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MCA_UNIT-4_Computer Oriented Numerical Statistical Methods

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Course: MCA
Subject: Computer Oriented Numerical Statistical Methods
Unit-4 Correlation
RAI UNIVERSITY, AHMEDABAD

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MCA_UNIT-4_Computer Oriented Numerical Statistical Methods

  1. 1. RAI UNIVERSITY, AHMEDABAD 1 Course: MCA Subject: Computer Oriented Numerical Statistical Methods Unit-4 RAI UNIVERSITY, AHMEDABAD
  2. 2. RAI UNIVERSITY, AHMEDABAD 2 Unit-IV-Correlation Sr. No. Name of the Topic Page No. 1 Introduction and Definition of Correlation 2 2 Types of Correlation and examples of correlation 3 3 Coefficient of Correlation, Methods for calculating coefficient of Correlation. 5 4 Scatter diagram method and its example, Merit and limitations of Scatter diagram method 6 5 Karl Pearson’s coefficient of correlation and examples based on its, Properties of Karl Pearson’s coefficient 12 6 Correlation Coefficient for Bivariate Frequency Distribution and its examples 16 7. Merits and limitations of Karl Pearson’s coefficient 21 8. Probable error and example based on it. 22 9. References 24 10 Exercise 25
  3. 3. RAI UNIVERSITY, AHMEDABAD 3 1.1 Introduction: In this Lesson, we have discussed about the correlation. In which, we have defined the measure of relationship between two variables... Suppose we have a set of 30 students in a class and we want to measure the heights and weights of all the students. We observe that each individual (unit) of the set assumes two values – one relating to the height and the other to the weight. Such a distribution in which each individual or unit of the set is made up of two values is called a bivariate distribution. Some examples of bivariate distribution are (i) In a class of 60 students the series of marks obtained in two subjects by all of them. (ii) The series of sales revenue and advertising expenditure of two companies in a particular year. (iii) The series of ages of husbands and wives in a sample of selected married couples. Thus in a bivariate distribution, we are given a set of pairs of observations, where in each pair represents the values of two variables. The concept of β€˜correlation’ is a statistical tool which studies the relationship between two variables and Correlation Analysis involves various methods and techniques used for studying and measuring the extent of the relationship between the two variables. 1.2 Definition: Two variables are said to be in correlation if the change in one of the variables results in a change in the other variable.
  4. 4. RAI UNIVERSITY, AHMEDABAD 4 2.1 Types of correlation. Correlation is classified by the following types and they are: 1. Positive and Negative 4. No correlation 2. Simple, Multiple and partial 5. Strong Correlation 3. Linear and non linear. 6. Weak Correlation 1. Positive and Negative Positive and negative correlation depends on the direction of change of the Variables. If move variables move in the same direction i.e., when there is an increase in The value one variable influenced by an increase in the value of other variable is called Positive correlation. If two variables tend to move in opposite directions so that an Increase in the values of one variable is influenced by decrease in the value of the other Variable, then the correlation is said to be negative correlation. 2.Simple and Multiple and Partial: When we study about only two variables then the correlation is said to be simple
  5. 5. RAI UNIVERSITY, AHMEDABAD 5 where as we study about more than two variables is called multiple correlation. In multiple variable environments, some variables excluded due to some reason then it is termed as partial correlation 3.Linear and Non-Linear: The distinction between linear and non-linear correlation is based upon the consistency of the ratio of change between the variables. If the amount of change in one variable tends to bear constant ratio to the amount of change in one variable then the correlation is said to be linear. Correlation would be called non-linear or curve linear if the amount of change in one variable does not bear a constant ratio to the amount of change in the other variable. For example, if we double the amount of rainfall, the production of rice or wheat, etc., would not necessary be doubled. 4. No Correlation No correlation occurs when there is no linear dependency between the variables.
