# Estimating the length of a line and the size of an angle.

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Introduction

## Introduction

This coursework is about estimating the length of a line and the size of an angle. The purpose of this is so that I could see which year group is better at estimating year 10 or year 11. The factors, which may influence people’s estimations are the following, firstly age. Year 11 students are one year older than year ten students so it is likely that year 11 students will know more about estimation than year 10. Secondly skills may effect people’s estimation because some people may have weak numerical skills while others may have good numerical skills. Thirdly the strategy of estimating can affect pupil’s estimation because different strategies of estimating can influence different estimations. Furthermore experience can affect people’s estimation because some people may already have done work on estimation and know a bit about it while others may have not. Plus education can affect people’s answers because some people may have started education at an earlier age than others so they will have more knowledge than the people who started education late so it can affect people’s estimations. Knowledge can cause variation in answers because some people may have more knowledge than others so they will no more and will able to estimate better than the others. In addition background can cause variation of peoples estimations because some people may have come from LEDC countries where education may be at low standards and their parents were not thought as well as the other pupils parents who lived in MEDC countries. So the genes the parents may have past on can affect the rate at which a child may learn so as a result some people will have a better and a closer estimation than others.

Middle

Hypothesis one

### To support my hypothesis one that year 11 is better than year 10 at estimating the length of a line and size of a angle. I will work out the mode, mean, median and range to support my hypothesis.

Year 10 results of females and males of estimation of length of a line.

7.5, 8, 8, 8.5, 9, 9, 9, 10, 10, 10, 10.25, 11, 11, 11, 11, 11.5, 11.5, 11.75, 12, 12, 12, 12, 12, 12.5, 13, 13, 13, 13.5, 14, 15

Mode = 12

Mean = 330.5 / 30 = 11

Median = 30 + 1 / 2 = 15.5 = 11+ 11.5 / 2 = 11.25

Range =15 - 7.5 = 6.5

Year 11 results of females and males of estimation of length of a line.

9, 9, 9, 9.5, 9.5, 10, 10, 10, 10, 10, 10.2, 10.4, 10.5, 10.75, 11, 11, 11, 11, 11, 11.5, 11.5, 12, 12, 12, 12, 12, 13, 13.5, 14, 15

Mode = 10, 11, 12

Mean = 331.35 / 30 = 11.1

Median = 30 + 1 / 2 = 15.5 = 11 +11 / 2 = 11

Range = 14 - 9 = 5

If you look at the data it shows that the mode for year 11 was more closer to the actual length of the line than the mode for year 11 was. This means that more people in year 11 estimated closer to the actual length than year 10. Which shows year 11 is better than year 10 at estimating. The mean for both sets of data show that the year 11 was slightly better than year 10 which means year 11 on average were better than year 10 at estimating. The median for year 10 is closer to the actual length than the median for year 11 which means year 10 was better than year 11. If you look at the range of the data you will notice that year 11’s range is smaller than year 10’s range this means that year 11’s estimations are less varied. This means it gives the best measure of location. This shows that overall year 11 is better than year 10 at estimating the length of a line which supports my hypotheisis.

Year 10 results of females and males of estimation of size of the angle

45, 49, 50, 52, 55.5, 57, 58, 59, 60, 60, 61, 62, 64, 65, 65, 65, 65, 70, 70, 70.1, 71, 72.2, 73, 74, 75, 75, 78, 78, 78.5, 80

Mode = 65

Mean = 1957.3 / 30 = 65.2

Median = 30 + 1 / 2 = 15.5 = 65 + 65 / 2 = 65

Range = 80 - 45 = 35

Year 11 results of females and males of estimation of size of the angle

54, 54.75, 55, 55.25, 55.5, 59, 60, 60, 60, 60, 61, 61, 62, 63, 64, 64.8, 65, 65, 67, 68.1, 68.5, 69, 69, 70.5, 71, 71, 72, 75, 77.5, 78

