INTENDED LEARNING OUTCOMES At the completion of this course, students are expected to:

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EDO UNIVERSITY IYAMHO COURSE CODE: ECO 113 COURSE TITLE: Introductory Statistics 1 NUMBER OF UNITS:2 Units COURSE DURATION: Two hours per week COURSE LECTURER: Dr.

differentiate and appraise Statistical scenarios Students should be able to apply knowledge of Statistics to solve problems in other disciplines Students should be able to gather data and develop

1 EDO UNIVERSITY IYAMHO COURSE CODE: ECO 113 COURSE TITLE: Introductory Statistics 1 NUMBER OF UNITS:2 Units COURSE DURATION: Two hours per week COURSE LECTURER: Dr. Sylvester Ohiomu INTENDED LEARNING OUTCOMES At the completion of this course, students are expected to: Students should be able to define and explain Statistics Students should be able to differentiate and appraise Statistical scenarios Students should be able to apply knowledge of Statistics to solve problems in other disciplines Students should be able to gather data and develop tables and models from observations Students should be able to apply statistics to daily problems, economic, national and global challenges Course Details Week 1: Week 2: Week 3: Week 4: The Meaning, Nature, Scope and Purpose of Statistics Sources of Statistical Data and Problems of data Collection Frequency Distribution Formats Pie Charts, Bar Charts, Quartiles and Percentiles Week 5: Measure of Central Tendency: Mean of Grouped and ungrouped Data Week 6: Median of Grouped and ungrouped Data Week 7: Mode of Grouped and ungrouped Data Week 8: Measure of Dispersion: Range and Mean Deviation

2 Week 9: Variance Week 10: Standard Deviation Week 11: Probability Distributions: Probability Theory Week 12: Probability Theorem Week 13: Revision Resources Lecturer s Office Hours Dr. Sylvester Ohiomu Wednesdays Pm Courseware: Books Anyiwe M.A. (2006). Applied Statistics in Social Sciences. Ethiope publishing Corp, Benin City, Elements of Statistics I: Descriptive Statistics and probability: Volume 1 (Schaum's Outline Series). Gupta, S.C.(1984), Statistical Methods, New Delhi; Sultan Chand & Sons Publishers Structure of the Programme/ Method of Grading Grading method is organized into two basic parts namely: Continuous assessment: 30% End of Semester Examination: 70% Assignments & Grading Academic Honesty: All classwork should be done independently, unless explicitly stated otherwise on the assignment handout. You may discuss general solution strategies, but must write up the solutions yourself. If you discuss any problem with anyone else, you must write their name at the top of your assignment, labeling them collaborators. NO LATE HOMEWORKS ACCEPTED Turn in what you have at the time it s due. All home works are due at the start of class. If you will be away, turn in the homework early.

3 Preamble Meaning and Nature of Statistics Statistics deals with quantitative data i.e. the collection, analysis, interpretation of data and drawing of necessary inferences Statistics is subdivided into three modern subgroups namely: Simple Statistics which deals with data collection Descriptive statistics which deals with methods of organizing, summarizing and presentation of data in a convenient and informative way Inferential or deductive statistics which deals with methods used to draw conclusion or inferences about the characteristics of populations based on sample data Data Collection Data is very important in any statistical investigation Sources of data collection include the following: Primary sources: questionnaire, personal interview, direct interview, registration of events, by telephone etc Secondary sources: regular records, periodic records and irregular records Problems of data collection in Nigeria include: secrecy, illiteracy, scanty records, under developed ICT and cumbersome nature of manual data storage Data Presentation and Organization Forms of presentation of data include the following Visual presentation Visual Presentation: Examples are (1) tabular presentation with practical demonstration the class (Frequency distribution table)

4 (2) Graphical presentation with practical demonstration on graphs (Histogram and frequency polygon) (3) Diagrams including bar charts and pie charts Multiple Bar Charts Series Series Series 1 0 Category Category Category Category Line Chart

5 Series Series Series 1 0 Category Category Category Category Pie Chart Sales

6 Area Chart Purpose and Importance of Statistics Statistical data are needed for business planning in organizations Statistics are needed in the appraisal and evaluation of business projects, Government budgetary policies and assessment Knowledge of statistics is needed for preparation of balance sheets

