4. parameter and statistic

Distinguish between Parameter and Statistic.
Calculate sample variance and sample standard deviation.
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1. Parameter and Statistic

2. Learning Competency The learner will be able to 1. Distinguish between parameter and statistic.

3. A parameter is a measure that describes a population . It is usually denoted by Greek letters. A statistic is a measure that describes a sample . It is usually denoted by Roman letters.

4. Examples Parameter Statistic

5. The population mean is the mean of the entire population. It is computed using the formula:

6. Example 1 . The numbers of workers in six outlets of a fast food restaurant are 12, 10, 11, 15, 12, and 14. Treating these data as a population, find the population mean .

7. Solution: The population mean is computed by summing up all the data and dividing the sum by the number of data N . All the data are used to find the population mean. Number x 1 12 2 10 3 11 4 15 5 12 6 14

8. The mean is affected by each datum. If the value of one datum is too small in comparison with the other data, it will lower the value of the mean. If a datum is too big compared with the other data, it will increase the value of the mean.

9. Example 2 . Assume that the last datum in Example 1 is 2 instead of 14. Number x 1 12 2 10 3 11 4 15 5 12 6 2 Solution:

10. Example 3 . Assume that the last datum in Example 1 is 30 instead of 14. Number x 1 12 2 10 3 11 4 15 5 12 6 30 Solution:

11. Population Variance and Population Standard Deviation

12. Variance and standard deviation are widely used measures of dispersion of data in research. The population variance is the sum of the squared deviations of each datum from the population mean divided by the population size N. The population standard deviation is the square root of the population variance .

13. Formula for the Population variance Formula for Population Standard Deviation

14. Example 4 . The following are the ages of 16 Teachers at USEP- Mintal . Compute the following: Population variance Population standard deviation 30 34 32 38 28 36 40 31 35 34 33 30 37 40 30 40

15. Solution: Step1. Compute the Step2. Compute Step3. Squared step 2 Then compute what is asked in a and b . 30 34 32 38 28 36 40 31 35 34 33 30 37 40 30 40

16. Teacher Age (x) 1 30 30-34.25= -4.25 18.0625 2 34 34-34.25= -0.25 0.0625 3 32 32-34.25= -2.25 5.0625 4 38 38-34.25= 3.75 14.0625 5 28 28-34.25= -6.25 39.0625 6 36 36-34.25= 1.75 3.0625 7 40 40-34.25=5.75 33.0625 8 31 31-34.25= -3.25 10.5625 9 35 35-34.25=0.75 0.5625 10 34 34-34.25= -0.25 0.0625 11 33 33-3425= -1.25 1.5625 12 30 30-34.25= -4.25 18.0625 13 37 37-34.25=2.75 7.5625 14 40 40-34.25=5.75 33.0625 15 30 30-34.25= -4.25 18.0625 16 40 40-34.25= 5.75 33.0625

17. a. b.

18. Sample Mean

19. The sample mean is the average of all the values randomly selected from the population. That is,

20. Example 5 . The following are the ages of 16 Teachers at USEP- Mintal . Assume that a researcher randomly selected 12 out of the 16 teachers. Assume that the encircled data below are those that were randomly selected. Compute the sample mean . 30 34 32 38 28 36 40 31 35 34 33 30 37 40 30 40 30 34 32 38 28 36 40 31 35 34 33 30 37 40 30 40

21. Teacher Population Age Sampled Age (x) 1 30 2 34 34 3 32 4 38 38 5 28 28 6 36 36 7 40 40 8 31 31 9 35 35 10 34 34 11 33 33 12 30 30 13 37 14 40 40 15 30 16 40 40 Solution:

22. Example 6 . Scores of the 15 Sampled Grade 9 students. Solution: Student Score 1 25 2 22 3 23 4 28 5 29 6 30 7 31 8 24 9 32 10 33 11 32 12 39 13 40 14 34 15 48

23. Sample Variance and Sample Standard Deviation

24. Sample Variance and Sample Standard Method 1 Method 2

25. Method 1 The sample variance is the sum of the squared deviation of each data from the sample mean divided by n-1. It uses the formula below. The sample standard deviation s is the square root of the sample variance given by the equation below.

