Assignment 1 Essay - 2845 Words
Assignment 1
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Compulsory school age begins at the start of the term following a child’s 5th birthday until the end of school year when the child turns 16.
The local authorities have a duty to provide a free school place for children of a c parents have a duty to make sure the child receives education during the compulsory age years. Under part of the Every Child Matters agenda and the Childcare act it’s an entitlement that all 3-4 years old in England receive up to 15 hours per week early years education for 38 weeks. This is funded by the local government.
The foundation curriculum is for children aged 3-5 and is used in reception classes and in school nurseries. 2.
There are 4 types of mainstream schools funded by the local authority and are classes as mainstream schools and they will follow the national curriculum.
Community schools – Run and owned by local authority that will support the school through looking to develop links with the local community. They may also lend out the school facilities to local groups for adult education or childcare.
Foundation and Trust Schools – Run by own governing body that determines its own admissions policy after consultation with the local government.
The school will have a charitable link with an outside partner such as a business. Voluntary Schools – There are two types – Voluntarily aided and voluntarily controlled. Voluntary aided – Although everyone can apply these are mainly faith or religious schools and is funded 3 ways, governing body, church or charity and Local Authority.
Voluntarily Controlled – Are run and funded by the local authority but run in the same way as the voluntarily aided. Specialist Schools – More often than not these are Secondary schools who apply for specialist status for one or more specialist subjects. They receive government funding for this, they can also apply for specialist SEN Status (Special Educational needs) Independent Schools – Different from local authority schools as they are funded privately by fees, investments or gifts and endowments. Most can have charitable status Academies – Historically academies were set up by sponsors from businesses although 5 years ago the government introduced more opportunities for communities’ to become involved in giving schools academy status.
There are 2 main choices for the over 16’s
Leave school and get a job, this used to be the traditional way as there weren’t many further education opportunities. There are now more apprenticeships which allow work and college and can be seen as a “win-win” situation you earn and learn at the same time. Or continue with their studies via apprenticeship or full or part time education..
This can be in school, college, independent college. Under the September guarantee the government says that by the end of Sept of the year teach person leaves compulsory education a further learning place will be available.
Governors – Usually a team of between 10-20 who have a responsibility for running the school, they will have links with the school and the local community. There should be at least one parent and one staff governor. Governors will work closely with the head teacher and Senior Management Team (SMT), to set aims and objectives for the school, adopt new policies for achieving these aims and objectives.
Senior Management Team – Made up of more experienced staff who may have management roles, usually deputy head teachers, year group leaders, SENCO’s (Special Education Needs Co-Ordinator) and Foundation Stage Leaders. They will meet on a regular basis to discuss the running of the school, and make any decisions which may affect the running of the school then are responsible for passing any new information onto the staff. SENCO – They are responsible for managing and monitoring the provision for those with special needs within the school. This can involve
Ensuring Liaison with parents and any other professionals ( Speech therapist for...
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a. Frequency Distribution Table
Class | Class Boundary | Frequency | Cumalative Frequency |
11 – 14 |
11.5 – 14.5 |
15.5 – 20.5
20 – 24 |
20.5 – 25.5
25.5 – 29.5 |
30.5 – 34.5 |
35 – 39 |
35.5 – 39.5 |
40 – 44 |
40.5 – 44.5 |
45 – 49 |
45.5 – 49.5 |
∑ ? = 36 |
b) . “ Less than or equal to”
& than 20 cars sold
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36-25 = 11
11/ 36 X 100 = 35.5 %
Discrete Quantitative variable |
Time ( H) | Frequency | Cum Frequency | Cum Freq % |
0-0.4 | 2 | 2 | 2.86 |
0.5-0.9 | 5 | 7 | 10.00 |
1.0-1.4 | 19 | 26 | 37.14 |
1.5-1.9 | 15 | 41 | 58.57 |
2.0-2.4 | 10 | 51 | 72.86 |
2.5-2.9 | 8 | 59 | 84.29 |
3.0-3.4 | 5 | 64 | 91.43 |
3.5-3.9 | 3 | 67 | 95.71 |
4.0-4.4 | 2 | 69 | 98.57 |
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Assignment 2 GDA
