The lengths of lines are easier to guess than angles. Also, that year 11s will be more accurate at estimating. Essay Example

The lengths of lines are easier to guess than angles. Also, that year 11s will be more accurate at estimating. Essay Example The lengths of lines are easier to guess than angles. Also, that year 11s will be more accurate at estimating. Essay The lengths of lines are easier to guess than angles. Also, that year 11s will be more accurate at estimating. Essay In this investigation, 3 year groups years 9, 10 and 11, were asked to estimate the lengths of some lines and angles, and the results that the pupils produced are going to be analysed to try and prove or disprove the hypothesis of:The lengths of lines are easier to guess than angles. Also, that year 11s will be more accurate at estimating.The reasons I think these things are because people are more used to seeing lines than they are angles, so this could mean that they are better at estimating the length of lines. The reason I think they year 11s will be more accurate is because they have done maths longer than the year 9s, so they have had more experience.I will be using an example of one line, and one angle, and the results of Year 9 and Year 11 estimates. This is secondary data which has been previously recorded, during a survey to find out the estimates that the pupils gave. This data is continuous as it is As there are 117 year 9s and 145 year 11s I will have to reduce the size of my sample as these numbers are too large to handle, so I will be using a stratified method to reduce the size of the samples as this method keeps the results for the year groups in proportion to each other.I am going to be sampling 60 people in total, out of the year 9s and year 11s, as this is a manageable amount, and it can represent the data from the two year groups accurately as a smaller number might not show the difference in results suitably.To choose my samples I am first going to add together the two total numbers of each year group, which is:145 + 117 = 262 (Year 11 / Year 9)Then I am going to do some calculations. For the year 11s I am going to do:(145 / 262) x 100 = 55.355.3 is about 55 %This means I need to have 55% of the sample of 60 from year 11s results. 55% of 60 is 33, so I need 33 samples to be Year 11 samples.For The Year 9s I am going to do:(117 / 262) x 100 = 44.644.6 is about 45%This means that 45% of the sample of 60 need to be Year 9 results. 45% of 60 is 27, so I need 27 Year 9 samples, which gives you the total of 60 samples.To get these 60 samples from the 262 results I am going to use a random systematic method. To do this I will use a random number generator to find a number from the year 11 and year 9 data, and I am then going to count down from that number, and every 7th piece of data I am going to use. (As 7 was the number that came up when I used a random number generator to find a number between 0 and 10.)Year 9 Random Number Generator91 was the number the generator produced for the year 9s, so I am going to use the 91st piece of data, and then every 7th piece of data after that I am going to use until I have my 27 pieces of data. So, the numbers I am going to use are:91, 98, 106, 113, 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, 94, 101, 108, 115, 5, 12, 19, 26, 33, 40Year 11 Random Number Generator127 was the number the generator produced, so I am going to start from the 127th piece of data and count down 6 piece s of data (as 6 was the number produced from the generator between 0 and 10) and then every 6th piece of data after that I am going to use until I have the 33 pieces of data I need for the year 11s. So the pieces of data are:127, 133, 139, 145, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 5, 11, 17, 23, 29Once I have collected my samples, I am going to draw some grouped frequency tables, which will also have frequency density on. These tables are there so that I can find the mean from grouped data. Also, because the data is put into groups it is easier to handle. I will also find out the spread of the data from the mean using standard deviation.Then, I am going to draw some histograms, using the frequency density from the grouped frequency tables. These will show the density of the data in certain groups. This shows which group had the most estimates in it.Next, I am going to draw some cumulative frequency tables and curves. These will show the inter-quartile range. This shows the range of the quartiles. Also, from the cumulative frequency curves I can draw some box plots. These will show the inter-quartile range, median and lower and upper quartiles in a more compact and easy to read way.Then, I am going to draw some percentage error tables. These will show the error of the estimates and if people estimated below or above the actual size or length of the line.I am then going to draw some scatter graphs showing the errors from the percentage error tables. From these you will be able to see if some guessed below the actual length of the line, and whether or not they guessed below the size of angle as well.Then I am going to draw stem and leaf diagrams for each year group. From these I will be able to find the median and mode. Stem and leaf diagrams show all the data in an easy to read way.Finally, I am going to find The Spearmans Coefficient of rank. This shows whether or not there will be negative or po sitive correlation in the scatter graphs which I will