Introduction

This dataset is called Anscombes Quartet. This is a synthetic dataset that was created to show the importance of looking outside pvalue and R squared value when doing analysis. Given that the model is synthetic there is no missing values and no data qaulity issues.

> head(four)

  x123   y1   y2    y3 x4   y4
1   10 8.04 9.14  7.46  8 6.58
2    8 6.95 8.14  6.77  8 5.76
3   13 7.58 8.74 12.74  8 7.71
4    9 8.81 8.77  7.11  8 8.84
5   11 8.33 9.26  7.81  8 8.47
6   14 9.96 8.10  8.84  8 7.04

Scatter Plots

These scatter plots are all positively correlated. It is important to notice that all of these visualizations look different but have the same R^2 and pvalue. This will be exemplified when linear regression is used.

> scatterplot(y1~x123, regLine=FALSE, smooth=FALSE, boxplots=FALSE, data=four)

> scatterplot(y2~x123, regLine=FALSE, smooth=FALSE, boxplots=FALSE, data=four)

> scatterplot(y3~x123, regLine=FALSE, smooth=FALSE, boxplots=FALSE, data=four)

> scatterplot(y4~x4, regLine=FALSE, smooth=FALSE, boxplots=FALSE, data=four)

Outliers

Y1

This value is normally distributed as the whiskers are approx the same size
There is no large outliers

> Boxplot( ~ y1, data=four, id=list(method="y"))

Y2

This value is not normally distributed as the whisker is much longer on the bottom than the top
There is one large outlier that 3
The distribution is negatively/left skewed

> Boxplot( ~ y2, data=four, id=list(method="y"))

[1] "8"

Y3

This value is normally distributed as the whiskers are approx the same size
There is one large outlier that 13

> Boxplot( ~ y3, data=four, id=list(method="y"))

[1] "3"

Y4

This value is normally distributed as the whiskers are approx the same size
There is one large outlier that 13

> Boxplot( ~ y4, data=four, id=list(method="y"))

[1] "8"

Summary of Variables

From the relationship between median and mean we can also determine distribution.

If \(mean > median = positively sknewed\) and if \(mean < median = negatively skewed\)

X123 is normally distributed as \(9=9\)
Y1 is negatively skewed as \(7.501<7.58\)
Y2 is negatively skewed as \(7.501<8.14\)
Y3 is positively skewed as \(7.5>7.11\)
X4 is positively skewed as \(9>8\)

> summary(four)

      x123            y1               y2              y3              x4    
 Min.   : 4.0   Min.   : 4.260   Min.   :3.100   Min.   : 5.39   Min.   : 8  
 1st Qu.: 6.5   1st Qu.: 6.315   1st Qu.:6.695   1st Qu.: 6.25   1st Qu.: 8  
 Median : 9.0   Median : 7.580   Median :8.140   Median : 7.11   Median : 8  
 Mean   : 9.0   Mean   : 7.501   Mean   :7.501   Mean   : 7.50   Mean   : 9  
 3rd Qu.:11.5   3rd Qu.: 8.570   3rd Qu.:8.950   3rd Qu.: 7.98   3rd Qu.: 8  
 Max.   :14.0   Max.   :10.840   Max.   :9.260   Max.   :12.74   Max.   :19  
       y4        
 Min.   : 5.250  
 1st Qu.: 6.170  
 Median : 7.040  
 Mean   : 7.501  
 3rd Qu.: 8.190  
 Max.   :12.500

Linear Regression

It is evident that the \(Pr(>|t|)\), \(Adjusted R^2\), \(Residual Std Error\) and \(p-values\) are the same in the four regression analysis done. This is important because

each of these variables has different distributions
all appear differently on scatter plots
it is important to remember these statistics are not nessicarily the end of the analysis and a well rounded analysis should always be done

The analysis the the linear regression outputs will include

Residuals
- difference between the prediction and the actual values to see how well the model fits
The Coefficient
- Calculate the linear expression for the regression model
- Standard error

The same calculate the linear expression for the regression model and std error can be used for all regression models

linear expression for the regression model is \[ y=0.5001x+3.0001\]
With std error the formula is
- max \[ y=0.618x+4.1257\]
- min \[ y=0.3822x+1.8763\]

Y1 and X123

Since the residual median is not 0 we can say that this model is not fully symmetrical. However, given the max and min residuals are similar it can be concluded that the model does fit somewhat.

> fourpt1.2 <- lm(y1~x123, data=four)
> summary(fourpt1.2)


Call:
lm(formula = y1 ~ x123, data = four)

Residuals:
     Min       1Q   Median       3Q      Max 
-1.92127 -0.45577 -0.04136  0.70941  1.83882 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   3.0001     1.1247   2.667  0.02573 * 
x123          0.5001     0.1179   4.241  0.00217 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.237 on 9 degrees of freedom
Multiple R-squared:  0.6665,    Adjusted R-squared:  0.6295 
F-statistic: 17.99 on 1 and 9 DF,  p-value: 0.00217

Y2 and X123

Since the residual median is not 0 we can say that this model is not fully symmetrical. However, given the max and min residuals are similar it can be concluded that the model does fit somewhat.

> RegModel.3 <- lm(y2~x123, data=four)
> summary(RegModel.3)


Call:
lm(formula = y2 ~ x123, data = four)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.9009 -0.7609  0.1291  0.9491  1.2691 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)    3.001      1.125   2.667  0.02576 * 
x123           0.500      0.118   4.239  0.00218 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.237 on 9 degrees of freedom
Multiple R-squared:  0.6662,    Adjusted R-squared:  0.6292 
F-statistic: 17.97 on 1 and 9 DF,  p-value: 0.002179

Y3 and X123

Since the residual median is not 0 we can say that this model is not fully symmetrical. Given the max and min residuals are not similar it can be concluded that the model does not fit well.

> fourpt2.4 <- lm(y3~x123, data=four)
> summary(fourpt2.4)


Call:
lm(formula = y3 ~ x123, data = four)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.1586 -0.6146 -0.2303  0.1540  3.2411 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   3.0025     1.1245   2.670  0.02562 * 
x123          0.4997     0.1179   4.239  0.00218 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.236 on 9 degrees of freedom
Multiple R-squared:  0.6663,    Adjusted R-squared:  0.6292 
F-statistic: 17.97 on 1 and 9 DF,  p-value: 0.002176

Y4 and X4

Since the residual median is 0 we can say that this model is fully symmetrical. Given the max and min residuals are similar it can be concluded that the model does fit somewhat.

> fourpt3.5 <- lm(y4~x4, data=four)
> summary(fourpt3.5)


Call:
lm(formula = y4 ~ x4, data = four)

Residuals:
   Min     1Q Median     3Q    Max 
-1.751 -0.831  0.000  0.809  1.839 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   3.0017     1.1239   2.671  0.02559 * 
x4            0.4999     0.1178   4.243  0.00216 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.236 on 9 degrees of freedom
Multiple R-squared:  0.6667,    Adjusted R-squared:  0.6297 
F-statistic:    18 on 1 and 9 DF,  p-value: 0.002165

Pop Quiz: Anscombes Quartet

Erin

2022-03-23