The length of the actual line is 11.5cm
The size of the actual angle is 61 degrees
What is the length of this line?
What is the size of this angle?
The results for the length of the line in centimetres for year 10.
The results for the size of the angle in degrees for year 10.
The results for the length of the line in centimetres for year 11.
The results for the size of the angle in degrees for year 11.
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.
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
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
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
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
I have found outlier s for year 11 estimations of a length of a line I will now find them for their estimations of a size of an angle.
The median for year 11 estimations of a length of a line is
½ (363 + 1) = 182nd value
= 55
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
= 45
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
= 60 + 60 /2
= 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
Which there are two outliers that I will eliminate
Now I will find any large outliers that distort the data. The large outliers are
UQ + 22.5 = 60 + 22.5
= 82.5
Which in the data there are 8
Comparing primary and secondary data
I will now compare the primary data estimations of the length of the line with the secondary data estimations of a length of a line, which I will do this by working out the relative percentage error. I will work out the relative percentage error for the lowest estimation (LE) in the primary data and the highest estimation (HE) in the primary data. Than I will repeat what I did for the primary data with the secondary data.
Primary data
The relative percentage error for the lowest estimation is
11.5 - LE / 11.5 * 100 = Relative percentage error
11.5 - 7.5 / 11.5 * 100 = 34.78%
The relative percentage e error for the highest estimation is
HE - 11.5 / 11.5 * 100 = Relative percentage error
15 - 11.5 / 11.5 * 100 = 30.43%
Secondary data
The relative percentage error for the lowest estimation is
4.6 - LE / 4.6 * 100 = Relative percentage error
4.6 – 2 / 4.6 * 100 = 56.52%
The relative percentage error for the highest estimation is
UE - 4.6 / 4.6 * 100 = Relative percentage error
180 - 4.6 / 4.6 * 100 = 3813.04%
By comparing the lowest and the highest estimation of the length of a line for the primary data with the lowest and highest estimation of the length of a line for the secondary data. I found out that the relative percentage error for the primary estimations have a smaller relative percentage error than the secondary estimations. This means that the primary estimations are more consistent than the secondary estimations. Which means that the primary data is more representative of the whole population than the secondary data.
Now I have compared the primary results of the estimations of the length of a line with the secondary results of the length of a line. I will now repeat what I did to compare the primary data with the secondary data of the estimation of the size of an angle.
Primary data
The relative percentage error for the lowest estimation is
61 - LE / 61 * 100 = Relative percentage error
61 - 45 / 61 * 100 = 26.23%
The relative percentage error for the highest estimation is
HE - 61 / 61 * 100 = Relative percentage error
80 - 61 / 61 * 100 = 31.15%
Secondary data
The relative percentage error for the lowest estimation is
56 - LE / 56 * 100 = Relative percentage error
56 - 6 / 56 * 100 = 89.29%
The relative percentage error for the highest estimation is
HE - 56 / 56 * 100 = Relative percentage error
105 - 56 / 56 * 100 = 87.5%
The results show that the relative percentage error for the lowest and highest estimation of the size of an angle for the primary data are smaller than the relative percentage error for the estimations of a size of an angle for secondary data. This means that the primary data is more consistent and that it has a smaller percentage error. This shows that the primary data is more representative of the whole population than the secondary data.
Hypothesis 2
My second hypothesis was that people who estimate the line accurately may not estimate the size of an angle accurately so my prediction would be the line of best fit is going to be negative as one estimation is high the other is low. To support this I will draw a scatter graph for both years. I will draw a scatter graph by plotting each person in the sample estimate of the line against their estimation of the angle. Which then draw a line of best fit.
I expect for year 11 that the data would be positively correlated. This is because they will have done a lot of work using lengths and angles. So they will no a fair bit about estimating and therefore their estimations will be closer to the actual estimation. While year 10 have just started their GCSE course and will not know a lot about estimating. So as result their estimations for the length of the line would be better than their estimation of the angle because estimating a line is easier than estimating the size of an angle. So I would expect their results to be negatively correlated.