then draw. These will show the estimates of the line for one individual person plotted against their estimate for the angle. From these scatter graphs you can see whether or not anybody guessed exactly the correct size or length.These things should help me prove or disprove my hypothesis.I have recorded the estimates in a table so as to know which pieces of data I am using they have been highlighted. I am using ICT to do parts of my work as it spreadsheets can work things out extremely quickly, but I will also check and record how I would work things out.First of all I am going to draw some grouped frequency tables, which also show frequency density. This will make the data easier to handle and will mean I can draw a histogram, and grouped frequency graphs. Also, from the frequency tables I am going to find the mean, and I am also going to use standard deviation to find the spread of the number from the mean. If the spread is sma ller, it means that the year group guessed closer to the mean value. An advantage of using standard deviation is that you use all of the data.This is a table to show the Year 9 estimates for the length of line 2.Estimate of length (cm)Frequency (f)Class Width (w)Frequency Density (f) / (w)Mid-Point (x)(f) x (x)3 ? cm ; 43133.510.54 ? cm ; 4.540.584.25174.5 ? cm ; 550.5104.7523.755 ? cm ; 68185.5446 ? cm ; 77176.545.57 ? cm ; 902080Total27140.75The average of results from the above table is the total (f) x (x) column divided by the total frequency. This is 140.75/27=5.21cm. This is 0.61cm longer than the actual length of the line. This is not very much, which means the year nines were quite accurate in estimating the length of the line.To find the spread of this data from the mean I am going to use the equation for standard deviation from grouped data, which is:Efxà ¯Ã‚ ¿Ã‚ ½ (mean)à ¯Ã‚ ¿Ã‚ ½EfSo, for the above table I would do:(33.5à ¯Ã‚ ¿Ã‚ ½)+(44.25à ¯Ã‚ ¿Ã‚ ½)+(54.75à ¯Ã‚ ¿Ã‚ ½)+(85.5à ¯Ã‚ ¿Ã‚ ½)+(76.5à ¯Ã‚ ¿Ã‚ ½)+(0x8à ¯Ã‚ ¿Ã‚ ½)27This equals 28.13194, which I will now subtract the meanà ¯Ã‚ ¿Ã‚ ½ from this. The meanà ¯Ã‚ ¿Ã‚ ½ is 5.21à ¯Ã‚ ¿Ã‚ ½, which gives 27.1441. I will subtract this to give me 0.987844, which I now find the square root of this answer, which is 0.9939. This is the spread of data from the mean. This is quite a low spread.This is a table to show the Year 9 estimates for the size of angle 6.Estimate for Angle sizeFrequency (f)Class Width (w)Frequency Density (f) / (w)Mid-Point (x)(f) x (x)20 ? à ¯Ã‚ ¿Ã‚ ½ ; 301100.124.524.530 ? à ¯Ã‚ ¿Ã‚ ½ ; 35350.6329635 ? à ¯Ã‚ ¿Ã‚ ½ ; 40751.43725940 ? à ¯Ã‚ ¿Ã‚ ½ ; 45751.44229445 ? à ¯Ã‚ ¿Ã‚ ½ ; 50450.84718850 ? à ¯Ã‚ ¿Ã‚ ½ ; 1005500.174.5372.5Total271234The mean of the above table is 1234/27=45.70à ¯Ã‚ ¿Ã‚ ½. This is 12.7à ¯Ã‚ ¿Ã‚ ½ bigger than the actual angle. This is quite a large amount which means that the year 9s were not very accurate in their estimates of angle 6. They we re better at estimating the length of the line.To find standard deviation from this I will do:(124.5à ¯Ã‚ ¿Ã‚ ½)+(332à ¯Ã‚ ¿Ã‚ ½)+(737à ¯Ã‚ ¿Ã‚ ½)+(742à ¯Ã‚ ¿Ã‚ ½)+(447à ¯Ã‚ ¿Ã‚ ½)+(574.5à ¯Ã‚ ¿Ã‚ ½)27This gives the answer of 2303.4, which I will now subtract 45.70à ¯Ã‚ ¿Ã‚ ½ from to give 214.91. I then have to find the square root of this answer. This gives 14.7. This spread is quite high, which means that the estimates given by the year 9s for the size of angle 6 was quite big. You can also see that the year nines had a lower spread of data for the length of line 2.This is a table to show the Year 11 estimates of the length of line 2.Estimate of length (cm)Frequency (f)Class Width (w)Frequency Density (f) / (w)Mid-Point (x)(f) x (x)3 ? cm ; 43133.510.54 ? cm ; 4.590.5184.2538.254.5 ? cm ; 530.564.7514.255 ? cm ; 6141145.5776 ? cm ; 72126.5137 ? cm ; 9221816Total33169The mean of the above table is 169/33=5.12cm. This is 0.52cm bigger than the actual length of the line. This is very low, which shows they were quite accurate in their estimates and were better than the year 9s.To find standard deviation from this I will do:(33.5à ¯Ã‚ ¿Ã‚ ½)+(94.25à ¯Ã‚ ¿Ã‚ ½)+(34.75à ¯Ã‚ ¿Ã‚ ½)+(145.5à ¯Ã‚ ¿Ã‚ ½)+(26.5à ¯Ã‚ ¿Ã‚ ½)+(28à ¯Ã‚ ¿Ã‚ ½)33This gives me an answer of 27.36, from which I will now subtract the meanà ¯Ã‚ ¿Ã‚ ½, which is 5.12à ¯Ã‚ ¿Ã‚ ½. This is 26.22, and subtracted from the precious calculation, the answer given is 1.14, which I then find the square root of, which gives me an answer of 1.07. This is a low spread of data, but not as low as the year 9 estimates of the length of line 2. The year nines estimates were therefore more close together.This table shoes the Year 11 estimates for the size of angle 6.Estimate for Angle sizeFrequency (f)Class Width (w)Frequency Density (f) / (w)Mid-Point (x)(f) x (x)20 ? à ¯Ã‚ ¿Ã‚ ½ 301100.124.524.530 ? à ¯Ã‚ ¿Ã‚ ½ 35650.83219235 ? à ¯Ã‚ ¿Ã‚ ½ 405513718540 ? à ¯Ã‚ ¿Ã‚ ½ 45850.64233645 ? à ¯Ã‚ ¿Ã‚ ½ 511360.547.5617.5Total331355The mean of the above table is 1355/33=41.06à ¯Ã‚ ¿Ã‚ ½. This is 8.06à ¯Ã‚ ¿Ã‚ ½ bigger than the actual angle, which shows that the year 11s again were better at estimating the size of the angle, the year 11s were better at estimating the length of the line though.I will now work out the standard deviation for this:(124.5à ¯Ã‚ ¿Ã‚ ½)+(632à ¯Ã‚ ¿Ã‚ ½)+(537à ¯Ã‚ ¿Ã‚ ½)+(842à ¯Ã‚ ¿Ã‚ ½)+(1347.5à ¯Ã‚ ¿Ã‚ ½)33This gives an answer of 1728.26, which I now subtract 41.06à ¯Ã‚ ¿Ã‚ ½ from to give 42.33. I now have to fins the square root of this to find the final