[Statistics Notes] Selected Topics for Statistical Calculation (Essence)

Selected Topics in Statistical Calculation

Question 1

The test scores of a class of students' physics courses are:

68 89 88 84 86 87 75 73 72 68

75 82 97 58 81 54 79 76 95 76

71 60 90 65 76 72 76 85 89 92

64 57 83 81 78 77 72 61 70 81

Grading requirements: 60 points or less is a failure; 60-70 points is a pass; 70-80 points are medium; 80-90 points are good, 90-100 points are excellent.

Claim:

(1) Divide the students who take the test into five groups according to the test scores: fail, pass, middle, good, and excellent and prepare a table of the number of assessment results;

(2) Point out the grouping sign and type and the grouping method adopted;

(3) Calculate the average score of students' physics course assessment

(4) According to the sequence of statistical variables after finishing, infer the degree range of all students' test scores with 95.45% probability guarantee degree.

(5) If other conditions remain unchanged, the allowable error range will be reduced by half, how many students should be scored?

answer:

First, by grouping statistics on the 40 scores of the students' physics course exams, as follows:

Achievement	Number of students	frequency(%)
60 points or less	3	7.5
60-70 points	6	15
70-80 points	15	37.5
80-90 points	12	30
90-100 points	4	10
total	40	100

Transform the above table and calculate:

Achievement	Group median $\large x$	Number of students $\large f$	frequency(%) $\large f/\sum f$	$\large \left ( x-\overline{x} \right )^{2}\times f$
60 points or less	55	3	7.5	$\large \left ( 55-77.0 \right )^{2}\times 3= 1452$
60-70 points	65	6	15	864
70-80 points	75	15	37.5	60
80-90 points	85	12	30	768
90-100 points	95	4	10	1296
total		40	100	4440

According to the table above: $\large \sum f = 40$

According to the calculation formula:

$\large \overline{x} = \sum x \tfrac{f}{\sum f}$

Thus: $\large \overline{x}$ = 55 × 7.5% + 65 × 15% + 75 × 37.5% + 85 × 30% + 95 × 10% = 77.0

That is, the average score of the student's physics course examination is: 77.0 points.

According to the formula:

$\large \sigma = \sqrt{\sum \left ( x-\overline{x} \right )^{2}\times f/\sum f}$

Thereby: $\large \sigma = \sqrt{\frac{4440}{40}}= 10.54$

That is, the standard deviation of the physics course examination results of the class students is ( $\large \sigma = 10.54$ )

and then, $\large \mu _{x} = \farc{\sigma }{\sqrt{n}}= 10.54 / \sqrt{40} = 1.67$

$\large \Delta _{x}= t\mu _x = 2\times 1.67 = 3.34$

The test score range of all students is:

Lower limit = $\large \overline{x} - \Delta _{x} = 77 - 3.34 = 73.66$

Upper limit = $\large \overline{x} + \Delta _{x} = 77 + 3.34 = 80.30$

That is, the score range of all student candidates is between 73.66 and 80.30 points.

If the allowable error range is reduced by half, the number of students that should be drawn is:

$\large n= \frac{t^{2}\sigma}{\Delta {x}^{2} } = \frac{2^{2} \times 10.54^{2}} {\left ( \frac{3.34}{2} \right )^{2}} \approx 159$

The answer is complete.

Question 2

There are two classes taking the statistical test. The average score on the deck is 75 points, and the standard deviation is 11.5 points.

Group by score (points)	Number of students (person)
60 points or less	2
60-70 points	5
70-80 points	8
80-90 points	6
90-100 points	4
total	25

Requirements: (1) Calculate the average score and standard deviation of class B; (2) Compare the average score of which class is more representative?

answer:

To calculate the average score of Class B, you need to perform some simple deformation calculations on the above table:

Group by score (points)	Group median (points) $\large \overline{x}$	Number of students (person) $\large f$	$\large \sum xf$
60 points or less	55	2	110
60-70 points	65	5	325
70-80 points	75	8	600
80-90 points	85	6	510
90-100 points	95	4	380
total		$\large \sum f = 25$	1925

The average grade of Class B is:

$\large \overline{x} = \sum xf / \sum f = 1925 / 25 = 77.0$

According to the formula:

$\large \sigma = \sqrt{\sum \left ( x-\overline{x} \right )^{2}\times f/\sum f}$

then: $\large \sigma = \sqrt{\frac{3400}{25}}= 11.66$

Calculate the coefficient of variation:

Class A: $\large \upsilon _{1} = \frac{\sigma }{\overline{x}} = 11.5 / 75 = 15.33$

Class B: $\large \upsilon _{2} = \frac{\sigma }{\overline{x}} = 11.66 / 77 = 15.14$

Because the standard deviation is greater than the coefficient A class standard deviation coefficient B class, the average score of the class B is more representative .

