懿说学区（11） | SPSS统计分析（21）配对样本T检验实例
Yishuo school district (11) | SPSS statistical analysis (21) paired sample t-test Example
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上一期我们进行了配对样本T检验的理论学习，了解了配对样本T检验的具体内容以及应用场景，本期，我们将要结合实例进行配对样本T检验的SPSS软件操作。
In the last issue, we learned the theory of paired sample t-test, and learned the specific contents and application scenarios of paired sample t-test. In this issue, we will conduct SPSS software operation of paired sample t-test with examples.
实例场景：以下是某大学15位跆拳道选手平衡训练的数据，结合实验前、后平衡训练成绩是否有差异。
Example scenario: the following is the data of balance training of 15 Taekwondo Players in a university, and whether there is any difference in balance training results before and after the experiment.
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第一步
第一步：建立SPSS数据文件，建立两个变量：“训练前”，“训练后”，录入相应数据，保存文件。
Step 1: create SPSS data file, create two variables: "before training" and "after training", input corresponding data and save the file.
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第二步
第二步：选择菜单“分析->比较平均值->成对样本T检验”，弹出对话框，确定要分析的变量，如下图所示进行设置。
Step 2: select the menu "analysis - > compare average value - > paired sample t-test" to pop up a dialog box, determine the variables to be analyzed, and set them as shown in the following figure.
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第三步
第三步：运行结果分析。
Step 3: operation result analysis.
结果分析
我们可以看到，在显著性水平为0.05的情况下，概率P值为0.132，大于0.05，接受原假设，可以认为训练前后的成绩没有明显的线性关系。
表中显示，训练前后配对样本的平均差值为5.933，差值的标准差是10.187，差值的均值标准误差为2.630，置信度为95%时，差值的置信下限和置信上限共同构成了该差值的置信区间，观测量的统计值t为-2.256，自由度df为14，显著性概率P值为双尾检测P值，在显著性水平为0.05时，由于概率P值为0.041，小于0.05，拒绝原假设，即认为μ1-μ2不等于0，故可以认为训练对成绩有显著效果。
We can see that when the significance level is 0.05, the probability p value is 0.132, greater than 0.05. Accepting the original hypothesis, it can be considered that there is no obvious linear relationship between the scores before and after training.
The table shows that the average difference of paired samples before and after training is 5.933, the standard deviation of the difference is 10.187, the mean standard error of the difference is 2.630, and the confidence level is 95%, the lower and upper confidence limits of the difference jointly constitute the confidence interval of the difference. The statistical value t of observation is -2.256, the degree of freedom DF is 14, and the significance probability p value is the two tailed detection p value. When the significance level is 0.05, because the probability p value is 0.041, less than 0.05, Reject the original assumption, i.e μ 1- μ 2 is not equal to 0, so it can be considered that training has a significant effect on performance.
下期预告：
下一节，
我们将会学习非参数检验的
概念体系和理论知识。
Next issue Preview: in the next section, we will learn the conceptual system and theoretical knowledge of nonparametric testing.
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That's all for today's sharing. If you have a unique idea about today's article, please leave us a message and let us meet tomorrow. I wish you a happy day today!
参考资料：百度百科，《00SPSS 23 统计分析实用教程》
翻译：百度翻译
本文由learningyard新学苑原创，部分文字图片来源于他处，如有侵权，请联系删除
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