26. Example 7 . Calculate the sample variance and sample standard deviation of the 12 randomly selected data in Example 5. Solution : Step1. Solve for the Step2. Solve for Step3. Solve for the square of step 2

27. Teacher Population Age Sampled Age (x) 1 30 2 34 34 -0.92 0.8464 3 32 4 38 38 3.08 9.4864 5 28 28 -6.92 47.8864 6 36 36 1.08 1.1664 7 40 40 5.08 25.8064 8 31 31 -3.92 15.3664 9 35 35 0.08 0.0064 10 34 34 -0.92 0.8464 11 33 33 -1.92 3.6864 12 30 30 -4.92 24.2064 13 37 14 40 40 5.08 25.8064 15 30 16 40 40 5.08 25.8064

28. Sample Variance Sample standard deviation

29. Example 8 . The following are the scores of 8 randomly selected students in Grade 10: C ompute the following: Sample mean Sample variance Sample standard deviation 7 8 12 15 10 11 9 and 14

30. Solution : Step1. Solve for the Step2. Solve for Step3. Solve for the square of step 2

31. Number x 1 7 -3.75 14.0625 2 8 -2.75 7.5625 3 12 1.25 1.5625 4 15 4.25 18.0625 5 10 -0.75 0.5625 6 11 0.25 0.0625 7 9 -1.75 3.0625 8 14 3.25 10.5625

32. Method 2 Sample Variance Sample Standard Deviation

33. Example 9 . Calculate the sample variance and the sample standard deviation of the data in Example 8 Method 2. Number x x 2 1 7 49 2 8 64 3 12 144 4 15 225 5 10 100 6 11 121 7 9 81 8 14 196 Sample Variance Sample standard deviation Method 2

34. Finding the Sample Mean from a Frequency Distribution

35. Sometimes, researchers organize and present their data using a frequency distribution table (FDT). Sample mean has two methods but still produce the same results.

36. Sample Mean Method 1 Method 2

37. Method 1 (Sample mean) Example 10 . Find the sample mean from the frequency distribution table below. Scores Frequency (f) 40-42 2 43-45 5 46-48 8 49-51 12 52-54 6 55-57 4 58-60 3

38. Solution: Scores f Class Mark (x) fx 40-42 2 41 82 43-45 5 44 220 46-48 8 47 376 49-51 12 50 600 52-54 6 53 318 55-57 4 56 224 58-60 3 59 177

39. Method 2 (sample Mean) Example 11 . Find the sample mean using Method 2 of example 10. Scores Frequency (f) 40-42 2 43-45 5 46-48 8 49-51 12 52-54 6 55-57 4 58-60 3

40. Solution: Scores Class Mark (x) d f fd 40-42 41 -3 2 -6 43-45 44 -2 5 -10 46-48 47 -1 8 -8 49-51 50 12 52-54 53 1 6 6 55-57 56 2 4 8 58-60 59 3 3 9

41. Method 2 (sample mean) Example 12 . Find the sample mean of the FDT using method 2. Ages of Teacher at USEP- Mintal Ages Frequency 21-25 4 26-30 7 31-35 6 36-40 10 41-45 14 46-50 9 51-55 8 56-60 2

42. Solution: Scores Class Mark (x) d f fd 21-25 23 -3 4 -12 26-30 28 -2 7 -14 31-35 33 -1 6 -6 36-40 38 10 41-45 43 1 14 14 46-50 48 2 9 18 51-55 53 3 8 24 56-60 58 4 2 8

43. Finding the Sample Variance & Sample Standard Deviation from a Frequency Distribution

44. Sample Variance & Sample Standard Deviation Method 1 Method 2

45. Method 1 Example 13. The scores in Chemistry of randomly selected Grade 10 students are shown below. Solve for Sample variance Standard deviation Scores Frequency 60-64 1 65-69 2 70-74 2 75-79 4 80-84 6 85-89 8 90-94 5 95-99 7

46. Solution: Scores f Class Mark (x) fx 60-64 1 62 62 -23 529 529 65-69 2 67 134 -18 324 648 70-74 2 72 144 -13 169 338 75-79 4 77 308 -8 64 256 80-84 6 82 492 -3 9 54 85-89 8 87 696 2 4 32 90-94 5 92 460 7 49 245 95-99 7 97 679 12 144 1008

47. Method 2 Example 14 . Solve for the Sample variance and Sample standard deviation of example 13 using Method 2. Scores f Class Mark (x) fx f x 2 60-64 1 62 62 3844 65-69 2 67 134 8978 70-74 2 72 144 10368 75-79 4 77 308 23716 80-84 6 82 492 40344 85-89 8 87 696 60552 90-94 5 92 460 42320 95-99 7 97 679 65863

48. Trivia The 12 th letter of the Greek alphabet was derived from the Egyptian hieroglyphic symbol for water (H 2 O). In the system of Greek numerals, it has a value of 40. In statistics, it represents the population mean.