Now I will draw another scatter graph but for year 11 estimations. I will do this by using the same procedure as I did for year 10 results.
By comparing both the scatter graphs for the estimations of the length of the line and size of an angle they both show a strong positive correlation. Which means that as one increase so does the other. So if someone estimates close to the length of a line then they will estimate close the size of the angle. While if someone does not estimates the length of a line accurately then they will not estimate the size of an angle accurately. Comparing my findings that some one who estimates the length of a line accurately will estimate the size of an angle accurately it shows that my hypothesis was wrong because if someone can estimate the length of a line accurately then they can estimate the size of an angle accurately. This would be because they have good estimating skills so if they can estimate one accurately then they are can estimate another thing accurately. While I was thinking that if someone can estimate the length of a line accurately couldn’t always estimate the size of the angle accurately because an angle is more difficult to estimate, but I was wrong. Due to this I thought the line of best fit would be negative because as one estimate is high the other estimate would be low but I was wrong because as one estimate is going to be to high so is the other estimation going to be to high.
The gradient of the line shows the steepness of the line. If you look at the scatter graph for year 10 and year 11 you will see that gradient for year 10 is steeper than the gradient for year 11. This means that year 10 estimations were further apart because they form a gradient that is steep. While the scatter graph for year 11 shows that the gradient is less steeper than the gradient for year 10 results. This means that as one estimation is high so is would be the other this shows that s someone estimates the line of a line to high would estimate the size of an angle to high. Y-intercept is the point that crosses the Y-axis. From the results you could see that the y-intercept for the line of best fit for year 10 results is lower to the actual length of the line than the y-intercept for the line of best fit for year 11 results which is closer to the actual estimation of length of the line. This shows that year 11 results are more accurate and reliable and show as one increase the other decreases.
To support my hypothesis even further I will use spearman’s coefficient of rank correlation. I will do this by ranking the estimations first than working out the spearman’s coefficient for the length of a line and size of an angle and use spearman’s coefficient of rank correlation to see if the results match up with the results of the scatter graphs. This will tell me that whether the scatter graphs were correct or not. My prediction is that the correlation between the length of the line and size of the angle is going to be positive. This is because I have drawn scatter graphs and found out that if someone estimates the length of a line accurately they will estimate the size of an angle accurately. So the spearman’s coefficient should tell me the same thing because I am going to use the same results as I did for the scatter graphs.
Year 10
Estimation of a length of a line = 10, 11, 12, 11, 8, 12.5, 9, 10, 13, 12, 11.75, 10, 11, 14, 13.5, 7.5, 11.5, 12, 13, 12, 11.5, 8, 9, 8.5, 9, 11, 12, 10.25, 13, 15
Rank –5, 7, 10, 7, 2, 11, 4, 5, 12, 10, 9, 5, 7, 14, 13, 1, 8, 10, 12, 10, 8, 2, 4, 3, 4, 7, 10, 6, 12, 15
Estimations of an size of angle = 70, 75, 62, 50, 45, 78.5, 70, 65, 70.1, 65, 75, 80, 64, 78, 52, 55.5, 73, 59, 60, 72.2, 71, 74, 57, 65, 58, 61, 49, 78, 60, 65
Rank –14, 20, 11, 3, 1, 22, 14, 13, 15, 13, 20, 23, 12, 21, 4, 5, 18, 8, 9, 17, 16, 19, 6, 13, 7, 10, 2, 21, 9, 13
d - (difference between the rank of the estimation of a length of a line and of an size of angle)
d = 9, 13, 1, 4, 1, 11, 10, 8, 3, 3, 11, 18, 5, 7, 9, 4, 10, 2, 3, 7, 8, 17, 2, 10, 3, 3, 8, 15, 3, 2,
d (squared) = 81, 169, 1, 16, 1, 121, 100, 64, 9, 9, 121, 324, 25, 49, 81, 16, 100, 4, 9, 49, 64, 289, 4, 100, 9, 9, 64, 225, 9, 4
Sum of d squared = 81 + 169 + 1 + 16 + 1 + 121 + 100 + 64 + 9 + 9 + 121 + 324 + 25 + 49 + 81 + 16 + 100 + 4 + 9 + 49 + 64 + 289 + 4 + 100 + 9 + 9 + 64 + 225 + 9 + 4
= 2126
Spearman’s coefficient is r = 1 - 6 * sum of d squared / n (n squared -1)
r = 1 - 2126 / 30 (30 squared -1)
r = 1 - 2126 / 30 * (900 –1)
r = 1 - 2126 / 26970
r = 1 – 0.078828327
r = 0.921171673
On the Spearman’s coefficient of rank correlation it shows that on Spearman’s coefficient it is strong positive correlation. This shows that estimating the length of a line and size of an angle are closely related. This means that if someone estimated the line inaccurately then there is a very high percentage that they will estimate the size of an angle in accurately. Relating my findings to my prediction it shows that my6 prediction was wrong. Now I will repeat the same procedure to find the correlation for year 11 data.