standard deviation. The square root is 6.5. This means the spread is smaller than the spread of the year 9s estimates of angle 6, but the year 11s had a lower spread of data for the line estimates. This means more year 11s guessed closer to the mean for the line. This was also the case for the year 9s.These are the histograms to show these details. The histograms show the spread and how densely populated each group of data is. To draw these histograms, I needed to find the frequency density. The grouped frequency tables on pages 3, 4 and 5 show this. To find the frequency density I did frequency divided by class width.From the histograms, you can see the dashed line. This line shows the actual length of the line, or size of the angle. From the year 11 and year 9 histograms from the angle data, you can see that the year 9s have a wider range of results because they have a very large group of data at the end of the histogram. The density of the group where the actual size of the angle is, is not very densely populated, which means not many people guessed within the correct class.From the histogram for the line data you can see that for the year 9 data the group where the actual length of the line was, had the highest frequency density, but was not the most densely populated. The year 11 data shows that not many people guessed in the correct group as it is not very dense.Cumu lative frequency tables group the data so you can see how much the data has gone up from group to group. A curve can then be drawn, and then a box plot can also be drawn.This is a cumulative frequency table to show the year 11 and the year 9 estimates for the length of line 2. These are drawn so I will be able to find the inter-quartile range of the data after a cumulative frequency curve has been drawn. The median can also be found.Estimated length; 4cm; 5cm; 6cm; 7cm; 8cm; 9cmCumulative Frequency Year 11s31529313133Cumulative Frequency Year 9s3122027This is a table to show the cumulative frequency of the year 11 and year 9 estimates of the size of angle 6.Estimated Size (à ¯Ã‚ ¿Ã‚ ½);30;40;45;50;60;70;100Cumulative Frequency Year 9s1111822252627Estimated size (à ¯Ã‚ ¿Ã‚ ½)3040455060Cumulative Frequency Year 11s112203233From these tables I was able to find the median and modal class intervals. The median class interval shows where the middle estimate is, and the modal class inter val shows which interval has the most estimates in it.These are the median class intervals:AngleYear 9: ;45 class interval. I found this by finding the middle of the total frequency and finding which interval that would be in.Year 11: ;45 class interval again. This shows the median of both results is in the same class interval.This is not the interval where the actual size of the angle would be. The actual size of the angle would be in the group ;40.LineYear 9: ;5 class intervalYear 11: ;5 class interval againThis is the correct group for where the actual length of the line is.The modal class intervals are:AngleYear 9: ;40 class interval. I found this by looking which interval had the most pieces of data in.Year 11: ;50 class interval.The year group who had the majority of estimates closest to the size of the angle were the year 9s, as their modal class was the correct class where the actual angle length could be found.LineYear 9: 6cm class intervalYear 11: 6cm class interval again. These show that the majority of people in both year groups guessed around the same number, but it was not the correct class interval of 5cm.These are the cumulative frequency curves to represent this data. Underneath each curve is a box plot. This shows the inter-quartile range of the data.This curve shows the cumulative frequency of the year 11 estimates for line 2.To work out the median from this curve, you add 1 to the total cumulative frequency and then divide by two. This means, for this particular curve I am going to do 33 + 1 = 34, so 34 / 2 = 17. This means I have to find the 17th piece of data. To do this I find 17 on the y axis, draw a line along until I meet the curve, then draw a line down the x axis, which gives me the median. Also, to find the Inter-quartile range I need to find the lower quartile and upper quartile. To do this I halve the median frequency, and follow along until I meet the curve, then draw a line down to the x axis. This gives me the Lower quartile. I then find the Upper quartile which means I add the lower quartile value from the y axis to the median value from the y axis which gives me the value for the y axis for the upper quartile. I then do as before and draw the lines.The LQ for this curve is approximately 4.5 cm.The median is approximately 5.1 cm.The UQ is approximately 5.7 cm.This means that the IQR is 5.7 4.5 which equals 1.2 cm.This next curve is for the year 9 estimates of line 2. The box plot on this curve actually shows that the lower quartile is below the actual length of the line. This is because people guessed a smaller length than the actual one of the line.The LQ for the last curve is approximately 4.5 cm.The median is approximately 5.2 cm.The UQ is approximately 6.1 cm.This means that the IQR is 6.1 4.5 which equals 1.6 cm.The next curve is for the year 11 estimates of angle 6.The LQ for this curve is approximately 38à ¯Ã‚ ¿Ã‚ ½.The median is approximately 43à ¯Ã‚ ¿Ã‚ ½.The UQ is approximately 47à ¯Ã‚ ¿Ã‚ ½.This means that the IQR is 47 38 which equals 9à ¯Ã‚ ¿Ã‚ ½.The next curve is for the year 9 estimates