Section: 手段的用法 | 9.6 手段方法

Section outline

Suppose you have taken several samples of 10 units each from a population of 500 students, and calculated the mean of each sample . How might you use the data you now have to estimate a mean for the entire population?
::假设您已经从500名学生中采集了10个单位的样本,并且计算了每个样本的平均值。您如何使用您现在必须使用的数据来估算整个人群的平均值?

The Mean of Means
::手段的用法

In statistics , you often need to take data from a small number of samples and use it to extrapolate an estimate of the parameters of the population the samples were pulled from. Since one of the more common parameters of interest is the mean, it is common to see a distribution of the means of a number of samples (I realize this may be confusing, “sample” here actually refers to the results of several individual samples) from the same population. This distribution is called, appropriately, the “ sampling distribution of the sample mean ”. We will be investigating the sampling distribution of the sample mean in more detail in the next lesson “The Central Limit Theorem”, but in essence it is simply a representation of the spread of the means of several samples.
::在统计中,通常需要从少量样本中提取数据,并用这些数据推断从这些样本提取的人口参数的估计数。由于一个更常见的利益参数是平均值,因此常见的做法是看到同一人群中若干样本方法的分布(我知道这可能令人困惑,“抽样”实际上是指若干单个样本的结果),这种分布恰当地称为“抽样平均值的抽样分布”。我们将在下一期“中央限制理论”课中更详细地调查样本平均值的抽样分布,但实质上它只是若干样本方法的分布。

Here we will be focusing on a single value in that sampling distribution, the “ mean of means ”. The mean of means is simply the mean of all of the means of several samples. By calculating the mean of the sample means, you have a single value that can help summarize a lot of data.
::在这里,我们将集中关注抽样分布中的单一价值,即“手段手段”这一“手段手段”。手段手段手段只是若干样品所有手段的平均值。通过计算抽样手段的平均值,你有一个单一价值,可以帮助总结大量数据。

The mean of means, notated here as $\begin{align*}\mu_{\overline{x}}\end{align*}$ , is actually a pretty straightforward calculation. Simply sum the means of all your samples and divide by the number of means.
::手段的用法,在这里以mx表示,其实是一个非常直截了当的计算。简单地总结一下所有样本的方法,并按手段的数量进行区分。

As a formula, this looks like:
::作为一种公式,这看起来像:

\begin{aligned} μ_{\bar{x}} = \frac{{\bar{x}}_{1} + {\bar{x}}_{2} + {\bar{x}}_{3} \dots + {\bar{x}}_{n}}{n} \end{aligned}

$\begin{align*}\mu_{\overline{x}}=\frac{\overline{x}_1+\overline{x}_2+\overline{x}_3\ldots +\overline{x}_n}{n}\end{align*}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The second common parameter used to define sampling distribution of the sample means is the “ standard deviation of the distribution of the sample means ”. The only significant difference between the standard deviation of a population and the standard deviation of sample means is that you need to divide the population standard deviation by the square root of the sample size .
::用于确定抽样手段抽样分布的第二个共同参数是“抽样手段分布的标准偏差”。人口标准偏差与抽样手段标准偏差的唯一显著区别是,你需要将人口标准偏差除以抽样规模的平方根。

As a formula, this looks like:
::作为一种公式,这看起来像:

\begin{aligned} σ_{\bar{x}} = \frac{σ}{\sqrt{n}} \end{aligned}

$\begin{align*}\sigma_{\overline{x}}=\frac{\sigma}{\sqrt{n}}\end{align*}$
::========================================================n

I recognize that the terminology in this lesson may be getting a bit scary, but the actual concept and the required calculations are actually not particularly difficult. Work your way through the examples below , and I think you will find that the hardest part of this lesson is getting past the wording!
::我承认,这一教训中的术语可能有点吓人,但实际概念和所需的计算实际上并不特别困难。尝试一下下面的例子,我想你会发现,这一教训中最难的部分已经超越了措辞!