Geographic Data Analysis 182/582VEGETATION SAMPLING(Assignment 2)Written by Agata Zabolotny Undergraduate Student25 October 2008 Curtin University of Technology Perth, Western AustraliaCurtin University-Agata ZabolotnyCONTENTSIntroduction:……………………………………………………………………………….…………1 Question 1…………..…..…………………………………………………………………….………2 Question 2 ………………………………….………………………………………………..….……6 Question 3………………………………………………………………………………….…………8 Question 4………………………………………….……………………………………………….…10 Conclusion……………………………………………………………………………..………………13 References…………………………………………………………………………….…………....…14Curtin University-Agata ZabolotnyGeographic Data Analysis 182/582INTRODUCTIONBy studying the region in Canning Regional Park, the purpose of this assignment was to analyse vegetation samples and interpret the results. This was to be completed using a sampling scheme to study the distribution of the vegetation south of the Canning River. The task was to then perform a quadrant analysis on different vegetation samples within the chosen grid area. Once collected, the data would need to be classified using various classification techniques. The classification methods would need to be compared and the best one utilise to make conclusions about the vegetation distribution in the area. The tasks assigned to complete were the following. 1. Complete an area-species curve where the number of species (y-axis) is plotted against the area (x-axis) using the &number of species& data. Explain what an appropriate quadrat size should be. 2. Perform a quadrat analysis on the three species, Tall Grass, Yellow and White flower, for which you did quadrat sampling. Explain the choice for the quadrat cell size and compare the results for the three species. 3. Perform two (2) different classifications. Explain your choice of classification method, and interpret the results of the classification. Decide and justify the number of classes to be used. Finally obtain a fragmentation index for each classification. Explain which classification worked better. 4. Analysis of a linear transect data collected. Focusing on any trends present in the growth abundance and height. The results were obtained using the ArcGIS and Excel programs. Therefore this report will detail those findings.Curtin University-Agata ZabolotnyQUESTION 1 To test all the regions we tested a series of areas and counted the specie number within each 1x1 cell. The total area we tested was an area of 10x10m. The results obtained are shown below. Column Dimensions .5x.5 .5x1 1x1 1x2 2x2 2x5 5x5 5x10 10x10 Number of Species 2 3 4 4 6 6 7 7 7Area 0.25 0.5 1 2 4 10 25 50 100From the information above we can see that there is a maximum of seven different vegetation species within the area. The number of species increased exponentially with the increment of the study area. Weather only two species were found in the first study region the number increased quickly with expansion. By plotting a specie area curve we can see this exponential relationship on a graph. The relationship however would not continue increasing. By judging the samples it is believed that a maximum number of species was recognised within the 5x5 area. This means that this would probably be the best cell area to choose as it represents the whole population of samples within the smallest area. The 2x2 area was too small and didn’t represent all the species while the 10x10 showed all the species but would be inefficient so sample as a smaller area is available for the same information. Therefore from the findings we can conclude that the best choice would be the 5m by 5m area.Figure 2: Number of species (y-axis) is plotted against the area (x-axis)Number of Species8 6 4 Number of Species 2 0 0.25 0.5 1 2 4 10 25 50 100Curtin University-Agata ZabolotnyQUESTION 2 By obtaining the vegetation samples of the 25m2 area, seven main species were identified. We divided the area into 1x1 m cells as it fit perfectly into the 5x5 quadrant. This cell size was most convenient as it was easy to work with and it fit perfectly within the 25m area. Therefore the cell size was set at 1x1. We focused on three major vegetation species within the area. The first was the Tall Grass, then the White and Yellow Flowers. To study their spatial distribution we had to count the number of species within each cell and perform various statistical operations on each. We staked out a 5x5 BraunBlanquet Scale to determine the coverage of each of the species in each of the cells. This table below shows the findings, using percentages. Part 2 Species 1White flower 10 20 5 3 3 3 5 5 5 10 Yellow flower 5 10 10 15 50 35 30 30 20 20 Tall grass 70 70 70 70 20 40 60 30 70 50% 10 5 20 15 1 % 15 15 2 30 60 % 70 50 60 20 202 10 3 15 210 15 15 20 10Species 230 5 50 10 2040 50 50 20 40Species 340 20 30 40 7030 20 20 10 20The data below represents the information for the Tall Grass.Curtin University-Agata ZabolotnyLevel of significance 0.05 degrees of freedom = (quads-1)25-1=24 VMR= Variance/Mean TABLE VALUE 36.415The null hypothesis we have chosen for this test was that ‘white flowers are randomly distributed within the quadrants’. This means that there is no clustering within any of the cells. We tested against this hypothesis. The Alternative hypothesis was that ‘there was clustering present within the quadrant’. The level of significance was set at 0.05 while the degrees of freedom were calculated that 25-1 was 24. We chose to test the Variance Mean Ration against the x2 test because it’s better for testing spatial relationships and