  6. 6. RAI UNIVERSITY, AHMEDABAD 6 5. Strong Correlation A correlation is stronger the closer the points are located to one another on the line. 6. Weak Correlation A correlation is weaker the farther apart the points are located to one another on the line. 2.2 Some examples of series of positive correlation are: (i) Heights and weights; (ii) Household income and expenditure; (iii) Price and supply of commodities; (iv) Amount of rainfall and yield of crops. 3.1 Coefficient of Correlation Correlation is statistical technique used for analyzing the behavior of two Variables. This analysis refers with the relationship between two or more variables. Statistical measures of correlation are co-variation between series, not of functional
  7. 7. RAI UNIVERSITY, AHMEDABAD 7 Relationship. It is not possible to obtain another variable, if the value of a one variable Known in one series. 3.2 Methods For calculating Coefficient of Correlation: 4.1 Scatter Diagram Method Scatter diagram is most popular and easy way of deciding the relation between two variables. It is a graphical method and ascertain the direction of correlation between two variables. To construct the scatter diagram take independent variable on X-Axis and dependent variable on Y Axis. Plot the graph of intersection points of two variables and decide the relation according to the scatter plot. 1. If all the points fall on a straight line moving from left lower corner to right upper corner then it is the perfect positive correlation.
  8. 8. RAI UNIVERSITY, AHMEDABAD 8 2. If all the points fall on a straight line moving from upper left corner to right lower corner then it is a perfect negative correlation. 3. If all the points scattered nearby around the line then this will be high or low degree of positive and negative correlation according to the direction of line.
  9. 9. RAI UNIVERSITY, AHMEDABAD 9 4. If all the points scattered every where on the graph and no pattern will be identified then there is no correlation between the variables.
  10. 10. RAI UNIVERSITY, AHMEDABAD 10 4.1.1 Exampleβ€”Company Ξ± decides to use scatter graph method to split its factory overhead (FOH) into variable and fixed components. Following is the data which is provided for the analysis. Mont h Units FOH 1 1,520 $36,375 2 1,250 38,000 3 1,750 41,750 4 1,600 42,360 5 2,350 55,080 6 2,100 48,100 7 3,000 59,000 8 2,750 56,800 Solution:
  11. 11. RAI UNIVERSITY, AHMEDABAD 11 Fixed Cost = y-intercept = $18,000 Variable Cost per Unit = Slope of Regression Line To calculate slop we will take two points on line: (0, 18000) and (3500, 68000) Variable Cost per Unit = (68000 βˆ’ 18000) Γ· (3500 βˆ’ 0) = $14.286 4.1.2 Example –Situation: The new commissioner of the American Basketball League wants to construct a scatter diagram to find out if there is any relationship between a player’s weight and her height. How should she go about making her scatter diagram?
  12. 12. RAI UNIVERSITY, AHMEDABAD 12 Collect the data (Remember to use 50-100 paired samples). Draw and label your x and y axes. Plot the data on the diagram.