Year 11
Estimation of a length of a line = 11, 10, 9, 10.5, 12, 14, 11, 11, 12, 15, 9, 9.5, 9, 10.2, 10, 11, 13.5, 11.5, 12, 11, 13, 9.5, 10, 12, 10.4, 10.75, 10, 11.5, 10, 12
Rank – 8, 3, 1, 6, 10, 13, 8, 8, 10, 14, 1, 2, 1, 4, 3, 8, 12, 9, 10, 8, 11, 2, 3, 10, 5, 7, 3, 9, 3, 10
Estimations of an size of angle = 72, 75, 64, 55, 55.25, 54, 60, 65, 64.8, 78, 60, 70.5, 65, 60, 62, 71, 54.75, 61, 63, 69, 71, 68.1, 77.5, 55.5, 59, 61, 67, 60, 68.5, 69
Rank – 20, 21, 11, 3, 4, 1, 7, 13, 12, 23, 7, 18, 13, 7, 9, 19, 2, 8, 10, 17, 19, 15, 22, 5, 6, 8, 14, 7, 16, 17
d - (difference between the rank of the estimation of a length of a line and of an size of angle)
d = 12, 18, 10, 3, 6, 12, 1, 5, 2, 9, 6, 16, 12, 3, 6, 11, 10, 1, 0, 9, 8, 13, 19, 5, 1, 1, 11, 2, 13, 7,
d (squared) = 144, 324, 100, 9, 36, 144, 1, 25, 4, 81, 36, 256, 144, 9, 36, 121, 100, 1, 0, 81, 64, 169, 361, 25, 1, 1, 121, 4, 169, 49
Sum of d squared = 144 + 324 + 100 + 9 + 36 + 144 + 1 + 25 + 4 + 81 + 36 + 256 + 144 + 9 + 36 + 121 + 100 + 1 + 0 + 81 + 64 + 169 + 361 + 25 + 1 + 1 + 121 + 4 + 169 + 49
= 1935.9
Spearman’s coefficient is r = 1 - 6 * sum of d squared / n (n squared -1)
r = 1 - 1935.9 / 30 (30 squared -1)
r = 1 - 1935.9 / 30 * (900 –1)
r = 1 – 1935.9 / 26970
r = 1 – 0.071779755
r = 0.928220245
The results on the Spearman’s coefficient of rank correlation show that there is strong positive correlation between estimating the length of a line and size of an angle. This means if someone estimates the size of an angle accurately then there is a very high chance that they will estimate the length of a line accurately. Comparing my findings to my prediction that someone who may estimate the length of line accurately may not estimate the size of an angle accurately is that my prediction is wrong.
Both the line of best fit and the Spearman’s coefficient show that if someone who estimates the length of a line inaccurately stands a very high chance of estimating the size of an angle accurately. This shows that my prediction was wrong.
