9.3 估计人口方法和比例-interactive

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• 选择节估计人口方法和比例

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  估计人口方法和比例

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  Lesson Objectives
  ::经验教训目标

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  • Use data from a sample survey to estimate a population mean or proportion.
    ::利用抽样调查的数据估计人口平均数或比例。
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  • Develop a margin of error through the use of simulation models for random sampling .
    ::通过使用模拟模型进行随机抽样,得出误差幅度。

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  Introduction: Apples to Apples
  ::导言:苹果苹果

  

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  Savannah owns an apple orchard and needs to estimate the value of the apples produced for insurance purposes. Savannah has 50 trees in her orchard in an arrangement displayed below. Savannah can't count every single apple, so she decides to pick 15 trees and count the apples on those trees. She will then use the results to make an estimate. Use the interactive below to take a cluster sample and answer the following questions:
  ::Savannah拥有一个苹果果园,需要估算为保险目的生产的苹果的价值。 Savannah在果园里有50棵树,在下面展示一个安排。 Savannah不能统计每个苹果。 Savannah不能统计每个苹果,所以她决定挑选15棵树,数下这些树上的苹果。然后,她将用结果来估算。用下面的交互作用来采集一个集束样本并回答下列问题:

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  Discussion Question: Which two numbers do you think the population mean is between? What percent of data values fall between your two numbers? How confident are you in your answer? What would make you more confident?
  ::讨论问题:你认为人口是指两个数字之间的两个数字?数据值中有多少百分比在两个数字之间?你对答案有多有信心?你认为什么能让你更有信心?

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  Activity 1: Margin of Error Intuition
  ::活动1:误差边缘

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  In the activity above, we explored the idea of being confident in our estimate of the population mean. Since estimates by their very nature are not expected to be exact, we need a way to quantify our level of confidence in our estimate. This problem is where the margin of error comes in. The margin of error is a number that represents the distance above and below the estimate that is likely to contain the parameter . In polling, you might see that a candidate is predicted to get 82% of the vote with a margin of error of 4%. This means that the candidate is expected to receive between 78% and 86% of the vote, which can also be written as 0.82 ± 0.04. In this scenario, we are confident that the population mean will be between 4% above 82% and 4% below 82%. 
  ::在上述活动中,我们探索了对人口平均估计有信心的想法。 由于估计本身的性质预计不准确, 我们需要一种方法来量化我们对估计的信任度。 问题在于误差的出处。 误差的差值是一个代表可能包含参数的距离大于或低于估计值的数字。 在投票中, 你可能会看到, 预计候选人获得82%的选票, 误差幅度为4%。 这意味着预计候选人将获得78%至86%的选票, 也可以写为0.82+0.04。 在这种情况下, 我们相信, 人口平均比例将在4 %以上82%到4 %以下82%之间。

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  One method for establishing the margin of error is by using the midrange . Recall that the midrange is half the difference between the greatest and least numbers in the set.
  ::确定误差幅度的一种方法是使用中程。提醒注意中程是集中最大和最小数字之间的一半差。

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  Example
  ::示例示例示例示例

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  Stan owns a gas station and wants to set his gas price slightly below the state average . Since asking every gas station their price for a gallon of gas is too time-consuming, Stan  chose a simple random sample of  20 gas stations. He obtained the following data :
  ::Stan拥有一个加油站,并希望将他的天然气价格略低于州平均水平。 由于问每个加油站的一加仑天然气价格太费时, Stan选择了20个加油站的简单随机抽样。 他获得了以下数据:

  2.69, 2.59, 2.87, 2.59, 2.69, 2.73, 2.39, 2.39, 2.54, 2.79, 2.49, 2.35, 2.39, 2.32, 2.41, 2.32, 2.33, 2.33, 2.30, 2.29

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  What is the sample mean ? Use the midrange to construct a  margin of error around the sample mean. Use the sample mean, with the margin of error, to estimate the state mean gas price.
  ::样本意味着什么? 使用中程图围绕样本平均值构建一个误差差差幅。 使用样本表示, 加上误差, 来估算平均天然气价格 。

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  To obtain the sample mean, we can add all values and divide by 20. 
  ::为了获得样本平均值,我们可以将所有数值加起来,再除以20。

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  $$\begin{align*}\overline{x}=2.49\end{align*}$$
  ::x=2.49

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  The midrange of the data can be found by taking half the difference between the least and greatest data values in the set.
  ::将数据集中最小数据值和最大数据值的差数减半,就可以找到数据中的中程距。

