Data Handling Checklist - Research Paper Example

Running Head. DATA HANDLING CHECKLIST Data Handling Checklist 1.1 Design and use questionnaire to collect data comprised of open and closed questions (p264 Chapter 28). Solution: Questionnaire to collect data about left-handedness 1. Please tick the appropriate box. I am right-handed. I am left-handed. 2. In your view weather left-handedness is related to any scientific, artistic or sporting ability Yes No. I have no idea. 3. As generally gadgets are design for right-handed people and it is bit difficult for left-handed people to use them. In your opinion, is it necessary to design more gadgets for left-handed people or not to overcome the difficulty Yes No Data obtained through Questionnaire Now suppose if the survey is conducted by taking the opinion of 100 people then initially the raw data can be put in the following form as given in table 1.1. Raw data results of survey to collect data about left-handedness Question 1 Question 2 Question 3 No. of right handed people No. of left-handed people No. of people say yes No. of people say No. No. of people say I have no idea No. of people say yes No. of people say No. 86 14 49 28 23 87 13 Table. 1.1 (Q. 1.1, page. 64, Chapter 28). 1.2 Use data collection sheets to record data from a practical exercise (e.g. measuring temperature change over time) (p264 Chapter 28). Solution: There are a number of methods available to record the raw data. Let here we use tally chart to record the data related to measure the temperature Raw Data about the Variation of Temperature over Time Time (hrs) 02 hrs 04 hrs 06 hrs 08 hrs 10 hrs 12 hrs 14 hrs 16 hrs 18 hrs 20 hrs 22 hrs 24 hrs Temperature Monday 22 21 21 25 29 33 36 39 37 35 29 26 (C) Tuesday 22 21 21 23 25 28 31 35 34 30 28 26 Wednesday 21 21 21 23 24 28 30 33 30 28 25 23 Thursday 25 24 21 23 24 27 35 38 35 31 28 27 Friday 23 24 25 25 25 27 29 39 35 27 26 26 Saturday 24 23 24 25 26 28 29 39 37 31 27 25 Sunday 22 21 21 25 26 26 29 39 36 31 26 23 Table. 1.2(a) Tally Chart for Variation of Temperature over Time Temperature (C) Tally Frequency 21 - 24 26 25 - 28 29 29 - 32 12 33 - 36 10 37 - 40 7 Total 84 Table. 1.2 (b) change over time. The data recorded in table 1.2(a) is the raw data about the variation of temperature over time for a hot summer week. 1.3 Identify strengths and weaknesses of different methods of data collection, e.g. tally chart - frequency table (p274 Chapter 28 submit Question 1). Answer. (Question 1, p274 Chapter 28). Tally Chart Strengths of Tally chart. A tally chart is a grid used to help clearly show information as it is collected. A good tally chart, shows the information clearly. have clear columns and headings. uses lines to show numbers and a total number to show the frequency. Weakness of Tally chart Tally chart is not a good option to use for a large complex data collection. Since Tally chart uses lines to show numbers, therefore for large data the Tally (lines) patterns are not only have an odd outlook but also cover a lot of space. Frequency Table When summarizing large masses of raw data it if often useful to distribute the data in classes or categories and to determine the number of individuals belonging to each class called the class frequency. A tabular arrangement of data by classes together with the corresponding class frequency is called a frequency distribution or frequency table. The data organized in frequency table is termed as grouped data. Strengths of Frequency Table A frequency table can split data into classes or categories. By frequency table the actual number of scores as well as the % age of scores in each interval can be displayed. A frequency table can be used to summaries categorical, nominal, and ordinal data. It may also be used to summaries continuous data and once the data set has been divided up into sensible groups. Answer (Question 1, Mixed Exercise, p. 274, Chapter 28: Collecting data) (a)Since the survey is conducted to find out; how many people use public transport, what type of people use the public transport, what type of public transport they use, how often and for what purpose they utilize the public transport. The questionnaire can be improve by putting, direct close question to know about the use of public or private transport instead of question no. 1, 3 & 6. open question giving the choice of preference about the type of public transport in a single question eliminating questions no. 3 & 6. Also to know about the type of people, the question no. 6 can be discarded because it does not give any information about the type of people. (b) Since the questionnaire is to know the use of public transport and their type so the best option is to conduct it in the town center. As second choice the survey can also be conducted outside cinema or outside the superstore. As survey already contains question to know about the type of transport, so (i) & (ii) options are least adoptable. 