Finding the Mean of Means
::寻求手段之方法

Given the following sample means, what is the mean of means?
::鉴于以下抽样方式,什么方式方式?

\begin{aligned} {\bar{x}}_{1} = 4.35, {\bar{x}}_{2} = 4.62, {\bar{x}}_{3} = 4.29, {\bar{x}}_{4} = 4.39, {\bar{x}}_{5} = 4.55 \end{aligned}

$\begin{align*}\overline{x}_1=4.35, \ \overline{x}_2=4.62, \ \overline{x}_3=4.29, \ \overline{x}_4=4.39, \ \overline{x}_5=4.55\end{align*}$
::-=================================================================================================================================================================================================================================

To calculate the mean of means, sum the sample means and divide by the number of samples:
::为计算手段的平均值,将抽样手段和样本数量除以:

\begin{aligned} μ_{\bar{x}} & = \frac{4.35 + 4.62 + 4.29 + 4.39 + 4.55}{5} \\ = \frac{22.20}{5} \\ μ_{\bar{x}} & = 4.44 \end{aligned}

$\begin{align*}\mu_{\overline{x}} & =\frac{4.35+4.62+4.29+4.39+4.55}{5} \\ & = \frac{22.20}{5}\\ \mu_{\overline{x}} & =4.44 \end{align*}$
::μx 4.35+4.62+4.29+4.39+4.555=22.205 μx 4.44

Real-World Application: Pizza
::真实世界应用:披萨

Brian works at a pizza restaurant, and has been carefully monitoring the weight of cheese he puts on each pizza for the past week. Each day, Brian tracks the weight of the cheese on each pizza he makes, and calculates the mean weight of cheese on each pizza for that day. If the weights below represent the mean weights for each day, what is the mean of means weight of cheese over the past week? If Brian makes 25 pizzas per day and knows the standard deviation of cheese weight per pizza is 0.5 oz, what is the standard deviation of the sample distribution of the sample means?
::布赖恩在一家比萨饼餐厅工作,并仔细监测他在过去一周里每次披萨上涂的奶酪重量。布莱恩每天追踪他做的每份比萨的奶酪重量,计算每天每份比萨的奶酪平均重量。如果以下重量代表每天的平均重量,那么上周奶酪的重量值是多少? 如果布莱恩每天做25份比萨,知道每份比萨饼的奶酪重量标准偏差是0.5oz,那么抽样抽样分布的标准偏差是多少?

DAY ::日间	WEIGHT (OZ) ::韦恩(OZ)
Monday ::星期一星期一	7.84
Tuesday ::星期二星期二星期二	7.93
Wednesday ::星期三	7.79
Thursday ::星期四星期四	8.03
Friday ::星期五,星期五	8.14
Saturday ::星期六星期六	8.09
Sunday ::星期日星期日	7.88

First calculate the mean of means by summing the mean from each day and dividing by the number of days:
::首先计算手段的平均值,方法是将每天的平均值打成平方,除以天数:

\begin{aligned} μ_{\bar{x}} & = \frac{7.84 + 7.93 + 7.79 + 8.03 + 8.14 + 8.09 + 7.88}{7} \\ = \frac{55.7}{7} \\ μ_{\bar{x}} & = 7.96 \end{aligned}

$\begin{align*}\mu_{\overline{x}} & = \frac{7.84+7.93+7.79+8.03+8.14+8.09+7.88}{7} \\ & = \frac{55.7}{7} \\ \mu_{\overline{x}} & =7.96\end{align*}$
::μx 7.84+7.93+7.79+8.03+8.14+8.09+7.887=55.77μx @7.96

Then use the formula to find the standard deviation of the sampling distribution of the sample means:
::然后使用公式找出样本抽样分布的标准偏差,系指:

\begin{aligned} σ_{\bar{x}} = \frac{σ}{\sqrt{n}} \end{aligned}

$\begin{align*}\sigma_{\overline{x}}=\frac{\sigma}{\sqrt{n}}\end{align*}$
::========================================================n