  13. 13. RAI UNIVERSITY, AHMEDABAD 13 According to this scatter diagram the new commissioner was right. There does seem to be a positive correlation between a player's weight and her height. In other words, the taller a player is the more she tends to weight. 4.2 Merits: (1) This is a simple method of studying correlation between two variables. (2)More mathematical knowledge is not required in this method. (3)If one of the pairs of values is extreme, it does not influence much in deriving the conclusion. (4)It is the first step in studying the relationship between two variables. 4.3 Limitations: (1)This method gives an idea about the direction and to some extent the degree of relationship between the variables, but does not give the exact measure of the relationship between the variables. 5.1 Karl Pearson’s Coefficient of Correlation The Karl Pearson’s method is popularly known as Pearson’s Coefficient of correlation. One of the most widely used statistics is the coefficient of correlationβ€˜π‘Ÿβ€™ which measures the degree of association between the two values of related variables given in the data set. The coefficient of correlation β€˜r’ is given by the formula π‘Ÿ = βˆ‘ π‘‹π‘Œ π‘›πœŽπ‘₯ 𝜎 𝑦 = βˆ‘ π‘‹π‘Œ βˆšβˆ‘ π‘₯2 βˆ‘ 𝑦2 [∡ 𝜎2 π‘₯ = βˆ‘ π‘₯2 𝑛 ; 𝜎2 𝑦 = βˆ‘ 𝑦2 𝑛 ]
  14. 14. RAI UNIVERSITY, AHMEDABAD 14 Here 𝑋 = (π‘₯ βˆ’ π‘₯Μ…); π‘Œ = (𝑦 βˆ’ 𝑦̅) 𝜎π‘₯ =Standard deviation of series π‘₯ 𝜎 𝑦 =Standard deviation of series 𝑦 𝑛 = Number of pairs of observations π‘Ÿ = The (product moment) correction coefficient This method is to be applied only where deviations of items are taken from actual mean and not from the assumed mean. The values of coefficient of correlation β€˜π‘Ÿβ€™ obtained from the above formula always lies between Β±1. When r = +1 it means there is a perfect positive correlation between the variables. When r = -1 it means there is a perfect negative correlation between the variables. However if r = 0 there is no relationship between the variables. 5.2 Steps: 1. Find out the mean of the two series 𝑋̅ π‘Žπ‘›π‘‘ π‘ŒΜ… 2. Take deviations of the two series from the respective means 𝑋̅ π‘Žπ‘›π‘‘ π‘ŒΜ… and denote x and y. 3. Square the deviations and find the sum of square and denote βˆ‘ π‘₯2 π‘Žπ‘›π‘‘ βˆ‘ 𝑦2 4. Multiply deviations of 𝑋 and π‘Œ i.e., βˆ‘ π‘‹π‘Œ 5. Substitute the values of βˆ‘ π‘‹π‘Œ, βˆ‘ π‘₯2 π‘Žπ‘›π‘‘ βˆ‘ 𝑦2 in the formula. 5.3 Properties of correlation Co-efficient: (1) The correlation coefficient lies between -1 and +1. (2)The correlation co-efficient is independent of change of origin and scale. (3)The correlation co-efficient is an absolute number and it is independent of units of measurement.
  15. 15. RAI UNIVERSITY, AHMEDABAD 15 (4) π‘Ÿ2 always lies between 0 and 1 i.e. 0 ≀ π‘Ÿ2 ≀ 1 5.1.1 Example:- Making use of the data summarized below, calculate the coefficient of correlation X 3 5 7 3 2 Y 2 4 6 2 1 Solution: First of All 𝑋̅ = βˆ‘ π‘₯ 𝑖 𝑛 and π‘ŒΜ… = βˆ‘ 𝑦 𝑖 𝑛 ∴ 𝑋̅ = 20 5 = 4 and π‘ŒΜ… = 15 5 = 3 𝒙 π’š 𝑿 = (𝒙 βˆ’ 𝑿)Μ…Μ…Μ…Μ… 𝒀 = (π’š βˆ’ 𝒀̅) π’™π’š 𝒙 𝟐 π’š 𝟐 3 2 -1 -1 1 1 1 5 4 1 1 1 1 1 7 6 3 3 9 9 9 3 2 -1 -1 1 1 1 2 1 -2 -2 4 4 4 16 16 16 π‘Ÿ = βˆ‘ π‘‹π‘Œ βˆšβˆ‘ π‘₯2 βˆ‘ 𝑦2 π‘Ÿ = 16 √16Γ—16 π‘Ÿ = 16 16 π‘Ÿ = 1 5.1.2Example:- Making use of the data summarized below, calculate the coefficient of correlation .
