相信许多留学生对数学代考都不陌生，国外许多大学都引进了网课的学习模式。网课学业有利有弊，学生不需要到固定的教室学习，只需要登录相应的网站研讨线上课程即可。但也正是其便利性，线上课程的数量往往比正常课程多得多。留学生课业深重，时刻名贵，既要学习知识，又要结束多种类型的课堂作业，physics作业代写，物理代写，论文写作等；网课考试很大程度增加了他们的负担。所以，您要是有这方面的困扰，不要犹疑，订购myassignments-help代考渠道的数学代考服务，价格合理，给你前所未有的学习体会。
我们的数学代考服务适用于那些对课程结束没有掌握，或许没有满足的时刻结束网课的同学。高度匹配专业科目，按需结束您的网课考试、数学代写需求。担保买卖支持，100%退款保证，免费赠送Turnitin检测报告。myassignments-help的Math作业代写服务，是你留学路上忠实可靠的小帮手！
电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|Inferring Population Parameters from Sample Parameters
Thus far, we have focused on statistics that describe a sample in various ways. A sample, however, is usually only a subset of the population. Given the statistics of a sample, what can we infer about the corresponding population parameters? If the sample is small or if the population is intrinsically highly variable, there is not much we can say about the population. However, if the sample is large, there is reason to hope that the sample statistics are a good approximation to the population parameters. We now quantify this intuition.
Our point of departure is the central limit theorem, which states that the sum of $n$ independent random variables, for large $n$, is approximately normally distributed (see Section 1.7.5). Suppose that we collect a set of $m$ samples, each with $n$ elements, from some population. (In the rest of the discussion, we will assume that $n$ is large enough that the central limit theorem applies.) If the elements of each sample are independently and randomly selected from the population, we can treat the sum of the elements of each sample as the sum of $n$ independent and identically distributed random variables $X_{1}, X_{2, \ldots}, X_{n}$. That is, the first element of the sample is the value assumed by the random variable $X_{1}$, the second element is the value assumed by the random variable $X_{2}$, and so on. From the central limit theorem, the sum of these random variables is normally distributed. The mean of each sample is the sum divided by a constant, so the mean of each sample is also normally distributed. This fact allows us to determine a range of values where, with high confidence, the population mean can be expected to lie.
To make this more concrete, refer to Figure $2.3$ and consider sample 1 . The mean of this sample is $\overline{x_{1}}=\frac{1}{n} \sum_{i} x_{1 i}$. Similarly, $\overline{x_{2}}=\frac{1}{n} \sum_{i} x_{2 i}$, and, in general, $\overline{x_{k}}=\frac{1}{n} \sum_{i} x_{k i}$.
Define the random variable $\bar{X}$ as taking on the values $\overline{x_{1}}, \overline{x_{2}}, \ldots, \overline{x_{n}}$. The distribution of $\bar{X}$ is called the sampling distribution of the mean. From the central limit theorem, $\bar{X}$ is approximately normally distributed. Moreover, if the elements are drawn from a population with mean $\mu$ and variance $\sigma^{2}$, we have already seen that $E(\bar{X})=\mu$ (Equation 2.6) and $V(\bar{X})=\sigma^{2} / n$ (Equation 2.9).
电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|Hypothesis Testing
Assertions about outcomes of an experiment can usually be reformulated in terms of testing a hypothesis: a speculative claim about the outcome of an experiment. The goal of an experiment is to show that either the hypothesis is unlikely to be true (i.e., we can reject the hypothesis), or the experiment is consistent with the hypothesis (i.e., the hypothesis need not be rejected).
This last statement bears some analysis. Suppose that we are asked to check whether a coin is biased. We will start with the tentative hypothesis that the coin is unbiased: $\mathrm{P}($ heads $)=\mathrm{P}($ tails $)=0.5$. Then, suppose we toss the coin three times and that we get three heads in a row. What does this say about our hypothesis? Conditional on the hypothesis being true, we have a probability of $0.5 * 0.5 * 0.5=12.5 \%$ that we obtain the observed outcome. This is not too unlikely, so perhaps the three heads in a row were simply due to chance. At this point, all we can state is that the experimental outcome is consistent with the hypothesis.
Now, suppose that we flip the coin ten times and see that it comes up heads nine times. If our hypothesis were true, the probability of getting nine heads in ten coin flips is given by the binomial distribution as $\left(\begin{array}{c}10 \ 1\end{array}\right) 0.5^{9} 0.5^{1}=10 * 0.5^{10}=10 / 1024<$ $1 \%$. Thus, if the hypothesis were true, this outcome is fairly unlikely: setting the bar for “unlikeliness” at $1 \%$. This is typically stated as: “We reject the hypothesis at the $1 \%$ confidence level.”