Where $\begin{align*}\sigma\end{align*}$ is the standard deviation of the population, and $\begin{align*}n\end{align*}$ is the number of data points in each sampling.
::何为人口标准偏差,何为每一抽样中的数据点数。

\begin{aligned} σ_{\bar{x}} & = \frac{.05 o z}{\sqrt{25}} \\ = \frac{.05}{5} \\ σ_{\bar{x}} & = .01 \end{aligned}

$\begin{align*}\sigma_{\overline{x}} & =\frac{.05 \ oz}{\sqrt{25}} \\ & = \frac{.05}{5}\\ \sigma_{\overline{x}} & =.01\end{align*}$
::= 05 oz25= 055 -31x 01

Brian’s research indicates that the cheese he uses per pizza has a mean weight of 7.96 oz, with a standard deviation of .01 oz.
::Brian的研究表明,他每吃一份披萨时使用的奶酪平均重量为7.96oz,标准偏差为.01oz。

Calculating the Mean
::计算平均值

Calculate $\begin{align*}\mu_{\overline{x}}\end{align*}$ , given the following:
::计算 mx, 给定如下 :

$\begin{align*}{\overline{x}_1}=352.7\end{align*}$
::x1=352.7

$\begin{align*}{\overline{x}_2}=351.9\end{align*}$
::x2=351.9

$\begin{align*}{\overline{x}_3}=349.97\end{align*}$
::=3=349.97 =3=349.97

$\begin{align*}{\overline{x}_4}=352.33\end{align*}$
::x'4=352.33

$\begin{align*}{\overline{x}_5}=353.1\end{align*}$
::x5=353.1

$\begin{align*}{\overline{x}_6}=349.63\end{align*}$
::x '6=349.63 = 349.63

The $\begin{align*}\mu_{\overline{x}}\end{align*}$ (mean of means) of the given data is:
::给定数据的mx'(手段手段)是:

\begin{aligned} μ_{\bar{x}} & = \frac{352.7 + 351.9 + 349.97 + 352.33 + 353.1 + 349.63}{6} \\ = \frac{2109.63}{6} \\ μ_{\bar{x}} & = 351.61 \end{aligned}

$\begin{align*}\mu_{\overline{x}} & = \frac{352.7+351.9+349.97+352.33+353.1+349.63}{6}\\ & = \frac{2109.63}{6}\\ \mu_{\overline{x}} & =351.61 \end{align*}$
::μx 352.7+351.9+349.97+352.33+353.1+349.636=2109.636 μx 351.61

Earlier Problem Revisited
::重审先前的问题

Suppose you have taken several samples of 10 units each from a population of 500 students, and calculated the mean of each sample. How might you use the data you now have to estimate a mean for the entire population?
::假设您已经从500名学生中采集了10个单位的样本,并且计算了每个样本的平均值。您如何使用您现在必须使用的数据来估算整个人群的平均值?

You could sum the means of each 10-unit sampling, and divide by the number of samples to get the mean of the means. You could further divide the standard deviation of the entire 500 students (if known) by $\begin{align*}\sqrt{10}\end{align*}$ (since each sampling contained 10 data points), to find the standard deviation of the distribution of the sample mean.
::您可以将每10个单位的抽样方法相加, 并用样本数量除以方法的平均值。您可以将全部500名学生( 如果已知的话)的标准差再除以10( 因为每个样本包含10个数据点), 以找到样本分布中值的标准差。

Examples
::实例

Example 1
::例1

Calculate $\begin{align*}\mu_{\overline{x}}\end{align*}$ and $\begin{align*}\sigma_{\overline{x}}\end{align*}$ , given the following data:
::以下列数据计算 =================================================================================================================================================

