# Labcoat Leni solutions Chapter 2

This document contains abridged sections from *Discovering Statistics Using R and RStudio* by Andy Field so there are some copyright considerations. You can use this material for teaching and non-profit activities but please do not meddle with it or claim it as your own work. See the full license terms at the bottom of the page.

## Is Friday 13th unlucky?

Let’s begin with accidents and poisoning on Friday the 6th. First, arrange the scores in ascending order: 1, 1, 4, 6, 9, 9.

The median will be the (*n* + 1)/2th score. There are 6 scores, so this will be the 7/2 = 3.5th. The 3.5th score in our ordered list is half way between the 3rd and 4th scores which is (4+6)/2= 5 accidents.

The mean is 5 accidents:

$$ \begin{align}
\bar{X} &= \frac{\sum_{i = 1}^{n}x_i}{n} \\

&= \frac{1 + 1 + 4 + 6 + 9 + 9}{6} \\

&= \frac{30}{6} \\

&= 5 \end{align} $$

The lower quartile is the median of the lower half of scores. If we split the data in half, there will be 3 scores in the bottom half (lowest scores) and 3 in the top half (highest scores). The median of the bottom half will be the (3+1)/2 = 2nd score below the mean. Therefore, the lower quartile is 1 accident.

The upper quartile is the median of the upper half of scores. If we again split the data in half and take the highest 3 scores, the median will be the (3+1)/2 = 2nd score above the mean. Therefore, the upper quartile is 9 accidents.

The interquartile range is the difference between the upper and lower quartiles: $ 9 - 1 = 8 $ accidents.

To calculate the sum of squares, first take the mean from each score, then square this difference, and finally, add up these squared values:

Score | Error (Score - Mean) | Error squared |
---|---|---|

1 | -4 | 16 |

1 | -4 | 16 |

4 | -1 | 1 |

6 | 1 | 1 |

9 | 4 | 16 |

9 | 4 | 16 |

So, the sum of squared errors is: 16 + 16 + 1 + 1 + 16 + 16 = 66.

The variance is the sum of squared errors divided by the degrees of freedom $ (N - 1) $:

$$ s^{2} = \frac{\text{sum of squares}}{N- 1} = \frac{66}{5} = 13.20 $$

The standard deviation is the square root of the variance:

$$ s = \sqrt{\text{variance}} = \sqrt{13.20} = 3.63 $$

Next let’s look at accidents and poisoning on Friday the 13th. First, arrange the scores in ascending order: 5, 5, 6, 6, 7, 7.

The median will be the (*n* + 1)/2th score. There are 6 scores, so this will be the 7/2 = 3.5th. The 3.5th score in our ordered list is half way between the 3rd and 4th scores which is (6+6)/2 = 6 accidents.

The mean is 6 accidents:

$$ \begin{align} \bar{X} &= \frac{\sum_{i = 1}^{n}x_{i}}{n} \\

&= \frac{5 + 5 + 6 + 6 + 7 + 7}{6} \\

&= \frac{36}{6} \\

&= 6 \end{align} $$

The lower quartile is the median of the lower half of scores. If we split the data in half, there will be 3 scores in the bottom half (lowest scores) and 3 in the top half (highest scores). The median of the bottom half will be the (3+1)/2 = 2nd score below the mean. Therefore, the lower quartile is 5 accidents.

The upper quartile is the median of the upper half of scores. If we again split the data in half and take the highest 3 scores, the median will be the (3+1)/2 = 2nd score above the mean. Therefore, the upper quartile is 7 accidents.

The interquartile range is the difference between the upper and lower quartiles: $ 7-5 = 2 $ accidents.

To calculate the sum of squares, first take the mean from each score, then square this difference, finally, add up these squared values:

Score | Error (Score - Mean) | Error squared |
---|---|---|

7 | 1 | 1 |

6 | 0 | 0 |

5 | -1 | 1 |

5 | -1 | 1 |

7 | 1 | 1 |

6 | 0 | 0 |

So, the sum of squared errors is: 1 + 0 + 1 + 1 + 1 + 0 = 4.

The variance is the sum of squared errors divided by the degrees of freedom $ (N - 1) $:

$$ s^{2} = \frac{\text{sum of squares}}{N - 1} = \frac{4}{5} = 0.8 $$

The standard deviation is the square root of the variance:

$$ s = \sqrt{\text{variance}} = \sqrt{0.8} = 0.894 $$

Next, let’s look at traffic accidents on Friday the 6th. First, arrange the scores in ascending order: 3, 5, 6, 9, 11, 11.