Hypothesis 3
My third hypothesis is that estimating the length of a line is easier than estimating the size of an angle to support this I will work out the standard deviation for the estimations of a length of a line and size of an angle for year 10 and year 11 estimations. I will use standard deviation because it uses all the data which means the results are very reliable. If the standard deviation of the estimations of a length of a line is smaller than the standard deviation for the size of an angle. Than it means that estimations of a length of a line is more narrower this means the results for estimations of a length of a line are more consistent and reliable which means they will be closer to the actual length of the line. This will show that estimating the length of a line is easier than estimating the size of an angle. This is because the smaller the standard deviation for either the estimations of a length of a line or of the size of an angle the more the people estimated closer to the actual length of a line or size of an angle which means it more easier than the other. This is because usually people estimated closer to the actual estimation for something that is easy to estimate.
The formula for standard deviation is the following:
s = Square root of sum of x squared / n – the mean squared
In order to work out the standard deviation I will have to work out the total of x, which is going to be the sum of the estimations of a length of a line and size of an angle and the x squared.
This is a table that shows part of the standard deviation is worked out for year 10 estimations of a length of a line.
Standard deviation for year 10 estimations of length of a line.
s = Square root of sum of x squared / n – the mean squared
s = Square root of 3777.62 / 30 – 122.4711111
= Square root of 125.9083333 – 122.4711111
= Square root of 3.4372222
= 1.853974703
= 1.85
Now I will use the same procedure to work out the standard deviation for year 10 estimations for estimations of a size of an angle.
Standard deviation for year 10 estimations of a size of an angle.
s = Square root of sum of x squared / n – the mean squared
s = Square root of 130318.4 / 30 – 65.24333333
= Square root of 4343.946667 – 65.24333333
= Square root of 4278.703334
= 65.41179813
= 65.41
The results from year 10 data show that the standard deviation for the length of a line is much smaller than the standard deviation for the size of an angle. This shows that the year 10 students gave closer estimations to the length of a line than they did for the size of an angle. This shows that estimating the length of a line is easier than estimating the size of an angle. Comparing this to my hypothesis three I was correct that estimating the length of a line is easier than estimating the size of an angle. This supports my prediction.
I will repeat what I did for year 10 to find the standard deviation from year 11 results.
This is a table that shows part of the working out of standard deviation from year 11 estimations of a length of a line.
Standard deviation for year 11 estimations of a length of a line.
s = Square root of sum of x squared / n – the mean squared
s = Square root of 3723.263 / 30 – 11.045
= Square root of 124.1087667 – 11.045
= Square root of 113.0637667
= 10.63314472
= 10.63
Now I will do the same to work out the standard deviation for year 11 estimations of size of an angle.
Standard deviation for year 11 estimations of a size of an angle.
s = Square root of sum of x squared / n – the mean squared
s = Square root of 126244.8 / 30 – 64.53
= Square root of 4208.16 – 64.53
= Square root of 4143.63
= 64.37103386
= 64.37
Now I have the standard deviation I will find the standardised scores which is the number of the standard deviations that a value lies above or below the mean. So the smaller the rage of the standardised scores the more consistent the results are which means that the more people estimated close to the actual answer which means it is easy. While the higher the range of the standardised score is the more varied it is. Which means that people estimated bit to high or a bit to low, which means that it is difficult because few people estimated close to the actual estimation.
The formula for standardised score is Value – Mean / Standard Deviation
The lowest standardised score for year 10 estimations of a length of a line is 7.5 – 11 / 1.85 = -1.89
The maximum standardised score for year 10 estimations of a length of a line is 15 – 11 / 1.85 = 2.16
The range of the results is 2.16 – -1.89 = 4.05
The lowest standardised score for year 10 estimations of a size of an angle is 45 – 65.2 / 65.41 = - 0.31
The maximum standardised score for year 10 estimations of a size of an angle is 80 –65.2 / 65.41 = 0.23
The range of the results is = -0.54
I will do the same for year 11 as I have done for year 10
The lowest standardised score for year 11 estimations of a length of a line is 9 – 11.1 / 10.63 = - 0.2
The maximum standardised score for year 11 estimations of a length of a line is 15 – 11.1 / 10.63 = 0.37
The range of the results is 0.37 – -0.2 = 0.57
The lowest standardised score for year 11 estimations of a size of an angle is 54 – 60.5 / 64.37 = -0.10
The maximum standardised score for year 11 estimations of a size of an angle is 78 - 60.5 / 64.37 = 0.27
The range of the results is 0.27 – -0.10 = 0.37
From the results of the standardised scores it shows that the standardised score for the length of a line for both year 10 and year 11 is bigger than the standardised scores for estimations of the size of an angle. This shows that the estimations for the length of a line were more varied than the estimations for the size of an angle. This shows that estimations for the size of an angle were more consistent which shows it was easier then estimating the length of a line. Comparing this to my hypothesis is that my hypothesis is wrong and that estimating the size of an angle is easier than estimating the length of a line.