Mode = 60

Mean = 1814.9 / 30 = 60.5

Median = 30 + 1 / 2 = 15.5 = 64 + 64.8 / 2 = 64.4

Range = 78 - 54 = 34

The results show that the mode for year 11 estimations of the size of the angle is closer to the actual size of the angle than the mode for year 10’s estimations. This shows that year 11 is better at estimating than year 10. In addition year 11’s mean is closer to the actual size of the angle than the mean for year 10, which means that year 11’s estimations are on average better than year 10’s estimation. Also the range of year 11’s results is smaller than the range of year 10 meaning that year 11 results are less varied and are more consistent. This gives the best measure of mode, mean and median. This consolidates my hypothesis that year 11 is better than year 10 at estimating.

#### Back to back stem and leaf diagram for the estimations of a length of a line.

Stem and leaf diagram for estimations of the length of a line

Key 0 /7/ means 7.0

Year 10 Year 11

5 7

9 0 0 8

0 0 0 9 0 0 0 5 5

25 0 0 0 10 0 0 0 0 0 2 4 5 75

75 5 5 0 0 0 0 11 0 0 0 0 0 5 5

5 0 0 0 0 0 12 0 0 0 0 0

5 0 0 0 13 0 5

0 14 0

0 15 0

Mode = 12 Mode = 10, 11, 12

Mean = 330.5 / 30 = 11 Mean = 331.35 / 30 = 11.1 Median = 30 + 1 / 2 = 15.5 Median = 30 + 1 / 2 = 15.5

= 11.5 + 11.5 / 2 = 11.5 = 11 + 11 / 2 = 11

Range = 14 - 9 = 5 Range = 15-7.5 = 6.5

The measures of location show that year 11 is better than year 10 at estimating and so does the range by their being a smaller range group so the estimations are more consistent which gives it the best measure of mode, mean and median.

#### Box and whisker diagrams

I will draw box and whisker diagrams for both year 10 and year 11 to support my hypothesis one, that year 11 is better than year 10 at estimating the length of a line. To this I will find the lowest value, lower quartile, median, upper quartile and the highest value for year 10 and year 11 results of estimating the length of a line and I will compare them to support my hypothesis.

Year 10 Year 11

LV = Lowest Value LV = Lowest Value

= 7.5 = 9

LQ = Lower Quartile LQ = Lower Quartile

= n + 1 / 4 = n + 1 / 4

= 30 + 1 / 4 = 30 + 1 / 4

= 7.5 term = 7.5 term

= 7 term + 8 term / 2 = 7 term + 8 term / 2

= 9 + 10 / 2 = 10 + 10 / 2

= 9.5 = 10

M = Median = n + 1 / 2 M = Median = n + 1 / 2

= (30 + 1 / 2) = (30 + 1 / 2)

= 31 / 2 = 31 / 2

= 15 .5 term = 15 .5 term

= 15 term + 16 term / 2 = 15 term + 16 term / 2

= 11 + 11.5 / 2 = 11 + 11 / 2

= 22.5/ 2 = 22/ 2

= 11.25 = 11

UQ = (n + 1 / 4) * 3 UQ = (n + 1 / 4) * 3

= (30 + 1 / 4) * 3 = (30 + 1 / 4) * 3

= 22.5 term = 22.5 term

= 22 term + 23 term / 2 = 22 term + 23 term / 2

= 12 + 12.5 / 2 = 12 + 12 / 2

= 24.5 / 2 = 24 / 2

= 12.25 = 12

HV = Highest Value HV = Highest Value

= 15 = 15

IQR = Interquartile range IQR = Interquartile range

= UQ – LQ = UQ – LQ

= 12.25 – 9.5 = 12 - 10

= 2.75 = 2

From year 10 estimations of a length of a line the LQ = 9.5, M = 11.25 and UQ = 12.25.