$\begin{align*}{\overline{x}_1}=251.6\end{align*}$
::x1=251.6

$\begin{align*}{\overline{x}_2}=242.8\end{align*}$
::x '2=242.8

$\begin{align*}{\overline{x}_3}=248.79\end{align*}$
::3=248.79 = 248.79

$\begin{align*}{\overline{x}_4}=245.33\end{align*}$
::x4=245.33

$\begin{align*}{\overline{x}_5}=253.21\end{align*}$
::x5=253.21

$\begin{align*}{\overline{x}_6}=256.31\end{align*}$
::x 6=256.31

$\begin{align*}\text{Sample size}=9, \ \sigma=5.8\end{align*}$
::样本规模=9,5.8

First calculate $\begin{align*}\mu_{\overline{x}}\end{align*}$ , using $\begin{align*}\mu_{\overline{x}}=\frac{\overline{x}_1+\overline{x}_2+\overline{x}_3 \ldots +\overline{x}_n}{n}\end{align*}$ :
::第一次计算,使用xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{aligned} \frac{251.6 + 242.8 + 248.79 + 245.33 + 253.21 + 256.31}{6} \\ \frac{1498.04}{6} \\ μ_{\bar{x}} = 249.67 \end{aligned}

$\begin{align*}&\frac{251.6+242.8+248.79+245.33+253.21+256.31}{6}\\ & \qquad \qquad \qquad \qquad \quad \frac{1498.04}{6}\\ & \qquad \qquad \qquad \qquad \ \ \mu_{\overline{x}}=249.67 \end{align*}$
::251.6+242.8+248.79+245.33+253.21+256.3161498.46 μx 249.67

Next calculate $\begin{align*}\sigma_{\overline{x}}\end{align*}$ , using $\begin{align*}\sigma_{\overline{x}}=\frac{\sigma}{\sqrt{n}}\end{align*}$ :
::下一个计算 ===================================================================================================

\begin{aligned} σ_{\bar{x}} & = \frac{5.8}{\sqrt{9}} \\ = \frac{5.8}{3} \\ σ_{\bar{x}} & = 1.93 \bar{3} \end{aligned}

$\begin{align*}\sigma_{\overline{x}} & =\frac{5.8}{\sqrt{9}} \\ & = \frac{5.8}{3}\\ \sigma_{\overline{x}} & =1.93\overline{3}\end{align*}$
::5点89分5分83分13分1分33分

Example 2
::例2

If the $\begin{align*}\sigma\end{align*}$ of a population is 2.94 and 25 samples of 12 samples each are taken, what is $\begin{align*}\sigma_{\overline{x}}\end{align*}$ ?
::如果每组人口有2.94个样本和25个样本,每组12个样本被采集,那么什么是xx?

To calculate $\begin{align*}\sigma_{\overline{x}}\end{align*}$ , use $\begin{align*}\sigma_{\overline{x}}=\frac{\sigma}{\sqrt{n}}\end{align*}$ :
::计算 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

\begin{aligned} σ_{\bar{x}} & = \frac{2.94}{\sqrt{12}} \\ = \frac{2.94}{4} \\ σ_{\bar{x}} & = .735 \end{aligned}

$\begin{align*}\sigma_{\overline{x}} & =\frac{2.94}{\sqrt{12}}\\ & = \frac{2.94}{4}\\ \sigma_{\overline{x}} & =.735\end{align*}$
::2.9412=2.944-13x 735

Example 3
::例3

Given the population [1, 2, 3, 4, 5], create a sampling distribution by finding the mean of all possible samples that include four units. How does $\begin{align*}\mu_{\overline{x}}\end{align*}$ compare to $\begin{align*}\mu\end{align*}$ ?
::鉴于人口[1、2、3、4、5],通过发现所有可能包括四个单位的样本的平均值,建立抽样分布。

This one requires a few steps, first we need to find the mean of each possible sample of four units:
::这需要采取几个步骤,首先,我们需要找到每个可能抽样的4个单元的平均值:

$\begin{align*}\overline{x}_1(1234)\end{align*}$ has a mean of 2.5
::x1(1234)的平均值为2.5

$\begin{align*}\overline{x}_2(1235)\end{align*}$ has a mean of 2.75
::x'[1235] 的平均值为2.75

