相关系数公式

相关系数用于计算两个变量之间连接的重要性。有不同类型的相关系数，其中最流行的是皮尔逊相关系数（也称为皮尔逊 R），它常用于线性回归。

公式类型

皮尔逊相关系数公式

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$ $\text{[math]}$

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样本相关系数公式

$r_{xy}= cov(x,y) / s_xs_y$ $\text{[math]}$

S_xy is the sample Covariance, and S_x and S_y are the sample standard deviations

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人口相关系数公式

$pxy= cov(X,Y) / σxσy$ $\text{[math]}$

It uses σ_x and σ_y as the population standard deviation and, σ_xy as the population Covariance.

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皮尔逊相关性

它是统计学中最常见的相关性。全称是Pearson's Product Moment correlation，简称PPMC。它显示了两组数据之间的线性关系。两个字母用于表示 Pearson 相关性：希腊字母 rho (ρ) 表示总体，字母“r”表示样本相关系数。

求 Pearson 相关系数的步骤

Step 1: Firstly make a chart with the given data like subject,x, and y and add three more columns in it xy, x² and y².

Step 2: Now multiply the x and y columns to fill the xy column. For example:- in x we have 24 and in y we have 65 so xy will be 24×65=1560.

Step 3: Now, take the square of the numbers in the x column and fill the x² column.

Step 4: Now, take the square of the numbers in the y column and fill the y² column.

Step 5: Now, add up all the values in the columns and put the result at the bottom. Greek letter sigma (Σ) is the short way of saying summation.

Step 6: Now, use the formula for Pearson’s correlation coefficient:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

To know which type of variable we have either positive or negative.

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线性相关系数

Pearson 相关系数是线性相关系数，它返回介于 -1 和 +1 之间的值。其中，-1 表示强负相关，+1 表示强正相关。如果它位于 0 则没有相关性。这也称为零相关。

使用 Pearson 相关性分析相关性稳定性的“粗略估计”：

r value	crude estimates
+.70 or higher	A very strong positive relationship
+.40 to +.69	Strong positive relationship
+.30 to +.39.	Moderate positive relationship
+.20 to +.29	weak positive relationship
+.01 to +.19	No or negligible relationship
0	No relationship [zero correlation]
-.01 to -.19	No or negligible relationship
-.20 to -.29	weak negative relationship
-.30 to -.39	Moderate negative relationship
-.40 to -.69	Strong negative relationship
-.70 or higher	The very strong negative relationship

Cramer 的 V 相关性

它与皮尔逊相关系数相似。它用于计算超过 2×2 行和列的相关性。 Cramer 的 V 相关性在 0 和 1 之间变化。接近零的值表示变量之间存在非常小的关联，如果接近 1，则表明关联非常强。

使用 Cramer 的 V 相关性解释相关性强度的“粗略估计”：

Cramer’s V	crude estimates
.25 or higher	Very strong relationship
.15 to .25	Strong relationship
.11 to .15	Moderate relationship
.06 to .10	weak relationship
.01 to .05	No or negligible relationship

示例问题

问题1：根据下表计算相关系数：

SUBJECT	AGE X	GLUCOSE LEVEL Y
1	42	98
2	23	68
3	22	73
4	47	79
5	50	88
6	60	82

解决方案：

Make a table from the given data and add three more columns of XY, X², and Y².

∑xy = 20379

∑x = 244

∑y = 488

∑x² = 11086

∑y² = 40266

n = 6.

Put all the values in the Pearson’s correlation coefficient formula:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

R = 6(20379) – (244)(488) / √[6(11086)-(244)²][6(40266)-(488)² ]

R = 3202 / √[6980][3452]

R = 3202/4972.238

R = 0.6439

It shows that the relationship between the variables of the data is a strong positive relationship.

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问题2：根据下表计算相关系数：

SUBJECT	AGE X	GLUCOSE LEVEL Y	XY	X²	Y²
1	42	98	4116	1764	9604
2	23	68	1564	529	4624
3	22	73	1606	484	5329
4	47	79	3713	2209	6241
5	50	88	4400	2500	7744
6	60	82	4980	3600	6724
∑	244	488	20379	11086	40266

解决方案：

Make a table from the given data and add three more columns of XY, X², and Y².

SUBJECT	AGE X	Weight Y	XY	X²	Y²
1	40	99	3960	1600	9801
2	25	79	1975	625	6241
3	22	69	1518	484	4761
4	54	89	4806	2916	7921
∑	151	336	12259	5625	28724

∑xy = 12258

∑x = 151

∑y = 336

∑x² = 5625

∑y² 28724

n = 4

Put all the values in the Pearson’s correlation coefficient formula:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

R = 4(12258) – (151)(336) / √[4(5625)-(151)²][4(28724)-(336)²]

R = -1704 / √[-301][-2000]

R=-1704/775.886

R=-2.1961

It shows that the relationship between the variables of the data is a very strong negative relationship.