Hypothesis 4
My prediction for hypothesis four is that girls and boys are about the same at estimating and one will not be better than the other. To support this I will draw a stem and leaf diagram.
In order to do this I will have to use stratified sampling to choose the number of people I will be needing from each strata. I will be using a sample of 16 pupils from each year .I will be using 16 pupils to represent the population of each year because 16 people will be an sufficient proportion to represent the sample and 16 is and evan number so I can choose equal number of girls and boys. I will not use the whole population to represent the sample this is because the numbers of girls are not proportional to the number of boys this will make the data biased.
I have now chose the sample needed from each strata to form my sample of 16 pupils from each year. Now using a scientific calculator I will choose 2 males and 2 females from each strata to form my sample. I will choose two males and two females from each strata because I have to choose 4 people from each strata in both year to make it fair I will choose two males and two females. I will choose two females and two males from each strata using a scientific calculator. I will do this by pressing shift and random and looking at the end two and three digits and if they are the numbers that are in the sample numbers of the 60 students I will record them and I will keep on pressing the equal button on the calculator until I got 32 random pupils which 16 are females and 16 are males.
Year 10
Year 11
Now I have all the pupils I need for my sample. I will know draw a stem and leaf diagram for year 10 and find out girls are better than boys at estimating. I will draw it by using their estimations of a length of a line and size of an angle.
Year 10 and 11
Key 7/5 = 7.5
Estimations of a length of a line
Female Males
5 7
5 8 0
5 0 9 0 0 5 0
0 0 0 10 0 25 5 0
0 0 0 5 11 0 0 5 5
0 0 0 0 12 0 0
0 13 0
Mode = 12 Mode = 9
Mean = 164 / 16 = 10.25 Mean = 167.25/ 16 = 10.45 Median = 16 + 1 / 2 = 8.5 Median = 16 + 1 / 2 = 8.5
= 11.5 + 11 / 2 = 11.25 = 10.5 + 10 / 2 = 10.25
Range = 13 - 7.5 = 5.5 Range = 13 - 8 = 5
If you look the mode for females is much closer to the actual length of the line than the mode for year 10. This shows that more girls had closer estimations to the actual length of a line than boys.
The mean for the girl’s estimations is closer to the actual length of a line than the mean for the boy’s estimations. Which shows that girls estimated better than the boys.
The median for year the girls are much closer to the actual length of the line than the median for the boys. This again clearly shows that girls estimate better than boys.
The range of the girl’s data is slightly less reliable than the range for the boy’s data. This shows that the boy’s results were slightly accurate than the girl’s results.
Overall this shows that the girls estimated better than the boys as shown by the measures of location. Comparing this to my prediction that there will be no such things girls are better than boys. I was wrong because the data shows that girls are better than boys at estimating.
Now I will use their estimations of a size of an angle to see whether girls are really better than boys at estimating. I will do this by using the same procedure as previously.
Year 10 and 11
Key 9/7 means 97
Estimations of a size of an angle.