$\begin{align*}\overline{x}_3(1245)\end{align*}$ has a mean of 3
::x 3( 1245) 的平均值为 3

$\begin{align*}\overline{x}_4(1345)\end{align*}$ has a mean of 3.25
::x 4(1345) 平均值为 3. 25

$\begin{align*}\overline{x}_5(2345)\end{align*}$ has a mean of 3.5
::x 5(2345) 平均值为3.5

Next we calculate the mean of means, $\begin{align*}\mu_{\overline{x}}\end{align*}$ , using $\begin{align*}\mu_{\overline{x}}=\frac{\overline{x}_1+\overline{x}_2+\overline{x}_3 \ldots +\overline{x}_n}{n}\end{align*}$ :
::接下来,我们计算手段的平均值, xx,使用 lxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{aligned} μ_{\bar{x}} & = \frac{2.5 + 2.75 + 3 + 3.25 + 3.5}{5} \\ = \frac{15}{5} \\ μ_{\bar{x}} & = 3 \end{aligned}

$\begin{align*}\mu_{\overline{x}} & =\frac{2.5+2.75+3+3.25+3.5}{5} \\ & = \frac{15}{5} \\ \mu_{\overline{x}} & =3\end{align*}$
::μx 2.5+2.75+3+3.25+3.55=155μx 3

Now we need to calculate $\begin{align*}\mu\end{align*}$ , using all the population data:
::现在我们需要使用所有的人口数据来计算微克:

\begin{aligned} μ & = \frac{1 + 2 + 3 + 4 + 5}{5} \\ = \frac{15}{5} \\ μ & = 3 \end{aligned}

$\begin{align*}\mu & =\frac{1+2+3+4+5}{5} \\ & = \frac{15}{5} \\ \mu & =3\end{align*}$

With the given data, $\begin{align*}\mu _{\overline{x}}=\mu\end{align*}$
::使用给定的数据, μx {_______________________________________________________

Review
::回顾

Find $\begin{align*}\mu_{\overline{x}}\end{align*}$ , given $\begin{align*}\overline{x}_1=21.0, \ \overline{x}_2=24.3, \ \overline{x}_3=25.0, \ \overline{x}_4=20.6, \ \overline{x}_5=22.3, \ \text{and} \ \overline{x}_6=22.3 \end{align*}$
::找到xxxxxxx1=21,0x2=24.3x3=25,0x4=20.6,x5=22.3和x6=22.3

Find $\begin{align*}\mu_{\overline{x}}\end{align*}$ , given $\begin{align*}\overline{x}_1=15.1,\ \overline{x}_2=15.77, \ \overline{x}_3=15.55,\ \overline{x}_4=15.99, \ \overline{x}_5=15.42, \ \text{and} \ \overline{x}_6=15.37\end{align*}$
::找到xxxxxx1=15.1,x2=15.77,x3=15.55,x4=15.99,x5=15.42,和x6=15.37

Find $\begin{align*}\mu_{\overline{x}}\end{align*}$ , given $\begin{align*}\overline{x}_1=341.52, \ \overline{x}_2=345.16, \ \overline{x}_3=343.66,\ \overline{x}_4=345.86, \ \text{and} \ \overline{x}_5=336.10 \end{align*}$
::找到xxxxxx341.52x345.16x3343.66x4=345.86x5=33610

Find $\begin{align*}\mu_{\overline{x}}\end{align*}$ , given $\begin{align*}\overline{x}_1=1.41,\ \overline{x}_2=0.59, \ \overline{x}_3=1.44,\ \overline{x}_4=0.93, \ \overline{x}_5=1.44, \ \overline{x}_6=1.01, \ \text{and} \ \overline{x}_7=0.74\end{align*}$
::找到xxxxxxxlx1=1.41x2=0.59x3=1.44x4x4=0.93x5=1.44x6=1.11和7=0.74

Find $\begin{align*}\mu_{\overline{x}}\end{align*}$ , given $\begin{align*}\overline{x}_1=218.19, \ \overline{x}_2=279.70, \ \overline{x}_3=262.86, \ \text{and} \ \overline{x}_4=243.88 \end{align*}$
::找xx,给x1=218.19,x2=279.70,x3=262.86,和x4=243.88

If the $\begin{align*}\sigma\end{align*}$ of a population is 292.66, and samples of 17 units each are taken, what is $\begin{align*}\sigma_{\overline{x}}\end{align*}$ ?
::如果人口是292.66,每个样本有17个单位,那么什么是xx?

