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Basic Statistics Presentation

1. 1. INTRODUCTION TO STATISTICSMd. Mortuza Ahmmed
2. 2. Applications of Statistics Agriculture Business and economics Marketing Research Education Medicine
3. 3. Variable Qualitative VariableIndependen Dependent t variable variable Discrete Continuous variable variable Quantitative Variable
4. 4. Scales of Measurement Nominal Ordinal scale Scale Ratio Interval scale scale
5. 5. FREQUENCY TABLERating of Relative Tally marks Frequency Drink Frequency P IIII 05 05 / 25 = 0.20 G IIII IIII II 12 12 / 25 = 0.48 E IIII III 08 08 / 25 = 0.32 Total 25 1.00
6. 6. SIMPLE BAR DIAGRAM160 150140120 100100 80 60 56 40 25 20 0 Muslim Hindu Christians Others
7. 7. COMPONENT BAR DIAGRAM300250200 Section D150 Section C Section B100 Section A 50 0 Male Female
8. 8. MULTIPLE BAR DIAGRAM100 90 80 70 60 Section A 50 Section B 40 Section C 30 Section D 20 10 0 Male Female
9. 9. PIE CHART Religion of studentsMuslim Hindu Christians Others 8% 15% 46% 31%
10. 10. LINE GRAPH Share price of BEXIMCO7000 64006000 56005000 5000 450040003000 300020001000 0 July August September October November
11. 11. HISTOGRAM20181614121086420
12. 12. BAR DIAGRAM VS. HISTOGRAM Histogram Bar diagramArea gives frequency Height gives frequencyBars are adjacent to Bars are not adjacent each others to each others Constructed for Constructed for quantitative data qualitative data
13. 13. STEM AND LEAF PLOT Stem Leaf 1 1479 2 13479 3 1379 4 1347 5 1349 6 1347
14. 14. SCATTER DIAGRAM 300 250 200Supply 150 100 50 0 0 5 10 15 20 25 30 Price
15. 15. COMPARISON AMONG THE GRAPHS Graph Advantages Disadvantages Shows percent of total Use only discrete data Pie chart for each category Can compare to normal curve Use only continuous data Histogram Compare 2 or 3 data sets Use only discrete dataBar diagram easily Compare 2 or 3 data sets Use only continuous data Line graph easily Shows a trend in the data Use only continuous data Scatter plot relationshipStem and Leaf Handle extremely large data Not visually appealing Plot sets
16. 16. MEASURES OF CENTRAL TENDENCYA measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data.  Arithmeticmean (AM)  Geometric mean (GM)  Harmonic mean (HM)  Median  Mode
17. 17. ARITHMETIC MEANIt is equal to the sum of all the values in the data set divided by the number of values in the data set.
18. 18. PROBLEMS Find the average of the values 5, 9, 12, 4, 5, 14, 19, 16, 3, 5, 7. The mean weight of three dogs is 38 pounds. One of the dogs weighs 46 pounds. The other two dogs, Eddie and Tommy, have the same weight. Find Tommy’s weight. On her first 5 math tests, Zany received scores 72, 86, 92, 63, and 77. What test score she must earn on her sixth test so that her average for all 6 tests will be 80?
19. 19. AFFECT OF EXTREME VALUES ON AMStaff 1 2 3 4 5 6 7 8 9 10Salary 15 18 16 14 15 15 12 17 90 95
20. 20. CALCULATION OF AM FOR GROUPED DATA x f f.x 0 05 00 1 10 10 2 05 10 3 10 30 4 05 20 10 02 20 Total N = 37 90AM = 90 / 37 = 2.43
21. 21. MEDIAN1 3 2 MEDIAN = 21 2 31 4 3 2 MEDIAN = (2 + 3) / 2 = 2.51 2 3 4
22. 22. MODE
23. 23. WHEN TO USE THE MEAN, MEDIAN AND MODE Best measure of central Type of Variable tendency Nominal Mode Ordinal Median Interval/Ratio (not Mean skewed)Interval/Ratio (skewed) Median
24. 24. WHEN WE ADD OR MULTIPLY EACH VALUE BY SAME AMOUNT Data Mean Mode Median Original 6, 7, 8, 10, 12, 14, 12.2 14 13 data Set 14, 15, 16, 20 Add 3 to 9, 10, 11, 13, 15, 17, 15.2 17 16each value 17, 18, 19, 23Multiply 2 12, 14, 16, 20, 24, 24.4 28 26 to each 28, 28, 30, 32, 40 value
25. 25. MEAN, MEDIAN AND MODE FOR SERIES DATA For a series 1, 2, 3 ….n, mean = median = mode = (n + 1) / 2 So, for a series 1, 2, 3 ….100, mean = median = mode = (100 + 1) / 2 = 50.5
26. 26. GEOMETRIC MEAN
27. 27. HARMONIC MEAN
28. 28. AM X HM = (GM) 2For any 2 numbers a AM X HM and b, = (a + b) / 2 . 2ab /AM = (a + b) / 2 (a + b)GM = (ab) ^ ½ = ab = (GM) 2 HM = 2 / (1 / a + 1 / b) = 2ab / (a + b)
29. 29. EXAMPLE For any two numbers, AM = 10 and GM = 8. Find out the numbers.(ab)^ ½ = 08 (a - b)2 = (a + b)2 – 4ab ab = 64 = (20)2 – 4 .64 = 144(a + b) / 2 = 10 a + b = 20 . . . . .(1) => a - b = 12 . . . .(2) Solving (1) and (2) (a, b) = (16, 4)
30. 30. EXAMPLEFor any two numbers, GM = 4√3 and HM = 6. Find out AM and the numbers. AM √ab = 4√3 (a - b)2= (GM)2/ HM =>ab = 48 = (a + b)2 – 4ab= (4√3) 2 / 6 = (16)2 – 4 . 48=8 (a + b) / 2 = 8 = 64 => a + b = 16 …(1) a - b = 8 ...(2) Solving (1) & (2) (a, b) = (12, 4)
31. 31. CRITERIA FOR GOOD MEASURES OF CENTRAL TENDENCY Clearly defined Readily comprehensible Based on all observations Easily calculated Less affected by extreme values Capable of further algebraic treatment
32. 32. AM ≥ GM ≥ HMFor any two numbers a & b (√a - √b) 2 ≥ 0AM = (a + b) / 2 a + b – 2(ab)^1/2 ≥ 0GM = (ab)^1/2 a+b ≥ 2(ab)^1/2HM = 2 / (1 / a + 1 / b) (a + b) / 2 ≥ (ab)^1/2 = 2ab / (a + b) => AM ≥ GM Multiplying both sides by 2(ab)^1/2 / (a + b) (ab)^1/2 ≥ 2ab / (a + b) GM ≥ HM So, AM ≥ GM ≥ HM