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问题3：计算以下数据的相关系数：

X = 7,9,14 和 Y = 17,19,21

解决方案：

Given variables are,

X = 7,9,14

and,

Y = 17,19,21

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula.

X	Y	XY	X²	Y²
7	17	119	49	36
9	19	171	81	361
14	21	294	196	441
∑ 30	∑ 57	∑ 584	∑ 326	∑ 838

∑xy = 584

∑x = 30

∑y = 57

∑x² = 326

∑y² = 838

n = 3

Put all the values in the Pearson’s correlation coefficient formula:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

R = 3(584) – (30)(57) / √[3(326)-(30)²][3(838)-(57)²]

R = 42 / √[78][-735]

R = 42/-239.43

R = -0.1754

It shows that the relationship between the variables of the data is negligible relationship

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问题4：计算以下数据的相关系数：

X = 21、31、25、40、47、38 和 Y = 70、55、60、78、66、80

解决方案：

Given variables are,

X = 21,31,25,40,47,38

And,

Y = 70,55,60,78,66,80

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula.

X	Y	XY	X²	Y²
21	70	1470	441	4900
31	55	1705	961	3025
25	60	1500	625	3600
40	78	3120	1600	6094
47	66	3102	2209	4356
38	80	3040	1444	6400
∑202	∑409	∑13937	∑7280	∑28265

∑xy = 13937

∑x = 202

∑y = 409

∑x² = 7280

∑y² = 28265

n = 6

Put all the values in the Pearson’s correlation coefficient formula:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

R = 6(13937) – (202)(409) / √[6(7280) – (202)²][6(28265) – (409)²]

R = 1004 /√[2876][2909]

R = 1004 / 2892.452938

R = 0.3471

It shows that the relationship between the variables of the data is a moderate positive relationship.

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问题5：计算以下数据的相关系数？

X = 5 ,9 ,14, 16 和 Y = 6, 10, 16, 20 。

解决方案：

Given variables are,

X = 5 ,9 ,14, 16

And

Y = 6, 10, 16, 20.

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula add all the values in the columns to get the values used in the formula

X	Y	XY	X²	Y²
5	6	30	25	36
9	10	90	81	100
14	16	224	196	256
16	20	320	256	400
∑44	∑52	∑664	∑558	∑792

∑xy = 664

∑x = 44

∑y = 52

∑x² = 558

∑y² = 792

n = 4

Put all the values in the Pearson’s correlation coefficient formula:

$R= n(∑xy) - (∑x)(∑y) / √[n∑x²-(∑x)²][n∑y²-(∑y)²$

R = 4(664) – (44)(52) / √[4(558) – (44)²][4(792) – (52)²]

R = 368 / √[296][464]

R = 368/370.599

R = 0.9930

It shows that the relationship between the variables of the data is a very strong positive relationship.

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问题 6：计算以下数据的相关系数：

X = 10, 13, 15 ,17 ,19 和 Y = 5,10,15,20,25。

解决方案：

Given variables are,

X = 10, 13, 15 ,17 ,19 and Y = 5, 10, 15, 20, 25.

To, find the correlation coefficient of the following variables Firstly a table is to be constructed as follows, to get the values required in the formula also add all the values in the columns to get the values used in formula,

X	Y	XY	X²	Y²
10	5	50	100	25
13	10	130	169	100
15	15	225	225	225
17	20	340	340	400
19	25	475	475	625
∑74	∑75	∑1103	∑1144	∑1375

∑xy = 1103

∑x = 74

∑y = 75

∑x² = 1144

∑y² = 1375

n = 5

Put all the values in the Pearson’s correlation coefficient formula:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

R = 5(1103) – (74)(75) / √ [5(1144) – (74)²][5(1375) – (75)²]

R = -35 / √[244][1250]

R = -35/552.26

R = 0.0633

It shows that the relationship between the variables of the data is a negligible relationship.

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习题7：计算下列数据的相关系数：

X = 12、10、42、27、35、56 和 Y = 13、15、56、34、65、26

解决方案：

Given variables are,

X = 12, 10, 42, 27, 35, 56 and Y = 13, 15, 56, 34, 65, 26

X	Y	XY	X²	Y²
12	13	156	144	169
10	15	150	100	225
42	56	2352	1764	3136
27	34	918	729	1156
35	65	2275	1225	4225
56	26	1456	3136	676
∑182	∑209	∑7307	∑7098	∑9587

∑xy = 7307

∑x = 182

∑y = 209

∑x² = 7098

∑y² = 9587

n = 6

Put all the values in the Pearson’s correlation coefficient formula:

$R= \frac{n(∑xy) - (∑x)(∑y)}{\sqrt{[n∑x²-(∑x)²][n∑y²-(∑y)²}}$

R = 6(7307) – (182)(209) / √ {[6(7098) – (182)²][6(9587)-(209)²]}

R = 5804 / √[9464][13841]

R = 5804/11445.139

R = 0.5071

It shows that the relationship between the variables of the data is a strong positive relationship.

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