of angle 6.The LQ for this curve is approximately 37à ¯Ã‚ ¿Ã‚ ½.The median is approximately 42à ¯Ã‚ ¿Ã‚ ½.The UQ is approximately 48.5à ¯Ã‚ ¿Ã‚ ½.This means that the IQR is 48.5 37 which equals 11.5à ¯Ã‚ ¿Ã‚ ½.From these curves I was able to find out that the IQR for the angles in both year groups was higher than the IQR for the line. Also, I found the year 11s had lower IQRs in both of the angle and the line estimates. This could suggest that the year 11s were better at estimating.Percentage error tables show what errors people made. To find the actual percentage error for the above table you have to do a calculation which is:(error/actual length of line) x 100This is a percentage error table for the year 9 estimates of the line and angle.LINE ANGLEEstimate (e)ErrorPercentage error (%)Estimate (e)ErrorPercentage error (%)61.430.43602781.826.21.634.78532060.614.5-0.1-2.173 526.0661.430.433526.0661.430.433526.0650.48.709562187.884-0.6-13.0440721.2150.48.70501751.5261.430.4340721.214-0.6-13.0430-3-9.094-0.6-13.0420-13-39.3950.48.70451236.3650.48.7040721.2150.48.703526.064-0.6-13.043526.0661.430.4340721.214.600.0030-3-9.0950.48.703526.064.5-0.1-2.1740721.214.5-0.1-2.17451236.363.5-1.1-23.91431030.303.5-1.1-23.91431030.303-1.6-34.78501751.524.5-0.1-2.17451236.3661.430.4330-3-9.0950.48.703526.0650.48.70451236.36Total6.6143.48Total238721.21The mean percentage error is worked out by dividing the total percentage error by how many pieces of data there are. So, for the mean of the year 9 line errors, the calculation would be 143.48/27 which equals 5.31%. This is quite a low percentage of error which means that the year 9s were quite good at estimating the line. For the angle it would be 721.21/27 which equals 26.71%. This percentage is so high because someone estimated 95à ¯Ã‚ ¿Ã‚ ½, which means that mean is made higher by this anomalous result. This shows th at the year 9s were better at estimating the length of the line.The next table is for the percentage error of year 11 estimates.LINE ANGLEEstimate (e)ErrorPercentage error (%)Estimate (e)ErrorPercentage error (%)50.48.7040721.2150.48.703526.0650.48.7040721.214-0.6-13.04451236.3650.48.70451236.3650.48.70451236.3650.48.70451236.363-1.6-34.7840721.2150.48.70461339.3950.48.7030-3-9.094-0.6-13.0430-3-9.093.5-1.1-23.9140721.214.5-0.1-2.1729-4-12.1250.48.70451236.364-0.6-13.0430-3-9.094.5-0.1-2.1730-3-9.093-1.6-34.7830-3-9.094-0.6-13.043526.0683.473.9140721.214-0.6-13.0430-3-9.0983.473.9140721.214-0.6-13.04451236.3661.430.43451236.3650.48.70451236.3650.48.703526.0650.48.70501751.524-0.6-13.04451236.364-0.6-13.043526.064-0.6-13.04451236.3650.48.703526.064.5-0.1-2.1740721.2151.430.43451236.3661.430.4340721.21Total6.2134.78Total206624.24The mean for the above table of the line errors would be 134.78/33 which equals 4.08%. This is a relative low percentage which means from the mean there doesn t seem to be much error for the line. For the angle the mean is 624.24/33 which equals 18.92%. This means that from the mean you can see that from the year 11 data, they were better at estimating the length of the line.From these calculations, I have found that the year 11s were better at estimating by finding the mean percentage error, as both of the mean percentage errors for the line and angle, were lower than the year 9 errors.From the percentage error tables I can plot a scatter graph to see how much error there was for the angle compared to the line for both year groups. These graphs were drawn from the actual error made. For example, if someones error for the line was -2, because they guessed 2 below the actual angle size, and their error for the angle was 3, because they guessed 3 above the actual length of the line, then the co-ordinate for their error would be (-2,3).This scatter graph shows the error of the year elevens. If someone had estimated exactly right and therefor e had no error then there mark would be at the point (0,0) on the scatter graph. The point circled in pink is an anomalous result. This is a result which is out of the pattern of the rest of the results.This scatter graph shows the error of the year 9 estimates. The pink circled result is again an anomalous result.Stem and leaf diagrams put the data in numerical order in an easy to read table. This is a stem and leaf diagram to show Angle estimates:Year 9ANGLE 6Year 110295,5,5,5,5,5,5,0,0,030,0,0,0,0,0,5,5,5,5,55,5,5,5,3,3,0,0,0,0,040,0,0,0,0,0,0,0,5,5,5,5,5,5,5,5,5,5,5,63,0,050067859KEY 0 2 9 means that the year 9 estimate was 20 and the year 11 estimate was 29.This is a stem and leaf diagram to show the Line estimates:YEAR 9LINE 2YEAR 115,5,030,0,56,5,5,5,5,0,0,0,040,0,0,0,0,0,0,0,0,5,5,50,0,0,0,0,0,0,050,0,0,0,0,0,0,0,0,0,0,0,0,02,0,0,0,0,0,060,080,0KEY 0 4 0 means the year 9 estimate was 4.0 and the year 11 estimate was 4.0 also.On the stem and leaf diagrams I have highlighted s ome of the values. These are the medians for each stem and leaf diagram. For the Year 11 line, the median is 50, for the year 11 angle it is 40. For the year 9 line the median is 50 and the median for the estimates of the angle by year 9s is 40.I can also find the mode for each group of data. The mode of the estimate of the year 9 line is 5.0 . The mode for the year 11 estimates for the line is also 5.0 . This shows that in both year groups most people guessed the same value, which means, from this mode, you cannot see who is better at estimating.The mode for the year 9 estimates of the angle is 