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Biology_A-level_Cie

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  1. 1-1-the-microscope-in-cell-studies
    5 主题
  2. 1-2-cells-as-the-basic-units-of-living-organisms
    5 主题
  3. 2-1-testing-for-biological-molecules
    3 主题
  4. 2-2-carbohydrates-and-lipids
    8 主题
  5. 2-3-proteins
    6 主题
  6. 2-4-water
    2 主题
  7. 3-1-mode-of-action-of-enzymes
    5 主题
  8. 3-2-factors-that-affect-enzyme-action
    8 主题
  9. 4-1-fluid-mosaic-membranes
    4 主题
  10. 4-2-movement-into-and-out-of-cells
    12 主题
  11. 5-1-replication-and-division-of-nuclei-and-cells
    6 主题
  12. 5-2-chromosome-behaviour-in-mitosis
    2 主题
  13. 6-1-structure-of-nucleic-acids-and-replication-of-dna
    4 主题
  14. 6-2-protein-synthesis
    5 主题
  15. 7-1-structure-of-transport-tissues
    4 主题
  16. 7-2-transport-mechanisms
    7 主题
  17. 8-1-the-circulatory-system
    7 主题
  18. 8-2-transport-of-oxygen-and-carbon-dioxide
    5 主题
  19. 8-3-the-heart
    4 主题
  20. 9-1-the-gas-exchange-system
    6 主题
  21. 10-1-infectious-diseases
    3 主题
  22. 10-2-antibiotics
    3 主题
  23. 11-1-the-immune-system
    4 主题
  24. 11-2-antibodies-and-vaccination
    6 主题
  25. 12-1-energy
    5 主题
  26. 12-2-respiration
    11 主题
  27. 13-1-photosynthesis-as-an-energy-transfer-process
    8 主题
  28. 13-2-investigation-of-limiting-factors
    2 主题
  29. 14-1-homeostasis-in-mammals
    8 主题
  30. 14-2-homeostasis-in-plants
    3 主题
  31. 15-1-control-and-coordination-in-mammals
    12 主题
  32. 15-2-control-and-coordination-in-plants
    3 主题
  33. 16-1-passage-of-information-from-parents-to-offspring
    5 主题
  34. 16-2-the-roles-of-genes-in-determining-the-phenotype
    7 主题
  35. 16-3-gene-control
    3 主题
  36. 17-1-variation
    4 主题
  37. 17-2-natural-and-artificial-selection
    7 主题
  38. 17-3-evolution
    2 主题
  39. 18-1-classification
    5 主题
  40. 18-2-biodiversity
    7 主题
  41. 18-3-conservation
    6 主题
  42. 19-1-principles-of-genetic-technology
    11 主题
  43. 19-2-genetic-technology-applied-to-medicine
    4 主题
  44. 19-3-genetically-modified-organisms-in-agriculture
    2 主题
  45. 1-1-the-microscope-in-cell-studies
  46. 1-2-cells-as-the-basic-units-of-living-organisms
  47. 2-1-testing-for-biological-molecules
  48. 2-2-carbohydrates-and-lipids
  49. 2-3-proteins
  50. 2-4-water
  51. 3-1-mode-of-action-of-enzymes
  52. 3-2-factors-that-affect-enzyme-action
  53. 4-1-fluid-mosaic-membranes
  54. 4-2-movement-into-and-out-of-cells
  55. 5-1-replication-and-division-of-nuclei-and-cells
  56. 5-2-chromosome-behaviour-in-mitosis
  57. 6-1-structure-of-nucleic-acids-and-replication-of-dna
  58. 6-2-protein-synthesis
  59. 7-1-structure-of-transport-tissues
  60. 7-2-transport-mechanisms
  61. 8-1-the-circulatory-system
  62. 8-2-transport-of-oxygen-and-carbon-dioxide
  63. 8-3-the-heart
  64. 9-1-the-gas-exchange-system
  65. 10-1-infectious-diseases
  66. 10-2-antibiotics
  67. 11-1-the-immune-system
  68. 11-2-antibodies-and-vaccination
  1. Biology_A-level_Cie
  2. 17-1-variation
  3. variation-t-test-worked-example
课 36, 主题 4
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variation-t-test-worked-example

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Variation: t-test worked example

Worked Example

The ear lengths of two populations of rabbits were measured.

Ear lengths of population A (mm):

62, 60, 59, 61, 60, 58, 59, 60, 57, 56, 59, 58, 60, 59, 57

Ear lengths of population B (mm):

58, 59, 57, 59, 59, 57, 55, 60, 57, 58, 59, 58, 57, 58, 59

Use the t-test to determine whether there is a significant difference in ear length between the two populations.

Solution

  • Null hypothesis: There is no significant difference between the ear lengths of the rabbits in populations A and B

  • Sample sizes:

    • Population A: n1 = 15

    • Population B: n2 = 15

    Step 1: Calculate the mean for each data set:

Mean for population A x̅1 = 885 ÷ 15 = 59 mm

Mean for population B x̅2 = 870 ÷ 15 = 58 mm

Step 2: Calculate the standard deviation (s) for each set of data

Population A

Population B

Difference between value and mean

(x – x̄)

Difference between value and mean

(x – x̄)2

Difference between value and mean

(x – x̄)

Difference between value and mean

(x – x̄)2

62 – 59 = 3

9

58 – 58 = 0

0

60 – 59 = 1

1

59 – 58 = 1

1

59 – 59 = 0

0

57 – 58 = -1

1

61 – 59 = 2

4

59 – 58 = 1

1

60 – 59 = 1

1

59 – 58 = 1

1

58 – 59 = -1

1

57 – 58 = -1

1

59 – 59 = 0

0

55 – 58 = -3

9

60 – 59 = 1

1

60 – 58 = 2

4

57 – 59 = -2

4

57 – 58 = -1

1

56 – 59 = -3

9

58 – 58 = 0

0

59 – 59 = 0

0

59 – 58 = 1

1

58 – 59 = -1

1

58 – 58 = 0

0

60 – 59 = 1

1

57 – 58 = -1

1

59 – 59 = 0

0

58 – 58 = 0

0

57 – 59 = -2

4

59 – 58 = 1

1

Total ∑(x – x̄)2

36

Total ∑(x – x̄)2

22

To find the standard deviations divide the sum of each square by n – 1 for each data set, and take the square root of each value

Population A (n1 = 15)

Population B (n2 = 15)

n1 – 1 = 14

n2 – 1 = 14

∑(x – x̄)2 = 36 

so 36 ÷ 14 = 2.57

∑(x – x̄)2 = 22 

so 36 ÷ 22 = 1.57

square root of 2.57 end root equals 1.60

square root of 1.57 end root equals 1.25

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