Females Males
4
5.5 4.75 7 5.5 5 0 8 5.25 4
4.8 5 5 2 6 4 5 3 8.1 9 8.5
7.5 1 8 2 5 3 0.1 0 7 0 2.2 4 8 0.5
8 0
Mode = 65 No Mode
Mean = 1066.15 / 16 = 66.63 Mean = 1059.55/ 16 = 66.22 Median = 16 + 1 / 2 = 8.5 Median = 16 + 1 / 2 = 8.5
= 64.8 + 70 / 2 = 67.4 = 68.1 + 69/ 2 = 68.55
Range = 77.5 – 55.5 = 22 Range = 80 – 50 = 30
The results show that the measures of location for the female’s estimations of a size of an angle is closer to the actual size than the measures of location for the males. In addition to that the female’s range is more consistent and reliable than the range of the Males results. This clearly shows that the Females are better than males at estimations. This shows that my theory is wrong and that girl’s are better than boys at estimating.
To support further that girls are better than boys at estimating I will draw a comparative pie chart to show that girls are better than boys at estimating the length of a line and size of an angle. To draw it for the estimations of a length of a line I will use all the estimations 1.5 cm in the region of 11.5. So I will use all the estimations between 10 and 13 for males and female’s because it will be sufficient proportion of estimations to use.
I will start with using a radius of 2 cm for the male’s estimations of a length of a line.
So the area of the circle will equal
3.14 * 2 squared
= 12.57cm squared
12.57 / 11 = 1.143 cm squared
So for representing the females the radius, which I will be using, is
1.143 * 12 = 13.72 cm squared
Pie r squared = 13.72
R squared = 13.72 / 3.14
4.369
sq root 4.369 = 2.1 cm
The percentage of estimations that are with in the region of 10 and 13 cm for the males are that 11 people out of 16 people estimations were in this region. The angle degrees of the people with in the region of 10 and 13 is the following
11/16 * 360 = 247.5degrees
This means that 112.5 degrees of the estimations out of 360 degrees were out of the region of 10 and 13.
While the percentage of estimations that are with in the region of 10 and 13cm for the females are 12 out of 16 pupils. This as a percentage is equivalent to
12/16 * 360 = 270 degrees
This means that 90 degrees of the estimations are out of the region while 270 degrees of the estimations are inside the region. This is shown by the comparative pie charts.
= 0.012 cm squared for each person
So the area for 16 pupils we need to use an area of 0.012 * 1059.55 = 12.71
12.71/3.14 = 4.05
Square root of 4.05 = 2.0
The comparative pie charts for the males and females estimations of a length of a line show that a higher proportion of students who are females had estimations within the region of 10 to 13 than males. This means that there were more girls that estimated closer to the actual length of a line than the boys. This shows that girls are better than boys at estimating. It supports my previous conclusion from my stem and leaf diagram. In addition it shows that my prediction was wrong.
I will now draw comparative pie charts for the estimations of the size of an angle. I will do it by the procedure I used to draw these comparative pie charts. Theactual size of the angle is 61 degrees. So I will use estimations from 56 degrees to 66 degrees, which is 5 cm away from the size of an angle from either size which gives me a range of 10 degrees. I will use this range because it is a big enough range to find wether girls are better than boys at estimating.
I will start with a radius of 2 for the males as I previously did
The area of the circle will equal
3.14 * 2 squared
= 12.57cm squared
12.57 / 4 = 3.146cm squared
To represent the females the radius, which I will be using, is
3.146 * 5 = 15.73 cm squared
Pie r squared = 15.73
R squared = 15.73 / 3.14
5.01 cm squared
sq root 5.01 = 2.2
The region of their estimations is going to be between 56 and 66 degrees. So the size of the angle for 4 males who estimated with in the region of 56 and 66 is
4/16 = 0.25 * 360 = 90 degrees
This means that 270 degrees of the estimates were more than 5cm from the actual size of the angle.
The size of the angle for the girl’s who estimated within a region of 10cm from the actual size of the angle is
5 / 16 * 360 = 112.5 degrees
This means that 247.5 degrees of the pie chart are of estimates that are out of the region of 10cm. Which is shown by the pie charts.