Question 3

A steel plant produces a certain type of steel pipe, and now a sample of 100 is selected from 500 products produced by the plant in a month. Knowing that the first-grade product rate is 60%, try to find the average sampling error of the first-grade product sample.

Solve:

It understood meaning of the questions: a product of 60%, i.e., p = 60%; capacity of 100 to extract a sample from the product 500, then: $\large N=500$ , $\large n=100$ ,

The answer is complete.

Question 4

The length of parts produced in a factory follows a normal distribution, and 25 parts are randomly selected from the parts produced in the factory. The average length of them is 30.2 cm. The overall standard deviation is known in $\large \sigma = 0.45$ centimeters.
Seeking: (1) Calculate the average sampling error and sampling tolerance. (2) Estimate the possible range of the average length of parts ( $\large \alpha = 0.05$ ).

answer:

From the meaning of the questions $\large X$ - $\large N\left ( \mu , 0.45^{2} \right )$ , $\large \overline{X}=30.2$ , $\large n=25$ , $\large 1-\alpha =0.95$

(1) sample average error is: $\large \sigma \left ( \overline{x} \right ) = \frac{\sigma }{\sqrt{n}} = \frac{0.45}{\sqrt{25}} = 0.09$ , standard normal distribution table found in check $\large \alpha =0.05$ when $\large z_{\alpha /2} = 1.96$ ,

So the sampling tolerance is: $\large \Delta _{\overline{x}} = z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}} = 1.96\times 0.09 = 0.1764$

(2) The confidence interval of the population mean is:

$\large \left ( \overline{X}-\frac{\sigma }{\sqrt{n}}, \overline{X}+\frac{\sigma }{\sqrt{n}}\right ) = \left ( \overline{X}-\Delta _{\overline{X}},\overline{X}+\Delta _{\overline{X}} \right )$

which is $\large \left ( 30.2-0.1764,30.2+0.1764 \right ) =\left ( 30.02,30.38 \right )$

即我们可以以95%的概率保证该厂零件平均长度在30.02厘米到30.38厘米之间。

解答完毕。

第 5 题

从某市高中生中按不重复抽样方法随机抽取25名调查每周收看电视的时间，分组资料见表：

要求：（1）计算抽样平均误差和抽样允许误差；（2）估计该市全体高中生每周平均看电视时间的置信区间（给定的显著性水平为0.05）。

解答：根据题目意思首先将上表做一个简单的处理，

每周看电视时间（小时）	组中值 $\large x$	学生人数（人） $\large f$	$\large \left ( x-\overline{x} \right ) ^{2}\times f$
2以下	1	2	32
2 ~ 4	3	6	24
4 ~ 6	5	8	0
6 ~ 8	7	8	32
8 ~ 10	9	1	16
合计		$\large \sum f=25$	104

学生看电视的平均值为： $\large \overline{x} = \frac{1\times 2+3 \times 6 + 5 \times 8 + 7 \times 8 + 9 \times 1}{25} = 5$ 小时，

样本方差为： $\large \sigma ^{2}= 104/\left ( 25-1 \right )=4.33$

查 $\large t$ 分布表知 $\large \alpha =0.05$ 时，临界值 $\large t_{\alpha /2}\left ( n-1 \right ) = t_{\alpha /2}\left ( 25-1 \right ) = 2.0639$

因此：

抽样平均误差为： $\large \sigma \left ( \overline{X} \right ) = \frac{s}{\sqrt{n}} = \frac{\sqrt{4.33}}{\sqrt{25}} = 0.416$

抽样允许误差为： $\large \Delta _{\overline{X}} = t_{\alpha /2} \times \frac{s}{\sqrt{n}} = 2.0639 \times 0.416 = 0.859$

总体均值置信度为95%的置信区间为： $\large \left ( \overline{X}-\Delta _{\overline{X}},\overline{X}+\Delta _{\overline{X}} \right )$ ，即 $\large \left ( 5-0.859,5+0.859 \right ) = \left ( 4.14,5.86 \right )$