  16. 16. RAI UNIVERSITY, AHMEDABAD 16 Case A B C D E F G H X 10 9 6 10 12 13 11 9 Y 9 4 6 9 11 13 8 4 Solution:- First of all we find π‘₯Μ… = βˆ‘π‘₯ 𝑛 = 80 8 = 10 , 𝑦̅ = βˆ‘π‘¦ 𝑛 = 64 8 = 8 Case 𝒙 𝒙 βˆ’ 𝟏𝟎 = 𝑿 𝑿 𝟐 π’š π’š βˆ’ πŸ– = 𝒀 𝒀 𝟐 𝑿𝒀 A 10 0 0 9 1 1 0 B 9 -4 16 4 -4 16 16 C 6 -1 1 6 -2 4 2 D 10 0 0 9 +1 1 0 E 12 +2 4 11 +3 9 6 F 13 +3 9 13 +5 25 15 G 11 +1 1 8 0 0 0 H 9 -1 1 4 -4 16 4 𝒏 = πŸ– βˆ‘π‘₯ = 80 βˆ‘π‘‹ = 0 βˆ‘π‘‹2 = 32 βˆ‘π‘¦ = 64 βˆ‘π‘Œ = 0 βˆ‘π‘Œ2 = 72 βˆ‘π‘‹π‘Œ = 43
  17. 17. RAI UNIVERSITY, AHMEDABAD 17 π‘Ÿ = βˆ‘ π‘‹π‘Œ βˆšβˆ‘ 𝑋2 βˆ‘ π‘Œ2 π‘Ÿ = 43 √32 Γ— 72 π‘Ÿ = 43 √2304 π‘Ÿ = 43 48 = +0.896 6.1 Correlation Coefficient for Bivariate frequency Distribution: When the number of observations is too large, the data are usually classified in to a two-way table known as a bivariate table. A bivariate frequency table is given below. 0-10 10-20 20-30 30-40 40-50 𝒇 π’š 0-10 1 3 5 2 1 12 10-20 2 4 6 8 2 22 20-30 3 0 7 9 4 23 30-40 4 7 10 7 20 48 40-50 1 2 5 3 4 15 𝒇 𝒙 11 16 33 29 31 120 In the above table a bivariate frequency distribution of marks of mathematics and marks of science of 120 students are shown. The marks of mathematics are denoted by π‘₯ and the marks of science are denoted by 𝑦.Both the variables π‘₯ and 𝑦 are classified into groups 0-10,10-20,…etc. The marks in science of different groups are represented in rows. In the first cell the frequency is 1, which indicates that there is 1student getting marks between 0-10 in science. The frequency of the cell of first row and second column is 3 which means there are 3 students getting π‘₯ 𝑦
  18. 18. RAI UNIVERSITY, AHMEDABAD 18 marks in mathematics between 10-20 and in science between 0-10.The different values of 𝑓𝑦 show the total number of students securing marks in science in the corresponding groups. Similarly the different values of 𝑓π‘₯ show the total number of students securing marks in mathematics in the corresponding groups.It is obvious that βˆ‘ 𝑓π‘₯ = βˆ‘ 𝑓𝑦 = 𝑛 (Total Frequency) The following formula is used for calculating correlation co-efficient for the data given in bivariate table. π‘Ÿ = 𝑛 βˆ‘ 𝑓𝑒𝑣 βˆ’ βˆ‘ 