The probability of an outcome assuming that a hypothesis is true is called its $p$-value. If the outcome of an experiment has a $p$-value less than $1 \%$ (or $5 \%$ ), we would interpret the experiment as grounds for rejecting a hypothesis at the $1 \%$ (or $5 \%$ ) level.
It is important to realize that the nonrejection of a hypothesis does not mean that the hypothesis is valid. For example, instead of starting with the hypothesis that the coin was unbiased, we could have made the hypothesis that the coin was biased, with $\mathrm{P}($ heads $)=0.9$. If we toss the coin three times and get three heads, the probability of that event, assuming that the hypothesis were true, would be $0.9 *$ $0.9 * 0.9=0.73$. So, we cannot reject the hypothesis that the coin is biased. Indeed, with such a small number of experiments, we cannot invalidate an infinite number of mutually incompatible hypotheses!
We are therefore led to two inescapable conclusions. First, even the most careful experiment may lead to an incorrect conclusion due to random errors. Such errors may result in rejection of a hypothesis, even though it ought not be rejected, or in nonrejection, when it should. Second, an experiment cannot result in the acceptance of a hypothesis but only in its rejection or nonrejection, which is not the same as acceptance. We deal with each conclusion in turn.
电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|Inferring Population Parameters from Sample Parameters
到目前为止，我们专注于以各种方式描述样本的统计数据。然而，样本通常只是总体的一个子集。给定样 本的统计数据，我们可以推断出相应的总体参数吗? 如果样本很小或者人口本质上是高度可变的，那么我 们对人口就无能为力了。但是，如果样本很大，则有理由希望样本统计量能够即好地近似总体参数。我们 现在量化这种直觉。
我们的出发点是中心极限定理，它指出 $n$ 独立随机变量，对于大 $n$, 近似正态分布（参见第 $1.7 .5$ 节) 。假 设我们收集了一组 $m$ 样本，每个都有 $n$ 元素，来自某些人群。（在接下来的讨论中，我们将假设 $n$ 大到足 以应用中心极限定理。）如果每个样本的元素是从总体中独立随机选择的，我们可以将每个样本的元素之 和视为 $n$ 独立同分布的随机变量 $X_{1}, X_{2, \ldots}, X_{n}$. 即样本的第一个元责是随机变量假设的值 $X_{1}$ ，第二个元 素是随机变量假设的值 $X_{2}$ ，等等。根据中心极限定理，这些随机变量的总和呈正态分布。每个样本的平 均值是总和除以一个常数，因此每个样本的平均值也是正态分布的。这一事实使我们能够以高置信度确定 可以预期总体平均值存在的一系列值。
为了更具体，请参阅图 $2.3$ 并考虑样本 1 。这个样本的平均值是 $\overline{x_{1}}=\frac{1}{n} \sum_{i} x_{1 i}$. 相似地， $\overline{x_{2}}=\frac{1}{n} \sum_{i} x_{2 i}$, 并且一般来说 $\overline{x_{k}}=\frac{1}{n} \sum_{i} x_{k i}$.
定义随机变量 $\bar{X}$ 接受价值观 $\overline{x_{1}}, \overline{x_{2}}, \ldots, \overline{x_{n}}$. 的分布 $\bar{X}$ 称为均值的抽样分布。从中心极限定理， $\bar{X}$ 近似正 态分布。此外，如果元素是从具有均值的总体中抽取的 $\mu$ 和方差 $\sigma^{2}$ ，我们已经看到 $E(\bar{X})=\mu$ (公式 2.6) 和 $V(\bar{X})=\sigma^{2} / n($ 公式 2.9）。
电子工程代写|计算数学基础代写Mathematical Foundations of Computing代考|Hypothesis Testing
