国家开放大学2023年秋季学期期末统一考试 社会统计学试题（含答案）

试卷代号：11318

2024年1月

学号：__________ 姓名：__________ 考点名称：__________

注意事项

1. 将你的学号、姓名及考点名称填写在试题和答题纸的规定栏内。考试结束后，把试题和答题纸放在桌上。试题和答题纸均不得带出考场。待监考人员收完试题和答题纸后方可离开考场。

2. 仔细阅读题目的说明，并按题目要求答题。所有答案必须写在答题纸的指定位置上，写在试题上的答案无效。

3. 用蓝、黑圆珠笔或钢笔（含签字笔）答题，使用铅笔答题无效。

一、单项选择题（每题只有一个正确答案，每题2分，共20分）

1. 某校社会工作专业共200名学生，其中男生120名，占总体60%，女生80名，占总体40%。为了了解该专业学生对社会统计学的掌握情况，需要抽取40人进行调查。该校按照男生抽取24人、女生抽取16人进行了抽样。这种抽样方法是()。

A.定额抽样

B.分层抽样

C.整群抽样

D.多阶段抽样

答案：A

2. 以下关于因变量与自变量的表述，正确的是()。

A.自变量的变化与因变量的变化互为前提

B.自变量是由于其他变量的变化而导致自身发生变化的变量

C.因变量是引起其他变量变化的变量

D.因变量的变化以自变量的变化为前提

答案：D

3. 某班级学生平均每学期的实习时长可以分为以下六组：1)50小时及以下：2)50-60小时；3)60-70小时；4)70-80小时；5)80-90小时；6)90小时及以上，则90小时及以上这一组的组中值近似为( )

A.95小时

B.100小时

C.95.5小时

D.100.5小时

答案：A

4. 中心极限定理认为不论总体分布是否服从正态分布，从均值为μ、方差为σ²的总体中，抽取容量为 n 的随机样本，当 n 充分大时（通常要求n≥30），样本均值的抽样分布近似服从均值为、方差为的正态分布。()

A.μ、σ²

B.μ/n、σ²/n

C.μ、σ²/n

D.μ/n、σ²

答案：C

5. 甲、乙两人同时打靶，各打10靶。甲平均每靶为8环，标准差为2；乙平均每靶9环，标准差为3。经过计算后发现，甲的稳定性要高于乙，得出该结论时使用的测量指标是()。

A.方差

B.标准差

C.均值

D.离散系数

答案：D

6. 以下四种不同特征的散点图，哪一个属于正线性相关()。

答案：A

7. 某单位对该厂第一加工车间残品率估计约为13%，该车间主任认为该比例不符合实际情况，如果要检验该说法是否正确，假设形式应该为()。

A.H₀:π≥0.13；H₁:π<0.13<>

B.H₀:π≤0.13；H₁:π>0.13

C.H₀:π=0.13；H₁:π≠0.13

D.H₀:π>0.13；H₁:π≤0.13

答案：C

8. 以下关于正态分布参数特点的描述，不正确的是()。

A.曲线的陡缓程度由σ决定

B.当均值μ相等时，标准差σ越大，峰值越小

C.当均值μ相等时，标准差σ越大，覆盖范围越大

D.当均值μ相等时，标准差σ越小，峰值越小

答案：D

9. 在回归方程中，若回归系数等于0，这表明()。

A.因变量 y 对自变量x的影响是不显著的

B.自变量x对因变量 y 的影响是不显著的

C.因变量y对自变量 x 的影响是显著的

D.自变量 x 对因变量y的影响是显著的

答案：B

10. 下表是某单位工作人员年龄分布表，该单位工作人员年龄的众数是( )。

组别 按年龄分组(岁) 工作人员数(人)