45 and the mode for the year 11s is 40. This shows that the mode for the year 11s is closer to the actual value of 33. This could suggest that the year 11s are better at estimating in this instance.I can also find the range from the stem and leaf diagrams, by subtracting the smallest value from the largest.The range for the year 9 angle estimates is 95 20 which equals 75, which is the range . For the year 9 line the range is 6.2 3.0 which gives you 3.2 as the range.The year 11 range for the angle is 50 29 which equals 21. For the line it is 8.0 3.0 which equals 5.0.From the ranges you can see that the year 9 angle estimates were more spread out than the year 11 angle estimates, but the year 11 line estimates were more spread out than the year 9 line estimates.These tables are made for use in with the scatter graphs. They show Spearmans Coefficient of Rank. This is basically to find the correlation of the data.This table is for the year 9 data.Year 9 Estimate of lineEstimate of angleLine RankingAngle RankingDifference (d)dà ¯Ã‚ ¿Ã‚ ½4.5608.526-17.5306.256.3532725244.5358.580.50.2553517.589.590.2543548-41659517.527-9.590.25440414-101006502523.51.52.2554017.5143.512.256302532248452017.5116.5272.254.6451120.5-9.590.2554017.5143.512.256352581728943548-41654017.517.50053017.5314.5210.2553517.589.590.254.5408.514-5.530.254.5458.520.5-1214454317.517.5003.543117.5-16.5272.2 5450423.5-19.5380.25445420.5-16.5272.2553017.5314.5210.2553517.589.590.2554517.520.5-39TOTALS-3.53494.25Now, to find the Spearmans coefficient of rank you have to use the formula:p = 1 6?dà ¯Ã‚ ¿Ã‚ ½n(nà ¯Ã‚ ¿Ã‚ ½ 1)In this formula, p is the Spearmans coefficient rank of correlation; d is the difference between one item of data and n is the number of items of data.So, for the year nine pieces of data, ?dà ¯Ã‚ ¿Ã‚ ½ = 12.25 and n = 27, so:1 6 x 3494.25 = 1 20965.5 = 1 1.066620879= -0.06662087927(729 1) 19656Therefore, this means that the spread of the year nine data will have almost no correlation at all when plotted on a scatter graph because if the Spearmans coefficient of rank is close to 1 then it means that the data will be strongly positively correlated. If it is close to 0 then it means there will be extremely little or no correlation, and if it is close to -1 it means that there will be very strong negative correlation. This scatter graph shows this:The thick, black lines on this graph show the actual length of the line and the actual size of the angle. If someone had guessed both pieces of data correctly, there estimate would be marked upon the intercept of those two lines. This graph shows that there is very little correlation, which means my Spearmans coefficient rank of correlation was correct. There is an anomalous result on this graph which is when someone has estimated the angle as around 90à ¯Ã‚ ¿Ã‚ ½. This was extremely far away from the actual size.The circled result is an anomalous result. It shows that someone guessed a very different angle estimate to the rest of the pupils. This person didnt really have a lot of error when estimating the line, they estimated only 0.4cm above the actual length but then estimated 57à ¯Ã‚ ¿Ã‚ ½ above the actual angle size, which again indicates that there is no correlation, showing that this person was not too bad at estimating the line length, but were quite bad at estimating angles. This shows th at this person found lines easier to estimate.This is the table for the year 11 data.Year 11 Estimate of lineEstimate of angleLine RankingAngle RankingDifference (d)dà ¯Ã‚ ¿Ã‚ ½54022.516.563653522.51012.5156.2554022.516.5636445827.8-19.8392.0454522.527.8-5.328.0954522.527.8-5.328.0954522.527.8-5.328.093401.516.5-1522554622.532-9.590.2553022.54.51832443084.53.512.253.540316.5-13.5182.254.5291411316954522.527.8-5.328.0943084.53.512.254.530144.59.590.253301.54.5-39435810-2484032.516.51625643084.53.512.2584032.516.516256445827.8-19.8392.0464530.527.82.77.2954522.527.8-5.328.0953522.51012.5156.2555022.533-10.5110.25445827.8-19.8392.04435810-24445827.8-19.8392.0453522.51012.5156.254.5401416.5-2.56.2554522.527.8-5.328.0964030.516.514196TOTALS-19.8392.04Now, to find out the correlation I will substitute the values for the year 11 data into the formula again:1 6 x 4243.74 = 1 25462.44 = 1 0.709181 = 0.29081933(1089-1) 35904This means that when these values are plotted on a scatter graph they will also have extremely little correlation. This shows that people who may have estimated the smallest value for the size of the angle, didnt necessarily estimate the smallest value for the length of the line.This graph shows the year 11 data.Again, the thick, black lines show the actual length of the line and size of the angle. From this graph you can see that the Spearmans coefficient rank was correct, and that the graph has very little correlation. Also, on this diagram, most of the points are relatively near to the line, but there is an anomalous result, which is at the estimate id 8 cm for the line. This persons estimate for the angle was not too inaccurate though. Again the pink circled point is an anomalous result.The lengths of lines are easier to guess than angles.This was my first hypothesis. I feel, through doing many calculations and graphs, that I was able to prove that this hypothesis was correct.Firstly, I drew some grouped frequency tables. From these I was abl e to see that for the year 9s, the mean for the estimate of the lines from the grouped frequency table was 5.21cm, this