2.1 Using information from tables, lists, computer databases construct frequency and grouped frequency tables (p275 Chapter 28 submit Questions 3&4). Solution. (Questions 3, p. 275, Chapter 28) Tally Chart Tally Chart for the No. of Hrs Sun Shines/day No. of Hrs Sun Shines Tally No. of Days (Frequency) 0 - 5 7 6 - 10 10 11 - 15 13 Total 30 Frequency Table Solution. (Questions 4, p. 275, Chapter 28) (a) Since there are 24 production results having 11 as its least value while 76 the maximum value. Following are required Tally chart & Frequency table. Frequency Table Frequency Table for the Hourly Production of Components No. of Components Produced No. of Hours (Frequency) 11 - 20 2 21 - 30 1 31 - 40 4 41 - 50 8 51 - 60 4 61- 70 3 71- 80 2 Total 24 (b) Since information about the type of components is provided so we assume that variation in production can be both due to size and complexity of components. 2.2 Using suitable examples, distinguish between discrete and continuous data (p. 275 Chapter 28 submit Question 2). Solution. (Question 2, p. 275, Chapter 28: Collecting Data) (a) The data obtained for heights measured in cm is a continuous numerical data. (b) The data obtained for weights measured in kg is continuous numerical data. (c) The data obtained for the days of week on which the students were born is a discrete numerical data. (d) The data obtained for the number of brothers & sisters the students have is a discrete numerical data. 3.1 Interpret critically statistical diagrams and graphs, explaining the main features of the data represented (p273, Chapter 28, Submit Question 3 & extension Question 273). Answer. Statistical Graphs and diagrams and graphs are graphical representation of data and can be used for two different purposes. They can be useful analytical tools for professional statisticians and they can be ways of displaying data to the public at large. There are a number of statistical diagrams and chart available including Time Charts, Log time charts, Bar charts, Scatter Diagrams and Histograms it can be expected that statisticians will be able to use them when appropriate and understand them without difficulty. Another important feature of statistical diagrams and graphs is that by using the statistical graphs and diagrams we can present data in a coherent and meaningful manner and there are a great many methods of doing so. One can hardly find an issue of a serious newspaper that does not include several graphs. A single such picture conveys - or can convey - the important features of the data more vividly and memorably than columns of data. Solution (p273, Chapter 28, Submit Question 3) Since the actual sale value is the product of RPI and total sale, therefore from given data table we have following actual sale values. Actual sale Value (million) Calculated by using RPI from the sale of a baby food Year 1993 1994 1995 1996 1997 1998 Retail Price Index (RPI) 100 105 110 113 122 131 Actual sale Value (million) 170 304.5 528 745.8 988.2 1271 Solution. (Extension Question 273, p273, Chapter 28). Following is the complete table for expected population of England in 2011. From expected table it is obvious that total number of women (26.1million) is greater than man population (25.9 million). (a) There are 9.03% male under age 15s. (b) There are 8.27% female under age 15s The % age difference between male and female under age 15s is 0.24%. Since there are 3.07% males and 4.62% female members who have age 75 or above, hence the %age difference between male and female members having age 75 or above is 1.55%. 3.2 Construct and use histograms with equal intervals from frequency distributions (p. 276 Chapter 29 submit Question 2 p283). Solution. (Question 2, p. 283, Chapter 29) (a) The required Tally Chart is given by table 3.2(a) on next page. (b) Table 3.2(b) is the required Frequency distribution Table. (c) Fig. 3.2(c1) and fig. 3.2(c2) be the required histogram and frequency polygon. Tally Chart for Sale Price of 30 Cars Price Range Frequency 0 - 999 0 1000 - 1999 5 2000 - 2999 8 3000 - 3999 5 4000 - 4999 8 5000 - 5999 4 Total 30 Table. 3.2(a) Frequency Distribution Table for Sale Price of 30 Cars Price Range Frequency 0 - 999 0 1000 - 1999 5 2000 - 2999 8 3000 - 3999 5 4000 - 4999 8 5000 - 5999 4 Total 30 Table. 3.2(b) Fig. 3.2(c1) For frequency polygon we have following frequency distribution table. Table. 3.2(c2) Fig. 3.2(c2) 3.3 Construct cumulative frequency diagram from selected data (p296 Chapter 31 submit Question 3 p299). Solution (p296 Chapter 31 submit Question 3 p299). For cumulative frequency polygon or ogive we have following cumulative frequency table. The cumulative-frequency polygon constructed from following table is given in fig. 3.3. Weight in KG Frequency (no. of babies) No. of babies have wt. less than 1.95 8 No. of babies have wt. less than 2.45 26 No. of babies have wt. less than 2.95 66 No. of babies have wt. less than 3.45 118 No. of babies have wt. less than 3.95 156 No. of babies have wt. less than 4.45 178 No. of babies have wt. less than 4.96 194 No. of babies have wt. less than 5.44 200 Table. 3.3(a) Fig. 3.3 Cumulative frequency Polygon. For median and inter-quartile range let we construct following table (3.3b). S.No. Weight in KG Class Mark 1 1.5 - 1.9 1.7 2 2.0 - 2.4 2.2 3 2.5 - 2.9 2.7 4 3.0 - 3.4 3.2 5 3.5 - 3.9 3.7 6 4.0 - 4.4 4.2 7 4.5 - 4.9 4.7 8 5.0 - 5.4 5.2 Table. 