7 Statistics is used in production forecasting for the future Statistics is used for pricing policy The knowledge of statistics is used in economics and econometric analysis Summary Statistics deals with the collection, analysis, interpretation of data and drawing of inferences Three subdivisions of statistics include: simple statistics, descriptive statistics and inferential statistics Data can be collected from primary and secondary sources Problems of data collection include: illiteracy, secrecy, scanty data, under developed ICT and challenges in manual data storage Data can be presented through verbal and visual modes Statistics is used for planning, forecasting, evaluation, assessment, pricing policy, economic and econometric analysis Measure of Central Tendency: Mean A measure of central tendency is any measure that indicates the centre of distribution which includes the mean, median, mode and Geometric mean The Mean (Arithmetic Mean) of a finite set of observations is the sum of their values divided by the number of observations Given a set of observations: x1, x2,, xn, the mean can be calculated as follows: Ẍ = x1 + x2 + + xn /n = n i Xi 1 /n Example 1; Given the following observations 43, 535, 70, 2779, 318. Find the mean Solution: Mean Ẍ = /5 = 3745/5 = 749 Calculation of mean from a Frequency Table FX / F Ẍ =

8 Where F = frequency or number of observations Fx = product of frequency and variable F = N = sum of frequencies Calculation of Mean for Grouped Data Explanation of steps involved Median The median is the value of the middle term or value of a series of observation when arrange either in ascending or descending order of magnitude Take the central number if odd and average of the two central numbers if even observations foe ungrouped data. For Grouped data, Median = L + w(n/2 Fc)/Fe Where L = Lower class boundary of class interval containing the median, n = total no of observations = f Fc = Cumulative frequency of the class preceding the median class Fe = Frequency of the class interval containing the median W = width or size of the median class For examples from ungrouped data and grouped data respectively refer to textbooks Mode The mode is defined as the observation(s) that occurs with the greatest frequency. It is possible to have more than one modal value appearing at a time. Thus: One modal value is called Unimodal

9 Two modal vales = bimodal Three modal values = trimodal Four modal values = quadrimodal Many modal values = multimodal For Grouped data, Mode = L1 + w{ (F1 F0)/[2F1 F0 F2]} Where L1 = Lower class limit of the modal class F1 = modal frequency F0 = Frequency of class immediately preceding the modal class F2 = Frequency of the class succeeding the modal class W = Size or width of the modal class For examples from ungrouped data and grouped data respectively refer to textbooks Measures of Dispersion (Variation) The spread of a figure also called the variation tells us how data are dispersed around the mean There are several measures of dispersion. These include the following: Range = Highest observation lowest observation Mean Deviation = average of the deviations of observations from the mean Variance = The square of Standard Deviation Standard Deviation = Coefficient of Variation For examples and exercises respectively refer to textbooks Coefficient of Variation

10 The main items of variation are the mean and standard deviation. It is the relative dispersion and expresses the standard deviation as a percentage of the mean. Coefficient of variation (CV) = (S.D./Mean) x 100 = Ẍ x 100% Coefficient of Mean Deviation = (M.D./Mean) x 100/1 Example: Given the wages of laborers and helpers in a construction firm below: Laborer Ẍ = = Solution; Helper Ẍ = = Determine the CV and interprete your result V1 = (S.D./Mean) x 100 = Ẍ x 100 = (408.60/ ) x 100 = 19.98% V2 = (S.D./Mean) x 100 = Ẍ x 100 = (32.40/154.25) x 100 = 21% Interpretation: From the result derived above, it is obvious that the wages of laborers are more stable than those of their Helpers since the former has less coefficient of variation (19.98%) than the latter (21.00%) Probability Theory Probability is a branch of statistics which aids or enables an investigator to generalize the character of a whole from the part. It helps to make inference about a population. Probability denotes that the occurrence or non-occurrence of an event is by chance. Probability = Total No of success/total No of outcome. * Example: When a coin is tossed what is the probability that the head will face up? Answer: ½ Types of Events/Rules of Probability A. Independent Events: Two or more events are said to be independent if the occurrence of one event has no effect on the occurrence of the other.