is 0.61cm longer than the actual length of the line, yet the mean for the year 9s angle estimates from grouped frequency was 45.70à ¯Ã‚ ¿Ã‚ ½, which is 12.7à ¯Ã‚ ¿Ã‚ ½ bigger than the actual size of the angle. There is a lot of difference between these means. Most obviously, the year 9s found it easier to estimate the line, as the difference between the actual length of the line and the mean is much lower. The difference between the size of the angle and the mean is probably higher as there is a result that is 90à ¯Ã‚ ¿Ã‚ ½, which makes the mean much higher. The year 11s mean for the line was 5.12, which is 0.52cm bigger than the actual length of the line. The mean for the year 11s estimates of the angle from grouped data was 41.06à ¯Ã‚ ¿Ã‚ ½, which is 8.06à ¯Ã‚ ¿Ã‚ ½ larger than the actual size of the angle. This shows again that the year 11s found the lines length easier to estimate than the size of the angle.I also did standard deviation which indicates how the data is spread from the mean. For the year 9 angle it was 14.7, which is quite a high spread, showing that the year 9 estimates for the angle size were quite spread apart. For the year 9 line estimates, was 0.99, which means that the results are very compact compared to the mean, which shows that many pupils estimated very close to the mean, showing that the year 9s were better at estimating the lines length.The standard deviation I did for the year 11s showed that the year 11 lines deviation from the mean was 1.07 which is very low, meaning that the results were quite compact. Then, for the angles the spread was 6.5 which is higher than the deviation for the line, showing that the year 11s were better at estimating the line.Next, I drew some histograms, and from these you can see that even though the most densely populated groups were not the groups of 4.5 5 for the line, or 30 35 for the angle, you can see that in the year 11 estimates for line 2, the most densely populated group was 4 4.5cm group. To find this I multiplied the sizes of the groups, in this case 0.5 by the frequency density, 18, which gives 9, which was the highest frequency. And for the angle the most densely populated group was 45 51à ¯Ã‚ ¿Ã‚ ½. Neither of the most populated groups contained the correct estimate for the line or the angle, so it is not possible to tell accurately which one was easier, but you can see that the year 11 estimate for the lines most densely populated group was the group next to the group containing the correct length, whereas this was not the case with the angle.For the year 9 histograms, you can see that the most densely populated group for the angle was joint between the groups of 35 40 à ¯Ã‚ ¿Ã‚ ½ and 40 45à ¯Ã‚ ¿Ã‚ ½ as they both had the highest frequency worked out form the histograms. For the line it was 4.5 5cm group, which is the correct gr oup that contains the actual length of the line. From this you can see that the year 9s were better at estimating the length of the line, which again proves my hypothesis.Next, I drew some cumulative frequency tables and curves. From the year 9 cumulative frequency curve I was able to find the median, which was 5.2cm, this is quite far away from the actual length of the line of 4.6cm. You can also see on the box plot that the actual length of the line is contained at the very edge of the box, nearer the edge of the lower quartile, which means that most year 9s guessed above the length of the line. For the year 9 angle estimates in the cumulative frequency curve you can see that the median is 42 which is very high compared to the actual size of the angle of 33à ¯Ã‚ ¿Ã‚ ½. This means that the year 9s box plot for the angle didnt contain the actual size of the angle in it. This is because someone estimated 90à ¯Ã‚ ¿Ã‚ ½, which is very high. This means that the year 9s, again found th e line easier to estimate.The cumulative frequency curves for the year 11s estimates of line 2 show that the median from the curve was 5cm. This is 0.4cm bigger than the actual length of the line. The actual length of the line is contained in the box, but at the very edge of the lower quartile. This again shows that most people estimated over the length of the line. For the year 11 angle cumulative frequency curve, the median was 43à ¯Ã‚ ¿Ã‚ ½, which is 10à ¯Ã‚ ¿Ã‚ ½ over the actual size of the angle. The actual size of the angle again is not in the box, showing that many people again guessed above it, which leads to show that the year 11s found the line easier to estimate.Next I drew some percentage error tables. For the year 9 line the mean error was 5.31%, which is relatively low, showing that there was not much error at all. The percentage of error for the angle though was 26.71% which is much higher than the percentage error for the line, meaning that the year 9s had less err or when estimating the length of the line.For the year 11 tables, I was able to find the mean percentage error of 4.08% for the estimating of the line, which is rather low showing that the year 11s did not make much error whilst estimating the line. However, their percentage error mean for the angle was 18.92 which is much higher than the line error percentage, showing that the year 11s were better at estimating the length of the line as they had less error.From the scatter graphs drawn by the error you can see that there is no correlation between the errors.From the stem and leaf diagrams, you can see that the median for