UQ – M = 12.25 – 11.25 M – LQ = 11.25 – 9.5

= 1 = 1.75

From year 11 estimations of the length of a line the LQ = 10, M = 11 and UQ = 12.

UQ – M = 12 – 11 M – LQ = 11 – 10

= 1 = 1

The results show that year median of year 10 estimations is not in the middle of the box but is closer to the upper quartile which means the it has a negative skew. While the median of year is in the centre of the box this means that year 11 estimations has a symmetrical skew.

The median for year 10 is closer to the actual estimation of the length of a line than the median for year 11 is.

Year 10 estimations for a length of a line have a wider box than year 11. Which means year 11 has a narrower interquartile range than year 10.

The whickers of the box for year 10 extend further than the whiskers of the box for year 11. So year 10 has a greater range of values than year 10.

This shows that the box and whiskers for year 10 is narrower than those for year 11. This means that year 11 estimations are more consistent than year 10. Which means that year 11 is better than year 10 at estimating the length of a line. This supports my hypothesis one.

Back to back stem and leaf diagram for the estimations of a size of an angle

Now I will draw a back-to-back stem and leaf diagram from both year 10 and year 11 using their estimations of a size of an angle.

Key 9/7 means 97

Year 10 Year 11

9 5 4

9 8 7 5.5 2 0 5 4 4.75 5 5.25 5.5 9

5 5 5 5 4 2 1 0 0 6 0 0 0 0 11 2 3 4 4.8 5 5 7 8.1 8.5 9 9 8.5 8 8 5 5 4 3 2.2 1 .1 0 0 7 0.5 1 1 2 5 7.5 8

0 8

Mode = 65 Mode = 60

Mean = 1957.3 / 30 = 65.2 Mean = 1933.2 / 30 = 64.44

Median = 30 + 1 / 2 = 15.5 Median = 30 + 1 / 2 = 15.5 = 65 + 65 = 65 = 64 + 64.8 / 2 = 64.4

Range = 80 - 45 = 35 Range = 78 - 54 = 24

The mode for year 11 is much closer to the actual size of the angle than the mode for year 10. This shows that more people in year 11 had closer estimations to the actual size of the angle than the people in year 10.

The mean shows the average, which the average of year 11 is closer to the actual estimation than the mean for year 10. Which shows that year 11 on average estimated closer to the actual estimation than year 10.

The median for year 11 is nearer the actual size of the angle than the median for year 10 this shows that year 11 estimated closer than year 10 to actual size of the angle.

The range of year 11 for estimations of the size of the angle is much smaller than the range for year 10 estimations of the size of an angle. This shows that year 10 estimations are more varied and less reliable than those in year 11. This shows that year 11 estimated better than year 10.

This clearly show that year 11 estimations are more accurate and reliable than estimations of year 10 this shows that year 11 is better than year 10 at estimating. This again supports my hypothesis.

Now I will draw a box and whisker diagram for the estimations of the size of an angle for year 10 and year 11. I will do this by following the same procedure as I did to draw the box and whisker diagrams for estimations of a length of a line.

Year 10 Year 11

LV = Lowest Value LV = Lowest Value

= 45 = 54

LQ = Lower Quartile LQ = Lower Quartile

= n + 1 / 4 = n + 1 / 4

= 30 + 1 / 4 = 30 + 1 / 4

= 7.5 term = 7.5 term

= 7 term + 8 term / 2 = 7 term + 8 term / 2

= 58 + 59 / 2 = 60 + 60 / 2

= 58.5 = 60

M = Median = n + 1 / 2 M = Median = n + 1 / 2

= (30 + 1 / 2) = (30 + 1 / 2)