11 In a specific term if A and B are two events; if the occurrence of A does not affect the occurrence of B then the events A and B are said to be independent events eg tossing two coins B. Dependent Events: Two or more events are said to be dependent if the occurrence of one event has effect on the occurrence of the other. C. Mutually Exclusive Events: If two events are dependent to the degree that the occurrence of one event excludes the other, the two events which can not take place simultaneously are called mutually exclusive events e.g. tossing a coin to have both head and tail. Rules of Probability 1, P(Q) = o Where Q is an empty set. 2. P(S) = 1 Where S is the sample size 3. P(A1) = 1 P(A) Where A1 is the complement of A 4. Additive Rule: Mutually exclusive events: Given any two mutually exclusive event A and B ie the outcomes can be either A or B but not both, then P(AuB) = P(a) + P(B) and P(AnB) = 0. While P(AuBuC = P(A) + P(B) + P(C) Non-Exclusive events P(AuB) = P(a) + P(B) - P(AnB) P(AuBuC) = P(a) + P(B) + P(C) - P(AnB) P(AnC) P(BnC) + P(AnBnC). Example: Given that there are 60 mangoes in a basket at random, 30 of them are unripe, 20 ripe and 10 soft. Find the probability of picking a ripe and soft mango at once. Solution,, Probability of picking a ripe mango (event A) is P(A) = 20/60 =1/3 Probability of picking a soft mango (event B) is

12 P(B) = 10/60 = 1/6 Given that events A and B are mutually exclusive, we apply the additive rule P(AuB) = P(A) + P(B) 1/3 +1/6 = 1/2 Conditional Probability. This is the probability of one event given that the other has already occurred. Probability of A given that B has already occurred ie P(A/B) = P(AnB)/P(B), P(B) =/ 0 or P(B/A) = P(AnB)/P(A), P(A) =/ 0 Multiplication Probability. The multiplication rule states that if two event are not independent, the probability of A and B occurring is given by the product of the probability of A, P(A) and the conditional probability of B occurring given that A has occurred. P(AnB) = P(A).P(B/A), given that P(A) =/ 0 or P(BnA) = P(B).P(A/B), given that P(A) =/ 0 Similarly, if A, B, and C are any 3 subsets of sample space, then P(AnBnC) =P(A). P(B/A).P(C/AnB) given that P(AnB) =/ 0 Example: If a basket of fruits contains 8 oranges and 6 apples, what is the probability that 3 fruits picked are, an orange, an apple and an orange in that order? Solution: P(AnBnC) =P(A). P(B/A).P(C/AnB) 8/14 x 6/13 x 7/12 = Example: A bag containing boxers has 10 red boxers, 6 white boxers and 4 blue boxers. Find the probability that the boxer drawn if not replaced is (i) red (ii) white (iii) blue in that order Solution: Let A,B,C, represent the events of drawing a red boxer, a white boxer and a blue boxer respectively (i) P(A) = 10/20 = ½ = 0.5

13 (ii) P(B/A) = 6/19 = (iii) P(C/AnB) = 4/18 = Thus, P(AnBnC) =P(A). P(B/A).P(C/AnB) 0.5 x x = Factorial, Permutation and Combination n! is defined as N! = nx(n-1)x(n-2)x x3x2x1 Thus 0! = 1 1! = 1 2! = 2x1 = 2 3! = 3x2x1 =6 4! = 4x3x2x1 = 24 Permutation: This is the ordering or arrangement of an object. A permutation of n different objects taken r at a time is the arrangement of r out of n objects. This can be written as npr = n!/(n-r)!, where r is less than or equal to n Example: Given that 5 officers of an institution are to be hosted in a meeting.(i) Determine the number of different sitting arrangements for the 5 of them (ii) Assume that only 2 of the principal officers were ask to represent the institution, determine the sitting arrangement this time. Solution: (i) Here the number of objects, n = 5 and n is taken at a time ie npn = 5P5 = 5!/(5-5)! = 5!/0! = 5x4x3x2x1/1 = 120 (ii) Here n=5 and r=2 ie npr= 5P2= 5!/(5-2)!= 5x4x3x2x1/3x2x1 = 20 Combination: This is concerned with the number of different groupings the object can occur rather than the ways it can be arranged or ordered as in the case of permutation. Thus, the number of combinations of n different objects taken r at a time is given as ncr = n!/(n-r)!r!, where r is less than or equal to n

14 Example: A school intends to hold her Inter-house sport Competition. Out of the 12 Prefects in the school, a committee of 4 is randomly selected, in how many ways can this committee be selected? Solution:12C4 = 12!/ (12-4)!4! = 12!/8!4! = 11880/24 = 495

MIDTERM EXAMINATION (Spring 2011) STA301- Statistics and Probability

STA301- Statistics and Probability Solved MCQS From Midterm Papers March 19,2012 MC100401285 Moaaz.pk@gmail.com Mc100401285@gmail.com PSMD01 MIDTERM EXAMINATION (Spring 2011) STA301- Statistics and Probability