the year 9s line estimate is 5.0cm, which is 0.4cm above the actual length of the line, and the year 11 median for the angle was 40. This is 7à ¯Ã‚ ¿Ã‚ ½ above the actual size of the angle, meaning that the year 11s were better at estimating the line. The numbers for the median of the year 9s from the box plots were the same as the year 11s showing that the year 9 s were better at estimating the length of the line also.I then did standard deviation to find out what correlation the scatter graphs would have when I drew them. From doing standard deviation I was able to see that for the year 9 scatter graph there would be extremely little correlation at all, as the answer I got after doing the equation for standard deviation, which is shown on page 17, was -0.066620879, this means that5 there would be VERY slight negative correlation, but hardly any at all, meaning that if a year 9 isnt very good at estimating the length of a line, then they are not necessarily bad at estimating the angle size.For the year 11s the answer I got from working out the standard deviation was 0.290819, which means that there would extremely slight positive correlation, but hardly any at all really. This means that if they were really good at estimating the line, they were not necessarily good at estimating the angle size. The scatter graphs show this data.Overall, I f eel I have managed to prove the hypothesis of the length of lines is easier to guess than angles as correct through many calculations and graphs.Year 11s will be more accurate at estimating.This was my second hypothesis, and through the calculations I have done I feel I have managed to come to a conclusion of proving or disproving this hypothesis.Firstly, I made some grouped frequency tables, which I then found the mean from. The mean for the year 11s estimates of the line from the grouped frequency was 5.12cm which is 0.52 above the actual length of the angle. The mean for the year 9 estimates was 5.21 which was 0.61cm above the actual length of the line. This shows that as the mean was closer to the actual length of the line, the year 11s were better at estimating the length of the line in this case. For the angle estimates from grouped frequency tables, I was able to see that the year 11 mean from the grouped frequency was 41.06à ¯Ã‚ ¿Ã‚ ½ which is 8.06à ¯Ã‚ ¿Ã‚ ½ above the act ual angle size, whereas the year 9 mean from the grouped frequency tables for the angle was 45.70à ¯Ã‚ ¿Ã‚ ½ which was 12.7à ¯Ã‚ ¿Ã‚ ½ higher than the actual size of the angle. This means that the year 11s were better at estimating the size of the angle than the year 9s were.Next, I found the standard deviation for both year groups. I used the equation on page 3 to do this. For the year 9 line estimates I found out that the spread of the data from the mean was 0.9939 which is quite a low spread, meaning the estimates were quite close together. For the year 11 estimates for the length of line 2, the deviation of the estimates was 1.07, which means that the year 9 estimates for the length of line 2 were less spread out than the year 11s.For the year 9 estimates of the angle, the spread from the mean was 14.7 which was quite high, showing that the results were quite spread out. For the year 11 estimates of the angle, the standard deviation was 6.5 which means there was quite a lot of spread, but less than the year 9s meaning that the year 11 results were less spread out in this case.After this, I drew some histograms. From the year 9 histograms you can see, for the line, that the most densely populated group was 5 6 cm which, is not the group where the actual length of the line was contained. For the year 11s, the most densely populated group was also the 5 6 cm group which was again not the correct group where the actual length was, showing that both the majority of the year 9s and year 11s guessed that the line was in the region of 5 to 6 cm.For the year 9 histogram for the angle, the most densely populated groups was joint between 35 40 and 40 45, which did not contain the actual size of the angle. For the year 11s the most densely populated group on the angle histogram was 35 40, which again did not contain the correct size of the angle, meaning that the majority of both year groups estimated in the incorrect group.I then drew some cumulative frequency curves. These showed me the frequency of the data in the form of a curve. For the year 9s, from the curve for the line estimates, I could see that the inter-quartile range of the data was 1.6cm and the median was 5.2cm. For the year 9s curve for the line the inter-quartile range was 1.2cm and the median was 5.1cm. From these results you can see that the year 11s had both a lower median and a lower inter-quartile range. This could suggest that they were better at estimating.For the angle cumulative frequency curve for the year 9s the inter-quartile range was 11.5à ¯Ã‚ ¿Ã‚ ½ and the median was 42à ¯Ã‚ ¿Ã‚ ½. For the year 11s the inter-quartile range was 9à ¯Ã‚ ¿Ã‚ ½ and the median was 43à ¯Ã‚ ¿Ã‚ ½, this means that the year 11s had a lower inter-quartile range value, but the year 9s had a lower median value.After drawing cumulative frequency curves, I was able to draw some box plots. From these I was able to see the inter-quartile range and median easily.For the year 9 box plot f or the line I was able to see that between the highest result and the upper-quartile there is very little difference. The upper quartile was 