𝑒𝑓𝑒. βˆ‘ 𝑣𝑓𝑣 βˆšπ‘› βˆ‘ 𝑒2 𝑓𝑒 βˆ’ (βˆ‘ 𝑒𝑓𝑒)2 Γ— βˆšπ‘› βˆ‘ 𝑣2 𝑓𝑣 βˆ’ (βˆ‘ 𝑣𝑓𝑣)2 The required values in the above formula can be obtained as follows. (i) Denote one variable by 𝑋 and another variable by π‘Œ. (ii) Denote mid-value of 𝑋 by π‘₯ and mid-value π‘Œ by 𝑦. (iii) Obtain 𝑒 and 𝑣 by the formula 𝑒 = π‘₯βˆ’π΄ 𝐢 π‘₯ 𝑣 = π‘¦βˆ’π΅ 𝐢 𝑦 Where 𝐴 and 𝐡 are assumed means of 𝑋 and π‘Œ respectively and 𝐢 π‘₯ and 𝐢 𝑦 are the class intervals of variables π‘₯ and 𝑦. It should be noted that frequency of π‘₯ and frequency of 𝑣 will be the same as frequency of 𝑦. Hence 𝑓π‘₯ will be 𝑓𝑒 and 𝑓𝑦 will be 𝑓𝑣. (iv) For each class of π‘₯ multiply 𝑒 and 𝑓𝑒 and obtain βˆ‘ 𝑒𝑓𝑒 similarly obtain βˆ‘ 𝑣𝑓𝑣. (v) For each group of π‘₯ multiply 𝑒𝑓𝑒 by 𝑒 and obtain βˆ‘ 𝑒2 𝑓𝑒. Similarly obtain βˆ‘ 𝑣2 𝑓𝑣. (vi) For each cell find the produced of 𝑓, 𝑒 π‘Žπ‘›π‘‘ 𝑣 and obtain 𝑓𝑒𝑣 for each cell. From them find βˆ‘ 𝑓𝑒𝑣. The value of π‘Ÿ can be obtained by substituting these values in the formula. 6.1.1 Exampleβ€”
  19. 19. RAI UNIVERSITY, AHMEDABAD 19 Find the coefficient of correlation from the following data: Age of Wife Solution: We Know that π‘Ÿ = 𝑛 βˆ‘ 𝑓𝑒𝑣 βˆ’ βˆ‘ 𝑒𝑓𝑒. βˆ‘ 𝑣𝑓𝑣 βˆšπ‘› βˆ‘ 𝑒2 𝑓𝑒 βˆ’ (βˆ‘ 𝑒𝑓𝑒)2 Γ— βˆšπ‘› βˆ‘ 𝑣2 𝑓𝑣 βˆ’ (βˆ‘ 𝑣𝑓𝑣)2 Here 𝑛 = βˆ‘ 𝑓π‘₯ = βˆ‘ 𝑓𝑦 = 140 𝑒 = π‘₯βˆ’25 10 and 𝑣 = π‘¦βˆ’25 10 All the values required in the formula can be available from the table. We have to obtain the values of 𝑓𝑒𝑣 for each cell. For the first cell 𝑓 = 20, 𝑒 = βˆ’1 and 𝑣 = βˆ’1. Hence the value of 𝑓𝑒𝑣 for the first cell is 20(-1)(-1)=20. Similarly for the last cell the values of 𝑓, 𝑒 π‘Žπ‘›π‘‘ 𝑣 are respectively 6,2 and 2. Therefore, 𝑓𝑒𝑣 = 6(2)(2) = 24. Thus the values of 𝑓𝑒𝑣 for each cell can be obtained. These values are shown in brackets in each cell. Age Of Husbands 10-20 20-30 30-40 40-50 10-20 20 26 - - 20-30 8 14 37 - 30-40 - 4 18 3 40-50 - - 4 6
