统计与优化之方差分析作业

- 日理万妓 2022-10-23 00:49 118阅读 0赞

2.抽查某地区三所小学五年级男学生的身高，测得数据如下表（单位：厘米）  
![加粗样式\*\*][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4MDQ4NzU2_size_16_color_FFFFFF_t_70]

对于这样一道题目，如果靠手机计算器去计算那些数据，简直能把人给累死，我使用numpy来对数据进行计算，具体辅助计算代码如下：  
import numpy as np

x = np.array(\[\[128.1, 134.1, 131.1, 138.9, 140.8, 127.4\],  
\[150.3, 147.9, 136.8, 126.0, 150.7, 155.8\],  
\[140.6, 143.1, 144.5, 143.7, 148.5, 146.4\]\])  
print(x.shape)

x\_mean\_1 = np.around(np.mean(x, axis=1), 2)  
x\_mean = np.around(x.mean(), 2)

print(x\_mean\_1)  
print(x\_mean)

sa = np.around(np.sum(np.square(x\_mean\_1 - x\_mean))\*6, 2)  
print(sa)

x\_mean\_1 = x\_mean\_1.reshape((3, 1))

se = np.around(np.sum(np.square(x - x\_mean\_1)), 2)  
print(se)

完整答案：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4MDQ4NzU2_size_16_color_FFFFFF_t_70 1]

3、 对四种不同的纺织机器采用三种不同的加压水平，对每种加压水平和每台机器作一  
次试验，测得纱支强度如下表  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4MDQ4NzU2_size_16_color_FFFFFF_t_70 2]

仍旧采用numpy辅助计算，如果用计算器的话，我同学说算了一小时没算完，下面是辅助代码：  
import numpy as np

x = np.array(\[\[1577, 1690, 1800, 1642\],  
\[1535, 1640, 1780, 1621\],  
\[1592, 1652, 1810, 1663\]\])  
a = x.shape\[0\]  
b = x.shape\[1\]  
print(a, b)

xipot\_mean = np.around(np.mean(x, axis=1), 2)  
xpotj\_mean = np.around(np.mean(x, axis=0), 2)  
x\_mean = np.around(np.mean(x), 2)  
print(xipot\_mean, xpotj\_mean, x\_mean)  
print(xpotj\_mean - x\_mean)

sa = np.around(b \* np.sum(np.square(xipot\_mean - x\_mean)), 2)  
sb = np.around(a \* np.sum(np.square(xpotj\_mean - x\_mean)), 2)  
print(sa, sb)

xipot\_mean.resize((len(xipot\_mean), 1))

se = np.sum(np.square(x - xipot\_mean - xpotj\_mean + x\_mean))  
print(np.around(se, 2))  
st = sa + sb + se  
print(st)  
s1\_square = np.around(sa/(a - 1), 2)  
s2\_square = np.around(sb/(b - 1), 2)  
s3\_square = np.around(se/(a\*b - a - b + 1), 2)

fa = np.around(s1\_square/s3\_square, 2)

fb = np.around(s2\_square/s3\_square, 2)  
print(s1\_square, s2\_square, s3\_square, fa, fb)  
完整的答案：

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4MDQ4NzU2_size_16_color_FFFFFF_t_70 3]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4MDQ4NzU2_size_16_color_FFFFFF_t_70 4]

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