= 31 / 2 = 31 / 2

= 15 .5 term = 15 .5 term

= 15 term + 16 term / 2 = 15 term + 16 term / 2

= 65 + 65 / 2 = 64 + 64.8 / 2

= 130 / 2 = 128.8 / 2

= 65 = 64.4

UQ = (n + 1 / 4) * 3 UQ = (n + 1 / 4) * 3

= (30 + 1 / 4) * 3 = (30 + 1 / 4) * 3

= 22.5 term = 22.5 term

= 22 term + 23 term / 2 = 22 term + 23 term / 2

= 72.2 + 73 / 2 = 69 + 69 / 2

= 145.2 / 2 = 138 / 2

= 72.6 = 69

HV = Highest Value HV = Highest Value

= 80 = 78

IQR = Interquartile range IQR = Interquartile range

= UQ – LQ = UQ – LQ

= 80 - 58.5 = 69 - 60

= 21.5 = 10

###### Box and whisker diagrams

Using these values I will draw a box and whisker diagram for year 10 and year 11 for their estimations of a size of an angle and I will compare them to prove my hypothesis one.

From year 10 estimations of a size of an angle the LQ = 58.5, M = 65 and UQ = 72.6.

UQ – M = 72.6 – 65 M – LQ = 65 – 58.5

= 7.6 = 6.5

From year 11 estimations of a size of an angle the LQ = 60, M = 64.4 and UQ = 69.

UQ – M = 69 – 64.4 M – LQ = 64.4 – 60

= 4.6 = 4.4

From the results the median of the results for year 10 estimations is not in the middle of the box but is closer to the lower quartile by 1.1 degrees from the centre. The median of year 11 is also not in the centre of the box but is closer to the lower quartile by 0.2 degrees from the centre. Both boxes show a negative skew but the box for year 10 shows a more negative skew than the one for year 11.

The median for year 11 is closer to the actual estimation of the size of an angle when comparing it to the median of year 10 box. This means year 11 estimations are more closer to the actual size of the angle than year 10 estimations.

Year 11 estimations for a size of an angle has a narrower box than year 11. Which means year 11 estimations are more consistent and reliable than year 10 results.

The whickers of the box for year 10 extend further than the whiskers of the box for year 11. This means that year 10 has a greater range of values than year 11.

This shows that the box and whiskers for year 11 is narrower than the box and whiskers for year 10. This means that year 11 estimations are more consistent than year 10. This means that year 11 is better than year 10 at estimating the size of the angle. This supports my hypothesis that year 11 is better than year 10 at estimating.

Bar chart

I will now draw a bar chart for year 10 and year 11 results for the estimations of a length of a line and size of an angle. This will show the comparison between year 10 and year 11 estimations.

From looking at both bar charts for year 10 and year 11 estimations of a length of a line I could see that the graph for year 10 has a higher and wider range of estimations for the length of a line than the graph for year 11 estimations of a length of a line. In addition the mode for year 11 is much closer to the actual length of the line than the mode for year 10. This clearly indicates that the results for your 10 are more spread out than the results for year 11. Which means year 11 results are more reliable than year 10. Which in turn means year 11 estimated closer to the actual length of the line than year 10, which shows that my hypothesis one was correct.

Now I will draw the bar chart for year 10 and year 11 estimations of a size of an angle. I will do this the same way that I used to draw a graph for the size of an angle.

If you compare both the graphs you will see that 11 has a narrower range of results than year 10. Also you could see the mode for year 11 is nearer to the actual size of the angle than year 10. This shows that the results for year 11 are more accurate because the measure of spread is closer to the actual size of the angle for year 11 than it is for year 10. This clearly shows that year 11 is better than year 10 at estimating. This supports my hypothesis one that year 11 is better than year 10 at estimating.

From the measures of spread and location, stem and leaf diagram, box and whisker diagrams and bar charts you could see that they all show that year 11 estimated closer to the actual length of the line and size of the angle than year 10. This supports my hypothesis that year 11 is better than year 10 at estimating.

#### Comparing with secondary data

#### I will compare my data with a secondary data which is from St.Bedes School which I downloaded it from the website. I am going to use the secondary data to see how representative it is of the whole population and because I want to compare our school estimations with their school estimations.