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  $$\begin{align*}\text{Midrange}=\frac{2.87-2.29}{2}=\frac{0.58}{2}=0.29 \end{align*}$$
  ::中程=2.87-2.292=0.582=0.29

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  Answer:  Our estimate of the state gas price is $2.49, with a margin of error of $0.29.
  ::答复:我们估计的国气价格为2.49美元,误差幅度为2.29美元。

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  This result means that we are confident that the mean gas price is between $2.20 and $2.78. Of all data values in the sample, 90% fall within our confidence interval . 
  ::这一结果意味着,我们相信平均天然气价格在2.20至2.78美元之间。 在抽样中的所有数据值中,90%在我们的置信区间之内。

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  Use a given sample from the apple orchard example above to answer the following questions.
  ::使用上文苹果果园实例的样本回答下列问题。

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  Discussion Question: Half the range is often used as a rough estimate for the margin of error. What are some cons to using the midrange as a margin of error?
  ::讨论问题:半数幅度通常用作误差幅度的粗略估计。使用中距作为误差幅度有什么缺点?

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  Activity 2: Margin of Error Formula
  ::活动2:误差边距公式

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  Another formula for the margin of error is $\begin{align*}2\cdot\frac{s}{\sqrt{n}},\end{align*}$   where s is the sample standard deviation and n is the sample size . We can represent the endpoints of the margin of error using the formula  $\begin{align*}\overline{x} ± 2\cdot\frac{s}{\sqrt{n}}.\end{align*}$
  ::误差幅度的另一个公式是 2sn, 其中 s 是样本标准偏差, n 是样本大小。 我们可以使用公式 'x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以代表误差差差差差差差的端点。

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  Example
  ::示例示例示例示例

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  Josiah and Mia are researching the working habits of high school students. Mia is researching the number of hours high schoolers with jobs spend at work in a week. She will only be focusing on high school students within her state between the ages of 16 and 19 since wages and working laws tend to vary from state to state. Mia reached out to randomly selected high school teachers and took a cluster sample using any classes willing to participate. The results below display the weekly working hours from one cluster of 60 high school students between the ages of 16 and 19.
  ::Josiah和Mia正在研究高中学生的工作习惯。Mia正在研究一周内从事工作的高中生工时数。她只关注她所在的16至19岁的高中学生,因为工资和工作法各州之间往往不同。Mia接触随机挑选的高中教师,利用愿意参加的任何班级进行分组抽样。下文显示16至19岁的60名高中学生的每周工作时间。

  

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  Use the sample mean to estimate the population mean with a margin of error.
  ::使用样本来估计人口值,误差幅度。

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  Using a calculator, she determined the sample mean , using the formula  $\begin{align*}\overline{x}=\frac{\sum x}{n},\end{align*}$    to be approximately 10.63 and the sample standard deviation, using the formula  $\begin{align*}s=\sqrt\frac{\sum(x-\overline{x})^2}{n-1},\end{align*}$ to be approximately 4.19. The sample size, n, is 60. This information is enough to use our margin of error formula:
  ::她使用计算器确定了样本平均值,使用公式xxxn,大约为10.63,而样本标准差则使用公式s{(x-x)2n-1),大约为4.19。样本大小n,60。这一信息足以使用我们的误差公式:

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  $$\begin{align*}ME =\:& 2\cdot\frac{s}{\sqrt{n}} \\ =\:& 2\cdot\frac{4.19}{\sqrt{60}}\\ =\:& 1.08\end{align*}$$
  ::ME=2sn=24.1960=1.08

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  Answer: $\begin{align*}\mu\approx 10.63 \pm 1.08\end{align*}$  
  ::答复:10.631.08

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  Based on our data, we can estimate the average high schooler works between 9.55 and 11.71 hours a week. Use the interactive below to explore this margin of error formula further.
  ::根据我们的数据,我们可以估计高中学生每周平均工作时间为9.55至11.71小时。用下面的交互方式进一步探讨这一误差公式。

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  Discussion Question: Use the formula $\begin{align*}2\cdot\frac{s}{\sqrt{n}}\end{align*}$   to find the margin of error from the apple orchard example. How does it compare to the margin of error obtained by using the midrange?
  ::讨论问题:使用公式2sn从苹果果园的例子中找到误差差。它如何与使用中间幅度获得的误差率相比较?