The pie charts show that more females estimated closer to the actual estimation by their estimations being in between the 10 degrees range. This means that more females estimated closer to the actual size of the angle which shows clearly that girls are better than boys at estimating. This again supports the previous findings that girls are better than boys at estimating linking it back to my prediction it shows by prediction is wrong because girls are better than boys.
Overall the results show that for hypothesis four that girl are better than boys at estimating as shown clearly by the results. Comparing this to my prediction that there would be no such thing as girls being better than boys. It shows I was wrong because the results show that girls are better than boys at estimating.
Conclusion
My hypothesis one was that year 11 students are better than year 10 students at estimating. I thought this because year 11 students have generally one year more knowledge and experience than year 10 students so year 11 estimations will be more accurate than year tens estimations. From the measures of location and the measures of spread I found out that year 11 was better than year 10 at estimating. From the stem and leaf diagrams for the length of a line and for the size of an angle I found out that the range of year 11 results was narrower than the range of year 10 results. This shows that year 11 results are more consistent and reliable. So it shows that year 11 students are better than year 10 students at estimating. The results of the box and whisker diagram for year 11 estimations of a length of a line and size of an angle was much narrower than the results of the box and whisker diagram for year 10 estimations of a length of a line and size of an angle. This supports the previous findings that year 11 is better than year 10 at estimating. The results of the bar charts for year 10 and year 11 estimations of a length of a line and size of an angle show that year 11’s estimations have a narrower range than year 10’s estimations. The bar charts also show that the mode for year 11 estimations of a length of a line and the size of an angle is closer to the actual length of the line and size of the angle than the mode for year 10’s estimations of a length of a line and size of an angle. This clearly shows that year 11 is better than year 10 at estimating. This supports my hypothesis that year 11 is better than year 10 at estimating
My hypothesis two was that people who estimate the length of a line accurately may not always estimate the size of an angle accurately. I thought this because estimating a length of a line is easier than estimating the size of an angle which can be tricky. From the scatter graphs for both year 10 and year 11 estimations of a length of a line and size of an angle it shows that people who estimate the length of a line accurately have a high chance of estimating the size of an angle accurately. The gradient of the line of the best fit for year 10’s estimations is steeper than the gradient of the line of best fit for year 11’s estimations. This shows that year 10’s estimations are more varied and less reliable. The Y-intercept for year 11 estimations is closer to the actual length of the line than the y-intercept for year 10 estimations. In addition the results of the Spearman’s coefficient rank correlation shows very strong positive correlation between the estimations of a length of a line and the size of an angle. Which shows that people who estimate the length of the line accurately have a very high chance of estimating the size of an angle accurately. This shows that people who estimate the length of a line accurately have a high chance of estimating the size of an angle accurately and visa versa. Comparing my findings to my hypothesis that people who estimate the length of a line accurately may not estimate the size of the angle accurately shows that my hypothesis is wrong.
My third hypothesis was that estimating the length of a line is easier than estimating the size of an angle. I thought this because the length of a line can be estimated accurately because most people use rulers frequently to measure things. So they will be better at estimating the length of a line than estimating the size of an angle which are angles are used less frequently in daily life because estimating angles can be tricky at times. The results of the standardised scores show that the standardised scores for the length of a line for both year 10 and 11 is bigger than the standardised scores of the size of the angle. This shows that estimations for the length of a line were more varied than the estimations of a size of an angle. Which means that estimations of a size of an angle were more consistent than the estimations of a length of a line. Which shows it was easier to estimate the size of an angle because the results were more consistent meaning they are more reliable. Comparing this to my hypothesis it shows my hypothesis was wrong and that estimating the size of an angle is easier than estimating the length of a line.