关于实验结果的断言通常可以重新表述为检验假设：关于实验结果的推测性主张。实验的目标是表明假设不太可能为真（即我们可以拒绝假设），或者实验与假设一致（即不需要拒绝假设）。
最后这句话有一些分析。假设我们被要求检查硬币是否有偏差。我们将从硬币无偏的假设开始：磷(头)=磷(尾巴)=0.5. 然后，假设我们掷硬币 3 次，连续出现 3 个正面。这说明我们的假设是什么？以假设为真为条件，我们有概率0.5∗0.5∗0.5=12.5%我们得到观察到的结果。这不是太不可能，所以也许连续三个头只是偶然的。在这一点上，我们只能说实验结果与假设一致。
现在，假设我们掷硬币十次，看到它出现九次正面。如果我们的假设是正确的，那么在十次抛硬币中出现九次正面的概率由二项分布给出：(10 1)0.590.51=10∗0.510=10/1024< 1%. 因此，如果假设是真的，那么这个结果是相当不可能的：将“不太可能”的标准设置为1%. 这通常被表述为：“我们在1%置信水平。”
假设一个假设为真的结果的概率称为它的p-价值。如果一个实验的结果有p-值小于1%（或者5%)，我们会将实验解释为在1%（或者5%） 等级。
重要的是要认识到一个假设的不被拒绝并不意味着该假设是有效的。例如，我们可以假设硬币是有偏见的，而不是从硬币无偏的假设开始，磷(头)=0.9. 如果我们掷硬币 3 次，得到 3 个正面，假设假设为真，该事件的概率为0.9∗ 0.9∗0.9=0.73. 因此，我们不能拒绝硬币有偏差的假设。确实，通过如此少量的实验，我们无法使无数相互不相容的假设无效！
因此，我们得出两个不可避免的结论。首先，即使是最仔细的实验也可能由于随机错误而导致错误的结论。这样的错误可能会导致假设被拒绝，即使它不应该被拒绝，或者在应该被拒绝时也不会被拒绝。其次，实验不能导致接受假设，而只能导致拒绝或不拒绝，这与接受不同。我们依次处理每个结论。
myassignments-help数学代考价格说明
1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。
2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。
3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。
Math作业代写、数学代写常见问题
留学生代写覆盖学科？
代写学科覆盖Math数学,经济代写，金融,计算机,生物信息，统计Statistics,Financial Engineering，Mathematical Finance，Quantitative Finance，Management Information Systems,Business Analytics,Data Science等。代写编程语言包括Python代写、Physics作业代写、物理代写、R语言代写、R代写、Matlab代写、C++代做、Java代做等。
数学作业代写会暴露客户的私密信息吗？
我们myassignments-help为了客户的信息泄露，采用的软件都是专业的防追踪的软件，保证安全隐私，绝对保密。您在我们平台订购的任何网课服务以及相关收费标准，都是公开透明，不存在任何针对性收费及差异化服务，我们随时欢迎选购的留学生朋友监督我们的服务，提出Math作业代写、数学代写修改建议。我们保障每一位客户的隐私安全。
留学生代写提供什么服务？
我们提供英语国家如美国、加拿大、英国、澳洲、新西兰、新加坡等华人留学生论文作业代写、物理代写、essay润色精修、课业辅导及网课代修代写、Quiz，Exam协助、期刊论文发表等学术服务，myassignments-help拥有的专业Math作业代写写手皆是精英学识修为精湛；实战经验丰富的学哥学姐！为你解决一切学术烦恼！
物理代考靠谱吗？
靠谱的数学代考听起来简单，但实际上不好甄别。我们能做到的靠谱，是把客户的网课当成自己的网课；把客户的作业当成自己的作业；并将这样的理念传达到全职写手和freelancer的日常培养中，坚决辞退糊弄、不守时、抄袭的写手！这就是我们要做的靠谱！
数学代考下单流程
提早与客服交流，处理你心中的顾虑。操作下单，上传你的数学代考/论文代写要求。专家结束论文，准时交给，在此过程中可与专家随时交流。后续互动批改
付款操作：我们数学代考服务正常多种支付方法，包含paypal，visa，mastercard，支付宝，union pay。下单后与专家直接互动。
售后服务：论文结束后保证完美经过turnitin查看，在线客服全天候在线为您服务。如果你觉得有需求批改的当地能够免费批改，直至您对论文满意为止。如果上交给教师后有需求批改的当地，只需求告诉您的批改要求或教师的comments，专家会据此批改。
保密服务：不需求提供真实的数学代考名字和电话号码，请提供其他牢靠的联系方法。我们有自己的工作准则，不会泄露您的个人信息。
myassignments-help擅长领域包含但不是全部:
myassignments-help服务请添加我们官网的客服或者微信/QQ，我们的服务覆盖：Assignment代写、Business商科代写、CS代考、Economics经济学代写、Essay代写、Finance金融代写、Math数学代写、report代写、R语言代考、Statistics统计学代写、物理代考、作业代写、加拿大代考、加拿大统计代写、北美代写、北美作业代写、北美统计代考、商科Essay代写、商科代考、数学代考、数学代写、数学作业代写、physics作业代写、物理代写、数据分析代写、新西兰代写、澳洲Essay代写、澳洲代写、澳洲作业代写、澳洲统计代写、澳洲金融代写、留学生课业指导、经济代写、统计代写、统计作业代写、美国Essay代写、美国代考、美国数学代写、美国统计代写、英国Essay代写、英国代考、英国作业代写、英国数学代写、英国统计代写、英国金融代写、论文代写、金融代考、金融作业代写。