6.1cm and the highest result was 6.2cm. This shows that the many people estimated higher up on the scale as between the lowest estimate of 3cm and the lower quartile of 4.5cm there is quite a large gap.For the year 11 box plot of the line, I was able to see that the data is quite evenly spread as the box plot is relatively in the middle of the lowest value of 3cm and the highest value of 8cm. The actual length of the line is contained with in the inter-quartile range box, but, towards the very end of the lower quartile side, meaning that most people estimated above it.For the year 9 box plot of the angle estimates I was able to see that the actual size of the angle is not contained within the inter-quartile range. This is because many people guessed above the correct size of 33à ¯Ã‚ ¿Ã‚ ½, meaning that it was not contained within the inter-qua rtile range.The year 11 box plot for the angle showed that many of them also estimated above the size of the angle of 33à ¯Ã‚ ¿Ã‚ ½, as it is again not contained in the inter-quartile range. After analysing these box plots, I do not feel that they have helped me to prove or disprove my hypothesis.Next, I drew some percentage error tables. From these I was able to see what error each individual made in their estimates. I then found how much error they made as a percentage. Overall, for the year 9 estimates of the line, the mean percentage error was 5.31%, which is quite low, means that they did not make too much of an error overall. For the year 11 estimates of line two, the mean percentage error was 4.08%. This is a lower percentage of error than the year 9s meaning that overall, the year 11s made less error in estimating the length of the line.For the year 9 estimates of the angle the mean percentage error was 26.71% which is quite high. This is because there was an anomalous resu lt of 95à ¯Ã‚ ¿Ã‚ ½, which means that the overall mean percentage error value will be higher. For the year 11 estimates of the angle, their mean percentage error was 18.92%, this is still quite high, but lower than the year 9s mean percentage error for the angle meaning that the year 11s had less error in estimating the angle overall again.Next, I drew some scatter graphs which showed how much error someone had made, for example if some had estimated 2à ¯Ã‚ ¿Ã‚ ½ above the actual angle size, and 0.3cm below the estimate of the line, their coordinates would be (-0.3,2). From the scatter graphs I was able to see that there was not really any correlation in either the year 11 or year 9 scatter graphs. Again, after analysing these I do not feel that they help me to prove or disprove my hypothesis.After this, I drew some stem and leaf diagrams. I was able to find the median and mode from these. They also showed me the data in a table, showing each and every result.For the year 9 line e stimates I was able to work out that the median was 5.0cm which was also the median for the year 11s. This means that when all the data is put in consecutive order, that both the year 9s and the year 11s had the same middle estimate. For the angles I was able to see that the median for the year 9s was 40à ¯Ã‚ ¿Ã‚ ½, and the median for the year 11s was also 40à ¯Ã‚ ¿Ã‚ ½. This means that again they had the same middle estimate. I do not feel that these have helped me to prove my hypothesis.Next, I did the spearmans coefficient rank. This shows what correlation the data will have when plotted on a scatter graph. For the year 11s I was able to see that after working out that the correlation would be hardly visible, but would be extremely slightly positive, as the answer I got after working out the spearmans coefficient rank was 0.290819. I then plotted my points on a scatter graph and found out that this was correct. I then worked out the spearmans coefficient of rank for the year 9s . This gave me an answer of -0.066620879, meaning that there would be a very slight negative correlation. I plotted my points on a graph and was able to see that again I was correct. I was shown that for both year groups the estimate for the line didnt necessarily correlate with the estimate for the angle. For example, my results did not show that if someone estimated high for the angle size, they did not necessarily estimate high for the line length.Additionally, from the scatter graphs I was able to see that there were some anomalous results. For example there was a year 9 who estimated the angle at being 95à ¯Ã‚ ¿Ã‚ ½, whereas they estimated the line at being 5cm which showed that they did not estimate too extremely for the length of the line. I do not feel these scatter graphs assisted me in proving my hypothesis.Overall, I feel I have been able to prove this hypothesis as correct through the calculations which referred to it.I think all of my calculations and diagrams were cor rect as they all led to the same conclusion, and through checking my answers I found that they were correct. I think there were not really any major anomalous results, apart from the ones previously mentioned and shown in my calculations.One problem I did have was finding the standard deviation, but I realised this was because I was trying to find it from grouped data, and therefore needed a different formula, and after getting the formula for grouped data I found it a lot easier to do.I managed to prove my hypotheses were correct in most instances and I managed to show many different types of calculations in proving my hypotheses.