1 20~24 6

2 25~29 14

3 30~34 24

4 35~39 18

5 40~44 12

6 45~49 18

7 50~54 14

8 55~59 6

合计 112

A.第二组

B.第三组

C.第四组

D.第七组

答案：B

二、名词解释（每小题5分，共20分）

11. 概率抽样

参考答案：按照随机原则进行的抽样，总体中每个个体都是有一定的、非零的概率入选样本，并且入选样本的概率都是已知的或可以计算的。

12. 中心趋势

参考答案：中心趋势也叫集中趋势，反映一组数据中各个数值向中心值集中的程度，是指一组数据向某一个中心值靠拢的趋势。

13. 区间估计

参考答案：区间估计是在估计总体参数时给出的不是一个数值，而是一个区间，是根据统计量的抽样分布的特点进行估计，同时给出总体参数落入这一区间的可能性大小。

14. 回归分析

参考答案：回归分析是通过一定的数学表达式将变量间的关系进行描述，确定一个变量或几个变量的变化对另一个特定变量的影响，是进行估计或预测的一种方法，侧重于考察变量之间的数量伴随关系。

三、简答题（每题10分，共30分）

15. 简述问卷调查的优缺点。

参考答案：

（1）优点：获得的资料便于进行定量分析；节约时间、人力和经费；应用范围广；结果具有较高的信度。（4分）

（2）不足：在探讨和分析变量间的因果关系方面相对较弱；对事物的理解和解释的深入性及效度相对较差；限制了被调查者对问题的回答，所得资料比较表面化、简单化。（6分）

16. 简述什么是简单一元线性回归分析? 其作用是什么?

参考答案：

简单一元线性回归分析，是对具有线性相关关系的两个变量（一个自变量、一个因变量）的数量变化规律进行分析，通过数学表达式描述两个变量的线性依存关系，以此实现变量预测与分析的统计方法。（4分）

主要作用：①从已知数据出发，确定变量之间的数学关系式；②对变量间的关系式进行统计检验，判断自变量对因变量的影响是否显著；③依托建立的关系式，根据自变量取值预测、估算因变量的取值。（6分）

17. 简述按照测量水平区分的四类变量，并举例说明。

参考答案：

（1）定类变量：变量值仅区分个体类别，无大小、程度差异。例如：性别、职业。（2.5分）

（2）定序变量：可区分类别，且能比较个体大小、优劣、高低等序次差异，无精确间隔。例如：学历、满意度等级。（2.5分）

（3）定距变量：可分类、排序，且能精准计算变量间的间隔差距，无绝对零点。例如：智商、温度。（2.5分）

（4）定比变量：具备前三类变量所有特征，有绝对零点，可计算变量比值。例如：收入、身高、体重。（2.5分）

四、计算题（每题15分，共30分）

18. 一项关于大学生体重状况的研究发现，男生的平均体重为60千克，标准差为5千克；女生的平均体重为50千克，标准差为5千克。请问：

(1) 是男生体重差异大还是女生体重差异大?为什么?

(2) 男生中有多少比重的人体重在55千克-65千克之间?

(3) 女生中有多少比重的人体重在40千克-60千克之间?

参考答案：

（1）通过离散系数对比：

男生离散系数：V男=5/60=0.083

女生离散系数：V女=5/50=0.1

V男＜V女，因此女生体重差异更大。（6分）

（2）P(55

Φ(1)=0.8413，代入得：2×0.8413-1=0.6826

即68.26%的男生体重在55-65千克之间。（4分）

（3）P(40

Φ(2)=0.9772，代入得：2×0.9772-1=0.9544

即95.44%的女生体重在40-60千克之间。（5分）

19. 某汽车生产商欲了解广告费用(万元)对销售量(辆)的影响。收集了过去12年的有关数据，通过分析得到：方程的截距为363，回归系数为1.42。

要求：

(1) 写出销售量 y 与广告费用x之间的线性回归方程。

(2) 假如明年计划投入广告费用为50万，根据回归方程估计明年汽车销售量。

参考答案：

（1）已知截距β₀=363，回归系数β₁=1.42

线性回归方程：y=363+1.42x（9分）

（2）当x=50时，y=363+1.42×50=434（辆）

预计明年汽车销售量为434辆。（6分）

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