  20. 20. RAI UNIVERSITY, AHMEDABAD 20 ↓ β†’ 𝒙 y 10-20 20-30 30-40 40-50 𝒇 π’š M.V. π’š 𝒗 𝒗𝒇 𝒗 𝒗 𝟐 𝒇 𝒗 𝒇𝒖𝒗 10-20 (20) 20 (0) 26 46 15 -1 -46 46 20 20-30 (0) 8 (0) 14 (0) 37 59 25 0 0 0 0 30-40 (0) 4 (18) 18 (6) 3 25 35 1 25 25 24 40-50 (8) 4 (24) 6 10 45 2 20 40 32 𝒇 𝒙 28 44 59 9 140 -1 111 76 M.V. 𝒙 15 25 35 45 𝒖 -1 0 1 22 𝒖𝒇 𝒖 -28 0 59 18 49 𝒖 𝟐 𝒇 𝒖 28 0 59 36 123 𝒇𝒖𝒗 20 0 26 30 76 Row- wise and column –wise totals of these values are shown in the last row and the last column respectively .From them βˆ‘ 𝑓𝑒𝑣 is obtained .The values of βˆ‘ 𝑓𝑒𝑣 obtained from rows and columns should be equal. In the given example βˆ‘ 𝑓𝑒𝑣 = 76. Now putting these values in the formula, we get π‘Ÿ = 140(76)βˆ’(49)(βˆ’1) √140(123)βˆ’(49)2Γ—βˆš140(111)βˆ’(βˆ’1)2 π‘Ÿ = 10689 √14819Γ—βˆš15539 π‘Ÿ = 0.70
  21. 21. RAI UNIVERSITY, AHMEDABAD 21 6.1.2 Exampleβ€” Find correlation coefficient from the following data: Scores 18 19 20 21 22 200-250 3 3 2 1 - 250-300 - 2 4 2 2 300-350 3 5 5 2 - 350-400 - 1 2 3 3 400-450 - - 2 4 1 Solution: ↓ β†’ 𝒙 y 18 19 20 21 22 𝒇 π’š M.V. π’š 𝒗 𝒗𝒇 𝒗 𝒗 𝟐 𝒇 𝒗 𝒇𝒖𝒗 200-250 (12) 3 (6) 3 (0) 2 (-2) 1 9 225 -2 -18 36 16 250-300 (2) 2 (0) 4 (-2) 2 (-4) 2 10 275 -1 -10 10 -4 300-350 (0) 3 (0) 5 (0) 5 15 325 0 0 0 0 350-400 (-1) 1 (0) 2 (3) 3 (6) 3 9 375 1 9 9 8 400-450 (0) 2 (8) 4 (4) 1 7 425 2 14 28 12 𝒇 𝒙 6 11 15 12 6 50 -5 83 32 M.V. 𝒙 18 19 20 21 22 𝒖 -2 -1 0 1 2
  22. 22. RAI UNIVERSITY, AHMEDABAD 22 𝒖𝒇 𝒖 -12 -11 0 12 12 1 𝒖 𝟐 𝒇 𝒖 24 11 0 12 24 71 𝒇𝒖𝒗 12 7 0 7 6 32 We shall denote age by 𝑦 and ages 18,19…shall be taken as mid values. π‘Ÿ = 𝑛 βˆ‘ π‘“π‘’π‘£βˆ’βˆ‘ 𝑒𝑓𝑒.βˆ‘ 𝑣𝑓𝑣 βˆšπ‘› βˆ‘ 𝑒2 π‘“π‘’βˆ’(βˆ‘ 𝑒𝑓𝑒)2Γ—βˆšπ‘› βˆ‘ 𝑣2 π‘“π‘£βˆ’(βˆ‘ 𝑣𝑓𝑣)2 π‘Ÿ = 50(32)βˆ’(1)(βˆ’5) √50(71)βˆ’(1)2Γ—βˆš50(83)βˆ’(βˆ’5)2 π‘Ÿ = 1605 √3549Γ—βˆš4125 π‘Ÿ = 0.42 7.1 Merits of Karl Pearson’s correlation coefficient: 1. Karl Pearson’s co-efficient of correlation is the best measure for representing the relationship between two variables. 2. The degree and direction of the relationship between the variables can be obtained by it. 7.2 Limitations of Karl Pearson’s correlation coefficient: 1. It is based on the assumption of linearity of relationship between the variables. 2. The computation by this method is difficult compared to other methods 3. The correlation co-efficient is highly influenced by extreme pairs of observations. 4. It is always difficult to interprete the correlation co-efficient, correctly.