3.3b Since there are eight given values so, Lower quartile Q1= Average of Class mark 2, 3 = 2.45 Also Q2 = Average of Class mark 4, 5 = 3.45 Upper quartile Q3 = Average of Class mark 6, 7 = 4.45 Hence, median = Q2 = 3.45 & Inter-quartile = Q3 - Q1 = 2 (c) Compare to national figure there is a bit difference in the median, lower and upper quartile weights for babies calculated on the basis of information provided by Avon ford General Infirmary. The differences are, Lower quartile Q1 is 0.45 less than National figure. Median Q2 is 0.20 greater than National figure 3.25, & Upper quartile Q3 is 0.85 greater than National figure. 3.5 By drawing a scatter diagram and a line of best fit by inspection, comment on the correlation of the data (p. 296 Chapter 31, Question 3 p. 303 & Question 3 p. 305). Solution (Question 3 p. 303, Chapter 31). Since the given data is shown in table 3.5. Table. 3.5 Fig. 3.5 (a) (a) The required scatter diagram for the record of birth-weight and length of baby born. From scatter diagram it is obvious that maximum babies have lengths in between 50 - 60 cm while the corresponding weights vary from 3.4 - 3.8 kg. The slope or rate change of trend length vs. weight can be calculated by considering any two points on trend line i.e; @ change of trend length vs. weight = = = 6 cm/kg Hence the height varies 6cm per kg weight of a new born baby. (b) In the scatter diagram the boys weights and heights are marked with red color while the girls data is marked with black dots (fig. 3.5b) By joining the dots we observe there is not any particular pattern that emerges although few spikes can be observed on the graph. Fig. 3.5b 4.1 Calculate mean, median and mode from lists, in grouped and grouped frequency tables (p. 286 Chapter 30 submit Question1 p. 294). Solution (Question.1 p. 294, Chapter 30). (a) Arithmetic Mean (AM) for give data without the given frequency be, AM = , (1-4.a) where X ans N stand for data and N for the total number of entries. Therefore, AM = AM = 104 Since the median of a given set of data is the middle value of given set of data, so for given set of data the median is 103 and also the required mode be 102. (b) Arithmetic Mean (AM) for give data with frequency be, AM = , (1-4.a) where f, X ans N stand for frequency data and N for the total number of entries. Therefore, AM = AM = 1.3 Also 2 is the required median and 0 is the required mode for given se of data. (c) As Arithmetic Mean (AM) for give data with frequency be, AM = , (1-4.a) where f, X ans N stand for frequency data and N for the total number of entries. Therefore, AM = AM = 1.4 The required median and mode for given se of data are 4 and 0 respectively. 5.1 Calculate the probability of a single event (p. 312 Chapter 33, Question 2 p. 313). Solution (Question 2, p. 313, Chapter 33,). (a) Since hearts appear on 13 cards, so the probability of having heart is 13/52 =1/4 = 0.25 (b) As there are 26 red cards in a pack so probability of having a red card is 26/52 = 1/2 = 0.5 (c) According to given condition, there can be 12 cards in a pack, so probability is 12/52. (d) Since there can be four chances to have an ace so probability of having an ace be 4/52. (e) As there is only a single queen of hearts in a pack 52 cards so probability of having a queen of hearts is 1/52. 5.2 Use the addition law for mutually exclusive events and multiplication law for independent events (p. 312 Chapter 33, Question 1 p. 317 & Question 1 p. 319). Solution (Question 1 p. 317, Chapter 33). The Page 317 was not found in scanned copy. Solution (Question 1 p. 319, Chapter 33). (a) Since the probability of having a spade for first card as spade is, P (X) = 13/52 = 1/4, Since the first card is replaced by second card so, P(Y) = 13/52 = 1/4, Now according to addition law for mutually exclusive events say X, &Y then: P(X or Y) = P(X) + P(Y), Therefore, P (XorY) = 1/4 +1/4 = 1/2 Also for multiplication law for independent events, P(X or Y) = P(X)P(Y) P(X or Y) = (1/4)(1/4) = 1/16. (b) Since the probability of having a spade for first card as spade is P (X) = 13/52 = 1/4, & the probability of having other than a spade is P (Y) = 39/52 = 3/4, Now according to addition law for mutually exclusive events P(X or Y) = P(X) + P(Y) = (1/4)+(3/4) = 1 Also, P(X or Y) = P(X) P(Y) = (1/4)(3/4) = 3/16 for multiplication law for independent events. 5.3 Draw and use a tree diagram to solve problems involving probability (p. 312 Chapter 33, Question 1 p. 321). Solution (Question 1, p. 321, Chapter 33) Let 'A' be the probability of students having bicycle but no CD player, 'B' be the probability of students having CD player but no bicycle 'C' be the probability of students neither have a bicycle nor CD player 'D' be the probability of students having both CD player and bicycle. From give information we can draw following tree diagram fig. 5.3. The probability of a randomly chosen student has is, P(x) = (3/5)(7/10)(1/10)(3/10) = 63/1000 Bibliography Hodder. (n. d. ). Mathematics GCSE in a Year. 2nd Edition. Read More