題（高精度），但是每個測量量值卻偏離真實(shí)值（低準(zhǔn)確度）。因此，準(zhǔn)確度和精密度之間并沒有關(guān)系。</p><p><b>　　2。數(shù)據(jù)分析的統(tǒng)計(jì)</b></p><p>　　常態(tài)分布。我們只要關(guān)注精密度，測量結(jié)果都會產(chǎn)生隨機(jī)誤差，不管準(zhǔn)確度是多少。我們看如下數(shù)據(jù)，通過用同一儀器測量金屬條的質(zhì)量24次得到的不同結(jié)果。</p><p>　

4、　通過數(shù)據(jù)直方圖，我們可以想象這種測量的重復(fù)性，說明測量值在這個集合中出現(xiàn)的頻率。這個實(shí)驗(yàn)的直方圖如下圖所示。</p><p>　　由圖可知，最常出現(xiàn)的測量值是8.148（4次），偏離8.148過多或過少的測量值很少。如果我們多次測量金屬條的質(zhì)量，我們將會得到一個形狀如鈴鐺的直方圖，如下圖所示。</p><p>　　這個鐘形曲線表示測量多次質(zhì)量的正態(tài)概率分布，正態(tài)分布曲線有兩個參數(shù)。即平均

5、值和標(biāo)準(zhǔn)差</p><p><b>　?。?）</b></p><p><b>　　（2）</b></p><p>　　對于多次測量來說，平均值是最有可能的值，如上圖所示，標(biāo)準(zhǔn)差是測量曲線的寬度：標(biāo)準(zhǔn)差大，分布更大。也就是說，測量精度越低，分布越寬，標(biāo)準(zhǔn)差值也就越大。</p><p>　　標(biāo)準(zhǔn)差的意

6、義如下：有68.30%的測量值介于和之間。也就是說，測量值中有68.30%都在平均值的標(biāo)準(zhǔn)差內(nèi)。舉例說明，如果=1.848g，s=0.004g，24次測量金屬條的質(zhì)量結(jié)果中有16次的測量值會介于1,844g—1,852g之間。參照上面描繪的正態(tài)分布，兩虛線之間鐘形曲線的區(qū)域占整個區(qū)域的68.30%。</p><p>　　人們對這個概率可能不會有深刻的印象。我們可以有不同的看法，但是，結(jié)果表明，在正態(tài)分布中，有95

7、.5%的測量值會在平均值1.884g的　=0.008g范圍內(nèi)。你相信，再一次測量金屬條的結(jié)果又95.5%的可能性落入1.8480.008g的范圍內(nèi)。這個概率決定你的實(shí)驗(yàn)的置信區(qū)間。你將公訴讀者，如果重新測量金屬條的質(zhì)量結(jié)果不在1.840-1.856g的范圍內(nèi)，人們有權(quán)懷疑測量結(jié)果的真實(shí)性。　</p><p>　　到目前為止,我們已經(jīng)處理了基于整個正態(tài)分布曲線的標(biāo)準(zhǔn)差的單個測量結(jié)果的精密度。平均值的精密度（不確定度

8、）是多少？標(biāo)準(zhǔn)偏差值由下式給出： </p><p><b>　?。?）</b></p><p>　　概率為95.5%的置信區(qū)間，平均值在的范圍內(nèi)。我們接著看我們的例子。如果=1.848g，=0.001g，平均值在1.8480.001g的范圍內(nèi)，也就是說，如果20名學(xué)生各自測量同一金屬條的質(zhì)量24次，20個測量平均值中有19個測量平均值介于1.847—1.849之間。&

9、lt;/p><p>　　處理數(shù)據(jù)集。在現(xiàn)實(shí)生活中，一個集合中很少有如此多的測量值，所有測量值的分布概率呈鐘形曲線。有可能你的測量值很少有如此多的測量值，所有測量值的分布概率呈鐘形曲線，有可能你的測量值不具有鐘形曲線的特性，在這種情況下，我們首先應(yīng)該通過式(1),(2)計(jì)算平均值和標(biāo)準(zhǔn)差，接著計(jì)算置信區(qū)間，置信區(qū)間決定平均值兩邊界的范圍，在給定的置信區(qū)間內(nèi)可以查出真實(shí)值</p><p><

10、b>　　(4)</b></p><p>　　t值是置信水平值，注意式（3）比式（4）少一個t變量，如下是一個簡明的t值表</p><p><b>　　3、誤差傳遞</b></p><p>　　糟糕的是無法消除隨機(jī)誤差。這就是為什么我們需要在實(shí)驗(yàn)室重復(fù)實(shí)驗(yàn)的原因，更糟糕的是當(dāng)你計(jì)算你的數(shù)值時，誤差發(fā)生傳遞，讓我們以測量密度為例，