If the $\begin{align*}\sigma\end{align*}$ of a population is 41.39, and 23 samples of 30 samples each are taken, what is $\begin{align*}\sigma_{\overline{x}}\end{align*}$ ?
::如果每组人口是 41.39 和23个样本每人30个样本被采集,那什么是xx?

If the $\begin{align*}\sigma\end{align*}$ of a population is 193.61, and samples of 19 units each are taken, what is $\begin{align*}\sigma_{\overline{x}}\end{align*}$ ?
::如果人口是193.61个单位,每个单位采集19个单位的样本,那么什么是xx?

If the $\begin{align*}\sigma\end{align*}$ of a population is 91.85, and 129 samples of 11 samples each are taken, what is $\begin{align*}\sigma_{\overline{x}}\end{align*}$ ?
::如果每组人口有 91.85 和129个样本分别采集11个样本, 那么这算什么?

If the $\begin{align*}\sigma\end{align*}$ of a population is 255.19, and 43 samples of 31 samples each are taken, what is $\begin{align*}\sigma_{\overline{x}}\end{align*}$ ?
::如果每组人口有 255.19 和43个样本分别采集31个样本

Given the population: {1, 2, 3, 4, 5, 6}, create a sampling distribution by finding the mean of all possible samples that include two units. How does $\begin{align*}\mu_{\overline{x}}\end{align*}$ compare to $\begin{align*}\mu\end{align*}$ ?
::考虑到人口总数:{1, 2, 3, 4, 5, 6}, 通过找到包括两个单位的所有可能的样本的平均值, 建立抽样分布。 x 如何与 μ 相比 ?

Given the population: {1, 2, 3, 4}, create a sampling distribution by finding the mean of all possible samples that include two units. What is $\begin{align*}\mu_{\overline{x}}\end{align*}$ ?
::考虑到人口:{1, 2, 3, 4}, 通过找到包括两个单位的所有可能样本的平均值, 建立抽样分布。什么是 ?

Given the population: {1, 2, 3, 4, 5}, create a sampling distribution by finding the mean of all possible samples that include two units. How does $\begin{align*}\mu_{\overline{x}}\end{align*}$ compare to $\begin{align*}\mu\end{align*}$ ?
::考虑到人口:{1, 2, 3, 4, 4, 5}, 通过找到包括两个单位的所有可能的样本的平均值, 建立抽样分布。 mx 如何与 μ 相比 ?

Given the population: {1, 2, 3, 4, 5}, create a sampling distribution by finding the mean of all possible samples that include three units. What is $\begin{align*}\mu_{\overline{x}}\end{align*}$ ?
::考虑到人口:{1, 2, 3, 4, 4, 5}, 找出包括三个单位的所有可能样本的平均值, 从而建立抽样分布。什么是 ?

Given the population: {1, 2, 3, 4}, create a sampling distribution by finding the mean of all possible samples that include three units. What is $\begin{align*}\mu_{\overline{x}}\end{align*}$ ?
::考虑到人口:{1, 2, 3, 4},通过寻找包括三个单位的所有可能样本的平均值, 建立抽样分布。什么是 ?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

The Mean of Means ::手段的用法

Finding the Mean of Means ::寻求手段之方法

Real-World Application: Pizza ::真实世界应用:披萨

Calculating the Mean ::计算平均值

Earlier Problem Revisited ::重审先前的问题

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)