From analysing the secondary data I could see that their size of the actual angle was 56 degrees and their actual length of the line was 4.6cm. The results for the estimations of the size of an angle are similar to the results I have got and their results of the length of a line more people estimated very close to then they estimated the length of my length of a line. This is because they chose a smaller length, which is much easier to estimate than a larger length. Evan thought their data is similar to the data I collected I could see that there are some outliers. Outliers are very small or very large values in a set of data I will find the outlier and ignore it because they can distort the data.

To find the value of the outliers I will need to find the interquartile range of the data. I will firstly find the median of the data and look at the data to the left of the median and find the position of the lower quartile and then the upper quartile. Then I will find the interquartile range by subtracting the lower quartile from the upper quartile. I will then draw box and whisker diagrams and analyse the secondary data for any outliers and I will eliminate them. This will leave me with reliable data.

The median for year 10 estimations of a length of a line is

½ (363 + 1) = 182nd value

= 5

The lower quartile for year 10 estimations of a length of a line is

½ (182 + 1) = 91½ th value

= 4.5 + 4.5 /2

= 4.5

The upper quartile for year 10 estimations of a length of a line is

½ (182 + 1) * 3 = 91½ th value

91½ th value * 3

= 274.5 the value

= 6 + 6 /2

= 6

So the interquartile range = UQ – LQ

= 6 – 4.5

=1.5

Interquartile range * 1.5

= 1.5 * 1.5

= 2.25

So any values that are below the lower quartile or the upper quartile by 2.25 are outliers.

LQ – 2.25 = 4.5 – 2.25

= 2.25

The small outliers are values that are below 2.25, which there are not any small outliers in the data.

Now I will find any large outliers that distort the data. The large outliers are

UQ + 2.25 = 6 + 2.25

= 8.25

any numbers that are bigger than 8.25, which in the results for year 10 there are 26 outliers, which is a lot, and I will get rid of them because they distort the data. I will get rid of them by eliminatating them from the data and not including them when drawing the box and whisker diagrams.

Number of times outliers occurs in the data | Outliers | Eliminating them |

3 | 9 | 9/4.6 = 1.96 |

1 | 9.2 | 9.2 / 4.6 = 2 |

9 | 10 | 10 / 4.6 = 2.17 |

1 | 12 | 12/4.6 = 2.61 |

2 | 13 | 13/4.6 = 2.83 |

1 | 15 | 15/4.6 = 3.26 |

1 | 18 | 18/4.6 = 3.91 |

2 | 20 | 20/4.6 = 4.35 |

1 | 30 | 30/4.6 = 6.52 |

1 | 57 | 57/4.6 = 12.39 |

1 | 74 | 74/4.6 = 16.09 |

1 | 180 | 180/4.6 = 39.13 |

Now using all the data except the outliers that distort the data I will work out draw a box and whisker diagram to show the results.

I will do the same as I did for the length of a line to find and eliminate the outliers for the estimations of the size of an angle for year 10.

The median for year 10 estimations of a size of an angle

½ (354 + 1) = 177.5th value

= 177.5

= 55 + 55 / 2

= 55

The lower quartile for year 10 estimations of a size of an angle

½ (177.5 + 1) = 89.25 th value

= 89th value

= 45

The upper quartile for year 10 estimations of a size of an angle

½ (177.5 + 1) = 89.25 th value

89.25 th value * 3

= 267.75 th value

= 268th value

= 60

So the interquartile range = UQ – LQ

= 60 – 45

=15

Interquartile range * 1.5

= 15 * 1.5

= 22.5

So any values that are below the lower quartile or the upper quartile by 22.5 are outliers.