11、為了測出物質(zhì)的密度，你必須測量兩個物理量-質(zhì)量和體積。每一個物理量都有不同的精密度。問題是：誤差與質(zhì)量和體積（即密度）的比率有什么關(guān)系。用簡單的方程計(jì)算兩個測量值的數(shù)學(xué)計(jì)算結(jié)果的不確定度。公式如下：此結(jié)果適用于標(biāo)準(zhǔn)差和均值標(biāo)準(zhǔn)差的傳遞。</p><p>　　加法和減法。如果你將和，測量結(jié)果的不確定度由下式給出：</p><p><b>　　(5)</b></p&

12、gt;<p>　　乘法和除法。如果你用乘以或者除以，測量結(jié)果的不確定度由下式給出：</p><p><b>　　(6)</b></p><p><b>　　(7)</b></p><p>　　自然對數(shù),和的不確定度由下式給出:</p><p><b>　　(8)</b&

13、gt;</p><p><b>　　(9)</b></p><p><b>　　(10)</b></p><p>　　由方程（5）-（10）描述的不同類型的誤差分析是勞動密集型行業(yè)，而且是必要的，它允許你在一項(xiàng)實(shí)驗(yàn)中確定一個薄弱環(huán)節(jié)，限制你的計(jì)算結(jié)果的可靠性，你應(yīng)該在你的實(shí)驗(yàn)報(bào)告中指明這一特定要求。</p>

14、<p><b>　　4 數(shù)據(jù)的不相容 </b></p><p>　　現(xiàn)在讓我們來考慮系統(tǒng)誤差。你經(jīng)常會碰到一個實(shí)驗(yàn)結(jié)果的數(shù)據(jù)并非都是合情合理的。例如，對金屬條進(jìn)行5次測量，你會得到如下數(shù)據(jù)（單位是克）：</p><p>　　8.148 8.145 8.156 8.149 8.177</p><p>　　最后的數(shù)據(jù)看起來有些不合

15、情理，單單從美學(xué)的角度看你可能試圖把它去掉。然而，除非你有數(shù)學(xué)上的統(tǒng)計(jì)或者化學(xué)上的原由，你才可以從數(shù)據(jù)集合中把它去掉。</p><p>　　從統(tǒng)計(jì)學(xué)上看，我們的問題是：8.177g這一測量結(jié)果是否與其它4個測量結(jié)果具有相同的正態(tài)分布？有兩種方法解決這一問題。</p><p>　　首先，我們采取一般方法來解決這個問題。由于你并未嚴(yán)格遵守實(shí)驗(yàn)指導(dǎo)書而形成的系統(tǒng)誤差，導(dǎo)致最后一個測量結(jié)果或許與其

16、它4個數(shù)據(jù)不屬于同一分布，也就是說，你知道你弄錯了。沒問題，我們所有的人都會遇到這種情況。在這種情況下，你沒有理由在你的數(shù)據(jù)集合中保留這個數(shù)據(jù)，即便你可能想采用另一種測量，以防萬一。然而，底線是你必須要對你的教授或管理者負(fù)責(zé)這次疏忽。這就是為什么你必須在實(shí)驗(yàn)報(bào)告的實(shí)驗(yàn)項(xiàng)目中完整而精確的記錄數(shù)據(jù)的原因之一。</p><p>　　在一些檢測工作完成以后，系統(tǒng)誤差或許是不那么顯而易見的。在這種情況下，用數(shù)理統(tǒng)計(jì)的方法去

17、掉一個測量值雖然并不明智，但卻是有可能的。</p><p>　　假如我們有一個非常大的數(shù)據(jù)集合，然后我們能夠計(jì)算出，進(jìn)而判斷落于置信區(qū)間外的有問題的數(shù)據(jù)。然而，我們的數(shù)據(jù)集是非常小的（N<10）,因此，標(biāo)準(zhǔn)偏差并不是一個好的判別準(zhǔn)則。</p><p>　　統(tǒng)計(jì)學(xué)家未檢測非隨機(jī)誤差已經(jīng)發(fā)明了很多實(shí)驗(yàn)，我們即將描述的僅有一個Q測試。在3<N<10情況下，這種方法的效果很好。&

18、lt;/p><p>　　為了測量8.177g的值，我們必須計(jì)算這一測量值所謂的值。一般而言，的值為：　</p><p>　　最接近的值與真值之間的差的絕對值∕值域 (11)</p><p>　　在我們所舉的例子中，為了計(jì)算的值，我們按遞增的方式排列這些數(shù)據(jù)，然后確定它的真值與它的最接近的值：</p><p>　　8.145 8.1