  23. 23. RAI UNIVERSITY, AHMEDABAD 23 8.1 Probable Error: Generally we obtain correlation co-efficient of a sample drawn from a bivariate population. If different samples of the same size are drawn from a given population, we get different values of π‘Ÿ. All these values of π‘Ÿ differ from the actual value of the population correlation co-efficient. The average of the absolute differences of correlation Co-efficients obtained from all possible samples and the population correlation co-efficient is known as probable error of the correlation Co-efficient. The value of probable error depends upon the size of the sample. If the sample is large, the value of probable error is small. From the value of the sample correlation co-efficient we can estimate the population correlation co-efficient, and with the help of probable error we can determine whether the correlation in the population is significant or not. If a sample of size 𝑛 is drawn from a bivariate population and if π‘Ÿ is its correlation co-efficient ,then the probable error (P.E) can be found out by the following formula: 𝑃. 𝐸. = 0.6745(1 βˆ’ π‘Ÿ2 ) √ 𝑛 The following rules can be applied to judge whether the correlation in the population is significant or not: (1)If π‘Ÿ < 𝑃. 𝐸., there is no evidence of correlation in the population. i.e. the correlation in the population is not significant. (2)If π‘Ÿ > 6(𝑃. 𝐸. ), there is evidence of significant correlation in the population. Moreover with the help of probable error we can determine the limits. Within which the population correlation co-efficient is expected to lie. The probable limits of the population correlation co-efficient are π‘Ÿ Β± 𝑃. 𝐸. 8.1.1 Exampleβ€”
  24. 24. RAI UNIVERSITY, AHMEDABAD 24 The correlation co-efficient obtained from 𝒂 sample of 16 pairs of observations drawn from a population is 0.7,calculate the probable error of the correlation co-efficient and interprete it. Also find the limits of the population correlation co-efficient. Solution: Here 𝑛 = 16 and π‘Ÿ = 0.7 ∴ Probable error = 0.6745(1βˆ’π‘Ÿ2) √ 𝑛 ∴ Probable error = 0.6745(1βˆ’(0.7)2) √16 ∴ Probable error = 0.086 Here π‘Ÿ = 0.7 π‘Žπ‘›π‘‘ 6(𝑃. 𝐸) =6 (0.086) = 0.516 i.e. π‘Ÿ > 6(𝑃. 𝐸. ) Thus there is significant correlation between the variables. The probable limit of the population co-efficient is π‘Ÿ Β± 𝑃. 𝐸. = 0.7 Β± 0.086 = 0.614 π‘‘π‘œ 0.786 This means that the population correlation co-efficient will most probably lie between 0.614 and 0.786.
  25. 25. RAI UNIVERSITY, AHMEDABAD 25 References and website Name: 1. Statistical Methods by S.P.Gupta 2. Business Statistics (B.S. Shah Prakashan) 3. http://www.tutorhelpdesk.com/homeworkhelp/Statistics-/Scatter-Diagram- Method-Assignment-Help.html 4. http://shakehandwithlife.blogspot.in/2014/09/measures-of-correlation.html 5. http://www.spcforexcel.com/files/images/positivecorrelation.png 6. http://www.spcforexcel.com/files/images/negativecorrelation.png 7. http://www.dhsbpsychology.co.uk/wp- content/uploads/2013/04/Correlations.jpg
  26. 26. RAI UNIVERSITY, AHMEDABAD 26 EXERCISE Q-1 Evaluate the following Questions: 1. Find Pearson’s correlation co-efficient. Wage 100 101 102 102 100 99 97 98 96 95 Cost of living Index 98 99 99 97 95 92 95 94 90 91 2. Find correlation Coefficient. x 300 350 400 450 500 550 600 650 700 y 800 900 1000 1100 1200 1300 1400 1500 1600 3. Find correlation co-efficient from the following data: Scores 18 19 20 21 22 200-250 3 3 2 1 - 250-300 - 2 4 2 2 300-350 3 5 5 2 - 350-400 - 1 2 3 3 400-450 - - 2 4 1 4. The Correlation co-efficient for a sample drawn form a bivariate population is 0.6 and its probable error is 0.05396.Find number of pairs of the sample. Also find the probable limits for the population correlation co-efficient. 5. Find number of pairs from the following data: π‘Ÿ = 0.5; βˆ‘ π‘₯𝑦 =120;βˆ‘ π‘₯2 = 90; 𝑆 𝑦 = 8.

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