LQ – 22.5 = 45 – 22.5

= 22.5

The small outliers are values that are below 22.5, which there are two small outliers. I will eliminate them

Number of times outliers occurs in the data | Outliers | Eliminating them |

1 | 6 | 6/45 = 0.13 |

1 | 10 | 10/45 = 0.22 |

Now I will find any large outliers that distort the data. The large outliers are

UQ + 22.5 = 60 + 22.5

= 80.5

Which in the data there are 8 theses are

Number of times outliers occurs in the data | Outliers | Eliminating them |

2 | 82 | 82 / 56 = 1.46 |

1 | 85 | 85/56 = 1.52 |

4 | 90 | 90 / 56 = 1.61 |

1 | 105 | 105/56 = 1.88 |

I have worked out the outliers for year 10 I will do the same and work out the outliers for year 11 data. Firstly I will workout the outliers for the length of a line.

The median for year 11 estimations of a length of a line is

½ (363 + 1) = 182nd value

= 5

The lower quartile for year 10 estimations of a length of a line is

½ (182 + 1) = 91½ th value

= 4.5 + 4.5 /2

= 4.5

The upper quartile for year 10 estimations of a length of a line is

½ (182 + 1) * 3 = 91½ th value

91½ th value * 3

= 274.5 the value

= 5.7 + 6 /2

= 5.85

So the interquartile range = UQ – LQ

= 5.85 – 4.5

=1.35

Interquartile range * 1.5

= 1.35 * 1.5

= 2.03

So any values that are below the lower quartile or the upper quartile by 2.03 are outliers.

LQ – 2.03 = 4.5 – 2.03

= 2.47

The small outliers are values that are below 2.47, which there is only one that is 2

Number of times outliers occurs in the data | Outliers | Eliminating them |

1 | 2 | 2/4.6 = 0.43 |

Now I will find any large outliers that distort the data. The large outliers are

UQ + 2.03 = 5.85 + 2.03

= 7.88

Which in the data there are 39 these are

Number of times outliers occurs in the data | Outliers | Eliminating them |

6 | 8 | 8 / 4.6 = 1.74 |

4 | 9 | 9/4.6 = 2.0 |

1 | 9.2 | 9.2 /4.6 =2 |

/9 | 10 | 10/4.6 = 2.17 |

1 | 11 | 11/4.6 = 2.39 |

2 | 12 | 12/4.6 = 2.61 |

2 | 13 | 13/4.6 = 2.83 |

4 | 15 | 15/4.6 = 3.26 |

1 | 17 | 17/4.6 = 3.7 |

1 | 18 | 18/4.6 =3.91 |

1 | 19 | 19/4.6 = 4.13 |

2 | 20 | 20/4.6 = 4.35 |

1 | 30 | 30/4.6 = 6.52 |

1 | 57 |
Conclusion
After that I will draw box and whisker diagrams for year 10 and year 11 estimations of a length of a line and size of an angle and I will compare year 10 results with year 11 results. Then I will support my finding even further by drawing a bar chart and support my hypothesis.Then I will correlate all my findings and come to a conclusion linking it to my hypothesis.When I have done that I will compare my results with another secondary school called St. Bedes and see if the results are similar to my results.The I will support hypothesis two that people who estimate the length of a line accurately may not estimate the size of an angle accurately. To support this I will draw scatter graphs and use spearman’s rank correlation to support my hypothesis.When I have done that I will come to a conclusion and again compare with my hypothesis.Then I will support hypothesis there by working out the standard deviation for year 10 and year 11 estimations of a length of a line and size of an angle and I will use it to work out the standardised scores to see whether estimating the length of a line is easier then estimating the size of an angle and then I will come to a finding to compare with my hypothesise.The I will support hypothesis four by using stratified and random sampling to choose 8 females and 8 males to represent my sample and use their estimations to draw a back-to-back stem and leaf diagram t compare the results. After that I draw comparative pie charts to show that girls are not better then boys at estimating but to show they are the same. After doing that I will correlate all my findings and compare it with my hypothesis.Then I will do a conclusion and evaluation for my project. ...read more.
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