19、48 8.149 8.156 8.177</p><p>　　(與之最接近的估計(jì)值) (真值)</p><p><b>　　然后,</b></p><p>　　現(xiàn)在我們比較與。如果>,則測量值可以被去掉。若<,無論你多想去掉這個數(shù)據(jù)也不可能去掉。的值取決于你的數(shù)組的置

20、信區(qū)間和測量值的數(shù)量，部分表如下： </p><p>　　回到我們的例子,N=5.我們可得： =0.66，=0.64大，因此，我們可以判定這一數(shù)據(jù)無效。但是，你必須在你的實(shí)驗(yàn)報(bào)告中指明，在90%的置信區(qū)間內(nèi)，可以運(yùn)用Q值測試方法。</p><p><b>　　附錄2 外文原文</b></p><p>　　EXPERIMENTAL ERRORS

21、AND DATA ANALYSIS</p><p>　　by J.C．de Paula</p><p>　　1. RANDOM AND SYSTEMATIC ERRORS</p><p>　　Every experimental result is subject to error.One can attempt to minimize errors but can

22、not eliminate them completely. Experimental errors fall under two categories: random or systematic.</p><p>　　Random errors arise from natural limitations of making physical measurements. For example, repeate

23、d measurements of the same property often differ even if they are performed on a single instrument that is calibrated and operated properly. Such variations establish the precision of the measurement. The precision is al

24、so referred to as the reproducibility.</p><p>　　Systematic errors arise from blunders in the measuring process. For example, an instrument that is not operating properly is likely to give erroneous results.

25、The measurement lacks accuracy. It is even possible that repeated measurements with this broken instrument will give reproducible results (high precision), but every one of them will deviate from the true value (low accu

26、racy). Thus, accuracy and precision are not related to each other.</p><p>　　2. STATISTICS OF DATA ANALYSIS</p><p>　　The Normal Distribution. Let us turn our attention to precision, since every s

27、et of measurements will be subject to random errors, no matter what the degree of accuracy. Let us consider the data below, obtained by measuring the mass of a metal strip 24 times with the same instrument.</p>&l

28、t;p>　　We can visualize the reproducibility of this set of measurements by plotting the data as a histogram, which shows how often (frequency) a given value was obtained in the set. The histogram for this experiment is

29、 shown below.</p><p>　　We see that the value that occurred most often was 8.148 (four times), and that values that deviated by too much or too little from 8.148 occurred infrequently. If we were to conduct a

30、 very large number of measurements on the metal strip, we would have obtained a histogram whose shape resembles a bell, as shown below:</p><p>　　This bell-shaped curve denotes the normal probability distribu

31、tion for a large number ofmass measurements. The exact shape of the normal distribution is characterized by two parameters:the mean value, , and the standard deviation, s:</p><p><b>　?。?）</b><

32、;/p><p><b>　　（2）</b></p><p>　　For a large number of measurements, the mean value is also the most probable value, as shown on the plot above. The standard deviation is a measure of the

33、breadth of the curve: the larger the standard deviation, the broader the distribution. Put differently, the less precise the measurement, the broader the distribution and the larger the standard deviation.The significanc

34、e of the standard deviation is as follows: the probability of finding a result xi falling betweenand is 68.30 %.Put differen</p><p>　　Your audience may not be impressed by these odds. We can state the resul

35、ts differently, however. It turns out that, in a normal distribution, 95.5 % of the observations will fall within the limits of 2s. In other words, the odds are 19 out of 20 that an individual measurement of the mass wil

36、l be within 2s = 0.008 grams of the mean value of 1.844 grams. You are 95.5 % confident that a new individual measurement of the mass will give a result in the range 1.848 ± 0.008 grams. This statement determ</p&

37、gt;<p>　　So far, we have dealt with the precision of a single measurement, based on the standarddeviation of the entire normal distribution curve. How about the precision (or uncertainty) of themean value x ? The

38、standard deviation of the mean value is given by</p><p><b>　　（3）</b></p><p>　　Using the 95.5 % confidence interval, the mean value should then be reported as.Let us go back to our ex

39、ample, where=1.848 grams and=0.001 grams. By reporting the average as 1.848 ± 0.001 grams, we are saying that if 20 different students were to measure the mass of the metal strip 24 times, 19 out 20 of the measured

40、mean values will fall within 1.847 and 1.849 grams.</p><p>　　Working with Small Data Sets. In real life, one seldom has so many measurements in a set as to see the bell-shaped curve in all of its glory. Ther