  

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  Activity 3: Estimating Proportions
  ::活动3:估计比例

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  We can also apply our margin of error formula to scenarios involving proportions. However, the standard deviation of a proportion is based on the binomial distribution standard deviation $\begin{align*}\sqrt{np(1-p)}.\end{align*}$  In this formula, n represents the number of independent trials , not the sample size. Since  the distribution represents one trial, n = 1, we can substitute  $\begin{align*}\sqrt{p(1-p)}\end{align*}$  into the formula from the previous activity for s and simplify it to obtain the margin of error formula for proportions. 
  ::我们还可以对涉及比例的假设情况适用误差公式。但是,一个比例的标准偏差是基于二元分布标准偏差 np(1-p)。在这个公式中, n 代表独立试验的数量,而不是样本大小。由于分布代表一个试验, n = 1, 我们可以将%p(1-p) 替换为先前活动的公式, 并简化该公式, 以获得比例误差公式。

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  However, since we do not know the population proportion , we will need to use the sample proportion to estimate. Using the formula $\begin{align*}ME = 2\cdot\frac{s}{\sqrt{n}},\end{align*}$ substitute the standard deviation  $\begin{align*}\sqrt{p(1-p)}\end{align*}$ for s to obtain the formula for the margin of error of a proportion. In this formula, the n does represent the sample size. The sample proportion will be notated as p̂.
  ::然而,由于我们不知道人口比例,我们需要使用样本比例来估算。使用 ME=2sn 公式,用标准偏差 p(1-p) 代替 s 来获取比例误差的公式。在这个公式中, n 代表样本大小。 样本比例将被称为 p 。

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  $$\begin{align*}\text{Margin of Error} = 2\cdot\frac{\sqrt{\hat{p}(1-\hat{p})}}{\sqrt{n}} = 2 \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\end{align*}$$
  ::错误边缘=2p(1p)n=2p(1p)n

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  Example
  ::示例示例示例示例

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  Josiah  wants to determine the proportion of high school students who have jobs. The cluster that  reported the working hours of the 60 high schoolers with jobs also reported that 60 out of 216 high school students between the ages of 16 and 19 were employed.  Use the sample mean to estimate the population proportion  with a margin of error.
  ::Josiah希望确定有工作的高中学生比例,报告有工作的60名高中学生的工作时间的群组也报告说,16至19岁的216名高中学生中有60名得到雇用,使用抽样来估计人口比例,误差幅度。

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  Using our sample data, we can obtain the  sample proportion:
  ::利用我们的样本数据,我们可以得到样本比例:

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  $$\begin{align*}\hat{p} = \frac{60}{216} \approx 0.278\end{align*}$$
  ::p=602160.278

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  Substituting this value along with n = 216 into our formula, we can calculate the margin of error.
  ::将此值与 n= 216 等值替换为我们的公式, 我们可以计算出误差幅度 。

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  $$\begin{align*}\text{Margin of Error} &= 2 \cdot \sqrt{\frac{0.278(1-0.278)}{216}}\\ &= 2 \cdot \sqrt{\frac{0.278(0.722)}{216}}\\ &\approx 0.06\end{align*}$$
  ::误差边距=20.278(1-0.2778216)=20.278216(0.278)(0.72216)

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  Answer:   $\begin{align*}p \approx 0.278\pm 0.06\end{align*}$
  ::答复:p 0.2780.06

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  We expect the proportion of working high schoolers to be between 0.216 and 0.338. Another way of stating this is that we are confident that between 21.6% and 33.8% of high schoolers between the ages of 16 and 19 in the state are working . 
  ::我们预计高中在校生的比例在0.216至0.338之间,另一种说法是,我们相信,16至19岁的高中生中,有21.6%至33.8%的人正在工作。

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  Discussion Question: What will happen to the margin of error if we increase the proportion? What if we increased the proportion from 0.2 to 0.5? From 0.2 to 0.8?
  ::讨论问题:如果我们增加比例,误差幅度会怎样?如果我们将比例从0.2增加到0.5,会怎样?从0.2增加到0.8?

  Summary

  播放段落 The margin of error is a number that represents the distance above and below the estimate that is likely to contain the parameter. ::误差差是一个表示可能包含参数的估计数上下距离的数值。 播放段落 The midrange of the data can be found by taking half the difference between the least and greatest data values in the set. ::将数据集中最小数据值和最大数据值的差数减半,就可以找到数据中的中程距。 播放段落 One formula for margin of error is $\begin{align*}2  \cdot \frac{s}{\sqrt{n}}.\end{align*}$   ::误差差率的一个公式是 2sn。 播放段落 Another formula for margin of error using proportions is $\begin{align*}\text{Margin of Error} = 2 \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\end{align*}$   ::使用比例表示误差差的另一种公式是误差的边距=2p(1p)n

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  Wrap-Up: Review Questions
  ::总结:审查问题