My fourth hypothesis was that females and males estimations would be about the same so one will not be better than the other. I thought this because in each year group females and males will be about the same age and will have started education approximately at the same time so both genders will have about an equal chance of an accurate estimation. Instead I would say it depends on how much each person knows and how much he or she have worked and whether or not any factors may have affected their estimations will finalise the accuracy of the results not if they are girl they are better at estimating than a boy. From the back too back stem and leaf diagrams of the length of a line show that girls are better than boys at estimating which are shown by the measures of location. In addition to that the results of the measures of location and measures of spread for the female’s estimations of a size of a angle is closer to the actual size of an angle than the measure of location and spread for the males estimations. This shows that females are better than males at estimating. Which supports the theory of females are better than males at estimating. The comparative pie charts for the males and females estimations of a length of a line show that higher proportion of students who are females had estimations within the region of 10 to 13 cm for the length of a line than males. This means that there were more female’s that estimated closer to the actual length of the line than males. Which shows that females are better males at estimating. This supports my previous conclusion from the stem and leaf diagram. The pie charts for the estimations of a size of an angle show that female’s estimations are closer to the actual size of the angle than the male’s estimations of the size of an angle. This is shown by a larger number of female’s estimations than male estimations being in between the 10 degrees range of the actual size of the angle. Which means that more females estimated closer to the actual size of the angle than the males. This shows that females are better than males at estimating. This again supports by previous findings that girls are better than boys at estimating. Linking this back to my prediction it shows that my prediction was wrong and girls are better than boys at estimating.
Evaluation
The things that I think are good in my projects are firstly the way I chose a fair sample and represented everyone fairly. Secondly I liked the way in which I used different methods of supporting my hypothesises. Thirdly I liked the size of the angle and the length of a line I chose because it was reasonable. In addition I liked the way I consolidated all my findings and linked them to my theory, which really strengthened my findings. I also liked the way I chose the sample of students from the quarters, which was quick and easy to do and makes a fairer sample. Finally I liked the way I solved the calculations and then interpreted on them which really supports my findings.
I think that I have justified every hypothesis well enough so that I could obtain findings. Also the sampling method I used was the best sampling method to use with this sort of data and was very accurate and fair by it represented all the sample fairly. I think that the use of graphs I used in my project was accurate and clearly shows the results, which were justified by the interpretations and by the previous calculations. I justified and thoroughly explained what each calculation means and linked it very well to my hypothesise which really strengthened my findings.
In the project I have not dealt with any unexpected problems but I have dealt with problems, which I was prepared for, and new what to do. An example of this was when I was pre testing the sheet two students were absent who did not come to school for several days and did cause me a little bit of hassle but even thought I chose another two pupils to represent the sample which was a backup plan for me.
I have used many facilities of the computer to draw bar charts, back-to-back stem and leaf diagrams, scatter graphs etc. I used these facilities because they were quick and easy to draw on the computer and were quite accurate. Also it could be interpreted properly and easily and there was not much need for constantly flicking back pages.
Evan thought my project has many good things it also has things that need improvement these are things like choosing a larger sample because 30 students may in my case may not have been enough to represent the whole population. Also in some of the hypothesise like hypothesis three I could have used more statistics to support my findings even further. In addition I could have investigate on how the factors can affect people’s estimations like a factor such as ethnicity and could have been represented fairly. This would have showed me how big affect the factor could have on people’s estimations.
Also I could have used a larger sample that would give me larger limitations and I could have used more statistics that could show I have further limitations that I consolidated my findings. Evan thought I did not include theses things if you look at my investigation it is how ever pretty reliable because I gave everyone a fair chance of being selected and I selected the number of male’s and females according to population and the sample size and I more or less have obtained what I needed to obtain in order to complete my project. Therefore it shows my project is pretty reliable.
Also I have compared my results with another secondary school called St. Bedes school to see how reliable are my results but from that I could see that my results are pretty reliable there results are similar to my results which shows my results are correct. While in their project there are more outliers in the data the main reason for this they used a very large population so more likely it is that the more outliers you will get because there are more people so more estimations could be estimated inaccurately.
Method
- After I have the sample I will collect the estimations of student during break, lunch and home time.
- I have collected my sample I will know use data to support my hypothesis. The first thing that I will use to support my hypothesis is the measure of spread and location and compare year 11 estimations of the length of the line and size of the angle with year 10 estimations.
- Then I will use back-to-back stem and leaf diagrams and see whether year 11 is better than year 10 at estimating.
- 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.