The median will be the (*n* + 1)/2th score. There are 6 scores, so this will be the 7/2 = 3.5th. The 3.5th score in our ordered list is half way between the 3rd and 4th scores. The 3rd score is 6 and the 4th score is 9. Therefore the 3.5th score is (6+9)/2 = 7.5 accidents.

The mean is 7.5 accidents:

$$ \begin{align} \bar{X} &= \frac{\sum_{i = 1}^{n}x_{i}}{n} \\

&= \frac{3 + 5 + 6 + 9 + 11 + 11}{6} \\

&= \frac{45}{6} \\

&= 7.5 \end{align} $$

The lower quartile is the median of the lower half of scores. If we split the data in half, there will be 3 scores in the bottom half (lowest scores) and 3 in the top half (highest scores). The median of the bottom half will be the (3+1)/2 = 2nd score below the mean. Therefore, the lower quartile is 5 accidents.

The upper quartile is the median of the upper half of scores. If we again split the data in half and take the highest 3 scores, the median will be the (3+1)/2 = 2nd score above the mean. Therefore, the upper quartile is 11 accidents.

The interquartile range is the difference between the upper and lower quartiles: $ 11 - 5 = 6 $ accidents.

To calculate the sum of squares, first take the mean from each score, then square this difference, finally, add up these squared values:

Score | Error (Score - Mean) | Error squared |
---|---|---|

9 | 1.5 | 2.25 |

6 | -1.5 | 2.25 |

11 | 3.5 | 12.25 |

11 | 3.5 | 12.25 |

3 | -4.5 | 20.25 |

5 | -2.5 | 6.25 |

So, the sum of squared errors is: 2.25 + 2.25 + 12.25 + 12.25 + 20.25 + 6.25 = 55.5.

The variance is the sum of squared errors divided by the degrees of freedom $ (N - 1) $:

$$ s^{2} = \frac{\text{sum of squares}}{N - 1} = \frac{55.5}{5} = 11.10 $$

The standard deviation is the square root of the variance:

$$ s = \sqrt{\text{variance}} = \sqrt{11.10} = 3.33 $$

Finally, let’s look at traffic accidents on Friday the 13th. First, arrange the scores in ascending order: 4, 10, 12, 12, 13, 14.

The median will be the (*n* + 1)/2th score. There are 6 scores, so this will be the 7/2 = 3.5th. The 3.5th score in our ordered list is half way between the 3rd and 4th scores. The 3rd score is 12 and the 4th score is 12. Therefore the 3.5th score is (12+12)/2= 12 accidents.

The mean is 10.83 accidents:

$$ \begin{align} \bar{X} &= \frac{\sum_{i = 1}^{n}x_{i}}{n} \\

&= \frac{4 + 10 + 12 + 12 + 13 + 14}{6} \\

&= \frac{65}{6} \\

&= 10.83 \end{align} $$

The lower quartile is the median of the lower half of scores. If we split the data in half, there will be 3 scores in the bottom half (lowest scores) and 3 in the top half (highest scores). The median of the bottom half will be the (3+1)/2 = 2nd score below the mean. Therefore, the lower quartile is 10 accidents.

The upper quartile is the median of the upper half of scores. If we again split the data in half and take the highest 3 scores, the median will be the (3+1)/2 = 2nd score above the mean. Therefore, the upper quartile is 13 accidents.

The interquartile range is the difference between the upper and lower quartile: $ 13-10 = 3 $ accidents.

To calculate the sum of squares, first take the mean from each score, then square this difference, finally, add up these squared values:

Score | Error (Score - Mean) | Error squared |
---|---|---|

4 | -6.83 | 46.65 |

10 | -0.83 | 0.69 |

12 | 1.17 | 1.37 |

12 | 1.17 | 1.37 |

13 | 2.17 | 4.71 |

14 | 3.17 | 10.05 |

So, the sum of squared errors is: 46.65 + 0.69 + 1.37 + 1.37 + 4.71 + 10.05 = 64.84.

The variance is the sum of squared errors divided by the degrees of freedom $ (N-1) $:

$$ s^{2} = \frac{\text{sum of squares}}{N- 1} = \frac{64.84}{5} = 12.97 $$

The standard deviation is the square root of the variance:

$$ s = \sqrt{\text{variance}} = \sqrt{12.97} = 3.6 $$