useful for comparing observed and expected“occurrences”. By using the values we are able to find out table value of 36.415 from the table. Weare then able to get the VMR. If the value for the VMR is less than one, that shows regularity in the distribution of species. If the VMR is larger than one that shows clustering and if the VMR is one that shows randomness. In this case the VMR was calculated to be 10.3 therefore we reject the null hypothesis. The value of over 1 indicates clustering therefore the Tall Grass is clustered. The following tables shows the information for the White Flower.Curtin University-Agata ZabolotnyFollowing the procedure explained above the VMR for the white flower was found to be also larger than one which means that the White Flower is also clustered. The following is a table showing the Statistics for the Yellow Flower.In this case the VMR was also found to be bigger than 1 therefore the Yellow flower is clustered also. From the results we can conclude that the Tall Grass is the most clustered, followed by the Yellow flower and finally by the White Flower. These results indicate strong clustering within the region.Curtin University-Agata ZabolotnyQUESTION 3 The two different classification schemes were chosen to represent the vegetation data. The while flower, yellow flower and the tall grass were all classified using the same number of classes but by different schemes. The choice to classify the information into 4 different classes was made in order to simplify the data. Four classes were chosen to make it easy for the visual representation. If a smaller number was chosen the data would all just merge into one and if a larger number was chosen the data would be less grouped and classified. Natural Breaks was chosen as one of the classification schemes. This scheme provides a unique classification type by judging the natural distribution of the data and classifying it into classes relevant to the breaks within. Natural breaks are useful as it reflects the nature of the spread of the data. It takes clustering into account when performing the classification. It divides data into classes based on the distribution of the data. It uses a formula (Jenks Optimisation) to judge grouping and reduce variance between individual samples to increase variance between groups. This method is useful for combining similar values and as we found out in part two of this report, all of the values are clustered thus this makes Natural Breaks a good classification scheme. The flowing were the results obtained from using natural breaks classification. The values in the table represent the percentages of the particular vegetation specie present in each cell. TALL GRASSCurtin University-Agata ZabolotnyYELLOW FLOWERWHITE FLOWERFrom the visual observation of the data we can see that where the white and yellow flower dominates, the tall grass does not occur often. Where the tall grass dominates the white and yellow flowers do not often occur. The presence of the species depends on where it grows within the quadrant. Considering that North is also towards the top on each picture, it is evident that Tall grass dominates the areas closer to the river which was located north of the quadrant. The yellow and white flowers however occur further away from the river and thus more southwards.Standard Deviation Classification was the second method used to classify the data. This classification finds the mean value of all of the data and separates the values using class breaks of 0.25, 0.5 or one standard deviation more from the mean until all of the data belongs to a class. Classes represent the statistical spread of the data. This was also useful because by classifying using 4 classes all of the data would be placed within one of the four quartiles. One standard deviationCurtin University-Agata Zabolotnyaway from the mean places the data within the first quintile. By basing the classes around the mean each cell represents how closely it correlates with the average value of the statistics. The following show the three vegetation species classed using standard deviation. YELLOW FLOWER The values on the left represent percentages and the values on the right represent standard deviations.WHITE FLOWERTALL GRASSCurtin University-Agata ZabolotnyOnce again we come to the same conclusion as with the natural breaks classification scheme. The tall grass species tends to dominate the areas near the river to the north of the quadrant. The white flower which is still quite a tall plant, dominates the central region of the sample area. The yellow flower which was the shortest and smallest plant out of the three, dominates the region away from the river. The yellow flower can thus be said to prefer drier soil while the taller specie of white flowers and tall grass dominate more moisture rich soils nearer to the Canning River. The type of classification can manipulate the spatial appearance of the data thus it is important to implement The Fragmentation index to see how fragmented the data is and remove bias. This index evaluates the results of the classification undertaken. It is calculated using the following formula. n-is the number of original map