即我们可以以95%的把握保证该市高中生每周平均看电视时间在4.14到5.86小时之间。

解答完毕。

第 6 题

某工厂对一批产成品按不重复抽样方法随机抽选200件进行质量检测，其中一等品160件，试以90%的概率估计一等品率的范围。

解答：

由题意已知：p = 160/200 = 80%； 1-α = 90% ；n=200；

查表知： $\large z_{\alpha /2} = 1.645$ ，计算得样本比例的抽样平均误差为：

$\large \sigma \left ( p \right ) = \sqrt{\frac{p\times \left ( 1-p \right )}{n}} = \sqrt{\frac{0.8\times \left ( 1-0.8 \right )}{200}} = 0.0283$

抽样极限误差为： $\large \Delta _{p} = z_{\alpha /2}\times \sigma \left ( p \right ) = 1.645 \times 0.0283 = 0.04655$ ，即 4.655%

所以，该批产品的一等品比例的置信区间为：80%-4.655% ~ 80%+4.655%，即 75.35% ~ 84.66% 之间。

解答完毕。

第 7 题

从某班学生中随机抽取16人，计算得语文平均成绩为75分，方差为25分。假定学生成绩服从正态分布，试求总体方差及标准差的置信区间（给定的显著性水平为0.05）。

解答：

有题目已知： $\large n=25$ ， $\large \alpha =0.05$ ，查 $\large \chi ^{2}$ 分布表确定两个临界值：

$\large \chi_{1-\alpha /2}^{2} \left ( n-1 \right )= \chi _{0.975}^{2}\left ( 16-1 \right ) = 6.262$ ， $\large \chi_{\alpha /2}^{2} \left ( n-1 \right )= \chi _{0.025}^{2}\left ( 16-1 \right ) = 27.488$

将临界值数字带入公式中，总体方差和标准差的置信度为 $\large 1-\alpha$ 的置信区间分别为：

$\large \left (\frac{\left ( 16-1 \right ) \times 25}{27.488}, \frac{\left ( 16-1 \right ) \times 25}{6.262} \right )$ ，即 $\large \left ( 13.64,59.89 \right )$

$\large \left ( \sqrt{13.64},\sqrt{59.89} \right ) = \left ( 3.69,7.74 \right )$

解答完毕。

第 8 题

1、某快餐店某天随机抽取49名顾客对其的平均花费进行抽样调查。调查结果为：平均花费8.6元，标准差2.8 元。试以95.45％的置信度估计：
（1）该快餐店顾客总体平均花费的置信区间及这天营业额的置信区间（假定当天顾客有2000人）；
（2）若其他条件不变，要将置信度提高到99.73％，至少应该抽取多少顾客进行调查？

（提示： $\large z{_{0.0455}} = 1.69$ ， $\large z{_{0.0455/2}} = 2$ ， $\large z{_{0.0027/2}} = 3$ ， $\large z{_{0.0027}} = 2.78$ ）

解答：

由题意值 $\large \overline{x} = 8.6$ ，标准差： $\large \sigma = 2.8$ ， $\large n=49$ ，

则： $\large \mu _{\overline{x}} = \frac{\sigma }{\sqrt{n}} = 2.8 / \sqrt{49} = 0.4$

由于以95.45%的置信度估计，则 $\large 1-0.9545 = 0.0455$

$\large \Delta _{x} = z_{\alpha /2}\times \mu _{\overline{x}} = z_{0.0455/2} \times 0.4 = 2\times 0.4 = 0.8$

总体均值的置信区间： $\large \left ( 8.6-0.8 , 8.6+0.8 \right )$ ，即 $\large \left ( 7.8, 9.4 \right )$

营业总额的置信区间： $\large \left ( 2000\times 7.8, 2000\times 9.4 \right )$ ，即 $\large \left ( 15600, 18800 \right )$

若其它条件不变，将置信度提高到 99.73%，至少应该抽取的顾客数量为：

$\large 1- 0.9973 = 0.0027$

必要的样本容量： $\large n = \frac{z_{\alpha /2 \times \sigma ^{2}}} {\Delta_{\overline{x}}^{2}} = \frac{3^{2} \times 2.8 ^{2}}{0.8^{2}} = 110.25 \approx 111$

解答完毕。

第 9 题

一所大学准备采取一项学生在宿舍上网收费的措施，为了解男女学生对这一措施的看法，分别抽取了150名男学生和120名女学生进行调查，得到的结果如下：

	男学生	女学生	合计
赞成	45	42	87
反对	105	78	183
合计	150	120	270

请检验男女学生对上网收费的看法是否相同。已知：显著性水平 $\large \alpha =0.05$ ， $\large \chi_{0.05}^{2} \left ( 1 \right )= 3.842$ ， $\large \chi_{0.05}^{2} \left ( 2 \right )= 5.992$ ， $\large \chi_{0.05}^{2} \left ( 4 \right )= 9.487$ 。