41、e is the possibility that your sample is not representative of the bell-shaped curve. In such instances, we first calculate a mean and a standard deviation according to Equations (1)-(2) and then calculate a confidence l

42、imit, ?. The confidence limit defines the range on both sides of the mean within which the true value can be expected to be found with</p><p><b>　　(4)</b></p><p>　　where the value of

43、 t depends on the confidence level. Note that Equations (3) and (4) differ only by the factor t. A brief table of t values </p><p>　　If we report a result as” ”, it is also necessary to report the confidence

44、 level and the number of measurements in the data set. It is not necessary to repeat units in the presentation of the result. For example, we write 1.848 ??0.001 grams (95%, N = 24) and not as 1.848 grams ??0.001 grams (

45、95%, N = 24).</p><p>　　In this course, you will often be asked to report your results as a mean value of two or more measurements with the all important uncertainty. Without the uncertainty, you cannot commu

46、nicate to the reader the degree of reproducibility of your experiment.</p><p>　　3. PROPAGATION OF ERRORS</p><p>　　It is bad enough that random errors cannot be eliminated and that is why we need

47、 to repeat experiments in the laboratory. Even worse is the fact that errors propagate as you perform your calculations. Let us consider the measurement of density as an example. In order to report the density of a subst

48、ance, you must make two measurements - one of mass and one of volume – each of which carries a different precision. The question is, what is the error associated with the ratio of mass over volume (i.</p><p>

49、;　　Addition and Subtraction. If you add or subtract the values and the uncertainty of the result is given by:</p><p><b>　　(5)</b></p><p>　　Products and Quotients. If you multiply o

50、r divideby,the uncertainties are given by:</p><p><b>　　(6)</b></p><p><b>　　(7)</b></p><p>　　Natural Logs. The uncertainties of ,,and are given by:</p>

51、<p><b>　　(8)</b></p><p><b>　　(9)</b></p><p><b>　　(10)</b></p><p>　　An error analysis of the type described by Equations (5)-(10) is labor i

52、ntensive but necessary. It allows you to determine the weak link in an experiment: the number that limits thereliability of your results. You should identify this specifically in your laboratory reports.</p><p

53、>　　4. REJECTION OF DATA</p><p>　　Let us now consider systematic errors. Every now and then, you will come across an experimental result that simply does not seem to make sense. For example, after making o

54、nly five measurements of our metal strip, you obtain the following results (in grams):</p><p>　　8.148 8.145 8.156 8.149 8.177</p><p>　　The last measurement seems a bit off, and you may be te

55、mpted to throw it out of the set on aesthetic grounds alone. However, you must never throw out a result from a data set unless you have a statistical or chemical reason to do so.</p><p>　　Statistically speak

56、ing, we are asking the question: does the measurement of 8.177 grams belong to the same normal distribution as the other four measurements? There are two ways to answer this question.</p><p>　　Let us first t

57、ake the “common sense” approach. The last measurement may not belong to the same distribution as the other four because of a systematic error that you tracked down upon close examination of your laboratory notebook. That

58、 is, you know you goofed! No problem, it happens to all of us! In that case, there is no reason for you to keep the number in your data set, although you may want to make another measurement in its stead just in case. Th

59、e bottom line is, however, that you still must</p><p>　　A systematic error may not be apparent even after some detective work. In such cases, it is possible, though not always advisable, to use statistical m

60、ethods to reject an observation. If we had a very large data set, then we could calculate,and then determine if the measurement in question falls outside the confidence interval. However, our data set is very small(N<

61、10), so that the standard deviation alone is not a good criterion for rejection.</p><p>　　Statisticians have devised many rejection tests for the detection of non-random errors. We will describe only one - t

62、he Q test - which works well in cases where 3 < N < 10.</p><p>　　In order to test the value of 1.877 grams, we must calculate the so-called Qalcc value for this observation. In general, the value of Qc

63、alc is given by</p><p><b>　　(11)</b></p><p>　　To calculate Qcalc in our example, we display the data in increasing order of numerical value, then identify the suspect value and the v

64、alue that is closest to it:</p><p>　　8.145 8.148 8.149 8.156 8.177</p><p>　　(value closest to suspect value) (suspect value)</p><p><b>　　Then,<

65、/b></p><p>　　We now compare this Qcalc with a critical value Qc. If Qcalc > Qc, then the observation</p><p>　　may be rejected. If Qcalc < Qc, then we must keep the observation no matter

66、 how tempted we may be to throw it out. The Qc value depends on the confidence level and the number of observations in your set. A partial list follows:</p><p>　　Returning to our example, where N = 5, we see