regions present on the map. m-is the number of contiguous map regions after the classification of the data. f- Is the fragmentation index The value of f ranges from 1 to 0. A value of 1 indicates perfect fragmentation and that every value on the map is a different region. A value of 0 indicates that there is no fragmentation and all the regions are grouped into 1. The values obtained for the equation are as follows. Tall Grass- Sd Model F= 14-1/25-1 = . Yellow Flower Sd Model F= 10-1/25-1= 0.375 White Flower Sd Model F= 15-1/25-1= .3333333 Tall Grass Natural Breaks Method F= 15-1/25-1= .333333 Yellow Flower Natural Breaks Method F= 14-1/ 25-1= .6666666 White Flower Natural Breaks Method F= 15-1/25-1= .333333 From these results we can conclude that overall the fragmentation index of the Standard deviation classification is lower. A lower value indicates that that the classes are less fragmented and thus better classified. The optimum result was gained from the Yellow Flower using standard deviation. The Rest of the results lied slightly over the 0.5 value of the index. This indicates that the data is more fragmented. Overall thus only the yellow flower produced a desirable value.Curtin University-Agata ZabolotnyQUESTION 4 Linear transect sampling could be used to perform a regression analysis on the type of vegetation present within the area and if any pattern is visible between the type of plants and their growth regions. The regression can be used to predict values and relationships of the values. The type of relationship expected is that taller grass would dominate the regions near the river while shorter species would be more abundant away from the river. By plotting Abundance and height of two different species and analysing the regression correlation it is possible to judge whether the trend really exists within the vegetation. The following is the data recorded for the Tall Grass species. Row 1 indicates the furthest cell from the river and row 11 indicates the closest to the river and thus the closest point north. ROW 1 2 3 4 5 6 7 8 9 10 11 Abundance 50 60 55 70 60 65 70 80 95 90 95 Height 20 20 30 20 45 50 70 70 85 90 100Tall Grass120 100 坐标轴标题 80 60 40 20 0 0 20 40 60 80 100 Height 线性 (Height) y = 1.671x - 65.47 R? = 0.793坐标轴标题From the graph above we can see that the value of the Regression is a very good positive number. The Regression line examines form of relationship between X and Y variables in this case being the abundance and height of the tall grass. The value in this instance is 0.7935 this is a very good regression indicating that there is a strong correlation between the two variables. However we cannot draw conclusions from one sample only. We thus also consider the yellow flower in the investigation also.Curtin University-Agata ZabolotnyThese are the results for the Yellow Flower.Abundance 10 10 20 30 15 20 15 10 5 0 0 Height 15 20 15 55 15 80 20 15 15 20 15 ROW 1 2 3 4 5 6 7 8 9 10 11Yellow Flower90 80 70 60 50 40 30 20 10 0 0坐标轴标题y = 1.408x + 8.625 R? = 0.353 Height 线性 (Height)1020 坐标轴标题3040From these results we can see that there is almost no correlation between the variables of abundance and height. The value of R2 is only 0.3535 which is quite a low number. This indicates no correlation and thus no relationship. From the results obtained we can conclude that there is a strong tendency for tall grass to grow in more abundance and height closer to the river thus preferring more moisture rich soils. The yellow Flower however shows no preferential growth pattern within the area.Curtin University-Agata ZabolotnyCONCLUSION Therefore this report has successfully conducted a classification of vegetation species within the Canning Regional Park area. There has been a test of species samples which provided vital statistical information for further classification. Two major classification schemes were tested against the data. The Standard Deviation was proved to be a more beneficial classification method compared to the Natural breaks. This was determined using the Fragmentation index values. Finally we tested for any relationships and spatial distributions along a linear transect and found that while some species of vegetation had no relationship in distribution, other tended to flourish more vigorously near the river.Curtin University-Agata ZabolotnyREFERENCESCoroner,R. 2008. Geographic Data Analysis 182/582: Module 3. Final Conclusion Lecture 6 (9): 1-56 http://webct.curtin.edu.au/SCRIPT/450_a/scripts/serve_home (accessed October 25, 2008).Coroner,R. 2008. Geographic Data Analysis 182/582: Module 3. Lecture 3: 1-56 http://webct.curtin.edu.au/SCRIPT/450_a/scripts/serve_home (accessed October 20, 2008).Truong, T. Comment on ‘Assignment 2’ 18-09-2008. http://webct.curtin.edu.au/SCRIPT/450_a/scripts/serve_home (accessed October 21, 2008).Bateson,G. Comment on ‘Assignment 2 ‘19-09-2008. http://webct.curtin.edu.au/SCRIPT/450_a/scripts/serve_home (accessed October 21, 2008).Coroner,R. 2008. Geographic Data Analysis 182/582: Assignment 2. http://webct.curtin.edu.au/SCRIPT/450_a/scripts/serve_home (accessed October 19, 2008).Curtin University-Agata Zabolotny
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