Download presentation

Presentation is loading. Please wait.

Published byJackson Thomas Modified over 3 years ago

1
Mean square deviation Root mean square deviation Variance Standard deviation

2
Choose one of the following: Example 1 : Method ARaw data using defintion Example 1 : Method A Example 1 : Method BRaw data using alternative Example 1 : Method B Example 2 : Method AFrequency distribution using defintion Example 2 : Method A Example 2 : Method BFrequency distribution using alternative Example 2 : Method B Summary of formulae Notes on this presentation

3
Example 1 : Method A Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Tabulate the values (x) and find the mean: The sample mean = = 5.9 x

4
Example 1 : Method A Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Find the deviation from the mean, for each x value, x – : = 5.9

5
Example 1 : Method A Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Square the deviations from the mean and find the sum of squares S xx : S xx = = 5.9 msd & rmsd Variance & standard deviation

6
Example 1 : Method A Mean square deviation Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n : Mean square deviation= = = 0.0844 (to 3 s.f.) = 5.9

7
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n and take the square root : Root mean square deviation (rmsd)= = = 0.291 (to 3 s.f.) RETURN = 5.9 Example 1 : Method A Root mean square deviation

8
Example 1 : Method A Variance Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n – 1 : Variance= = = 0.095 = 5.9

9
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n – 1 and take the square root : Standard deviation s= = = 0.308 (to 3 s.f.) RETURN = 5.9 Example 1 : Method A Standard deviation

10
Example 1 : Method B Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Tabulate the values (x) and find the mean: The sample mean = = 5.9 x

11
Example 1 : Method B Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Tabulate the squares of the values (x 2 ) and find the total: x 2

12
S xx = = 314.05 – 9 5.9 2 = 0.76 Example 1 : Method B Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Find the sum of squares S xx by subtracting from x 2 : = 5.9 x 2 msd & rmsd Variance & standard deviation

13
Example 1 : Method B Mean square deviation Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n : Mean square deviation= = = 0.0844 (to 3 s.f.) = 5.9 x 2

14
Example 1 : Method B Root mean square deviation Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n and take the square root : Root mean square deviation (rmsd)= = = 0.291 (to 3 s.f.) = 5.9 x 2 RETURN

15
Example 1 : Method B Variance Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n – 1 : Variance= = = 0.095 = 5.9 x 2

16
Example 1 : Method B Standard deviation Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows: 6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9 Divide the sum of squares S xx by n – 1 and take the square root : Standard deviation s= = = 0.308 (to 3 s.f.) = 5.9 x 2 RETURN

17
Example 2 : Method A The number of children per family, x, for a random selection of 100 families, is given by the following table: The sample mean = = 2.1 xf First tabulate the xf values and find their mean: No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090 2124 323570 431957 54832 65315 76212 Total 21100210 n = f

18
Example 2 : Method A The number of children per family, x, for a random selection of 100 families, is given by the following table: Find the deviation from the mean, for each x value, : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.1 2124 -1.1 323570-0.1 4319570.9 548321.9 653152.9 762123.9 Total 21100210 = 2.1

19
Example 2 : Method A The number of children per family, x, for a random selection of 100 families, is given by the following table: Square the deviations from the mean : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.14.41 2124 -1.11.21 323570-0.10.01 4319570.90.81 548321.93.61 653152.98.41 762123.915.21 Total 21100210 = 2.1

20
Example 2 : Method A The number of children per family, x, for a random selection of 100 families, is given by the following table: Find the sum of squares S xx, the sum of : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.14.4139.69 2124 -1.11.2129.04 323570-0.10.010.35 4319570.90.8115.39 548321.93.6128.88 653152.98.4125.23 762123.915.2130.42 Total 21100210 169 = 2.1 msd & rmsd Variance & standard deviation

21
Example 2 : Method A Mean square deviation The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.14.4139.69 2124 -1.11.2129.04 323570-0.10.010.35 4319570.90.8115.39 548321.93.6128.88 653152.98.4125.23 762123.915.2130.42 Total 21100210 169 Mean square deviation= = = 1.69

22
Example 2 : Method A Root mean square deviation The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n and take the square root : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.14.4139.69 2124 -1.11.2129.04 323570-0.10.010.35 4319570.90.8115.39 548321.93.6128.88 653152.98.4125.23 762123.915.2130.42 Total 21100210 169 Root mean square deviation= = = 1.3 RETURN

23
Example 2 : Method A Variance The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n – 1 : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.14.4139.69 2124 -1.11.2129.04 323570-0.10.010.35 4319570.90.8115.39 548321.93.6128.88 653152.98.4125.23 762123.915.2130.42 Total 21100210 169 Variance= = = 1.71 (to 3 s.f.)

24
Example 2 : Method A Standard deviation The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n – 1 and take the square root : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090-2.14.4139.69 2124 -1.11.2129.04 323570-0.10.010.35 4319570.90.8115.39 548321.93.6128.88 653152.98.4125.23 762123.915.2130.42 Total 21100210 169 Standard deviation s= = = 1.31 (to 3 s.f.) RETURN

25
Example 2 : Method B The number of children per family, x, for a random selection of 100 families, is given by the following table: The sample mean = = 2.1 xf Tabulate the xf values and find their mean: No. of children, x 0123456>6 Frequency, f 92435198320 xfxf 1090 2124 323570 431957 54832 65315 76212 Total 21100210 n = f

26
Example 2 : Method B The number of children per family, x, for a random selection of 100 families, is given by the following table: x 2 f Tabulate the squares of the values (x 2 f) and find the total: No. of children, x 0123456>6 Frequency, f 92435198320 n = f xfxf x2fx2f 10900 2124 323570140 431957171 54832128 6531575 7621272 Total 21100210610

27
Example 2 : Method B The number of children per family, x, for a random selection of 100 families, is given by the following table: Find the sum of squares by subtracting from x 2 f : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf x2fx2f 10900 2124 323570140 431957171 54832128 6531575 7621272 Total 21100210610 S xx = = 610 – 100 2.1 2 = 169 = 2.1 n = f x 2 f msd & rmsd Variance & standard deviation

28
Example 2 : Method B Mean square deviation The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf x2fx2f 10900 2124 323570140 431957171 54832128 6531575 7621272 Total 21100210610 = 2.1 n = f Mean square deviation= = = 1.69 x 2 f

29
Example 2 : Method B Root mean square deviation The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n and take the square root : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf x2fx2f 10900 2124 323570140 431957171 54832128 6531575 7621272 Total 21100210610 = 2.1 n = f Root mean square deviation= = = 1.3 x 2 f RETURN

30
Example 2 : Method B Variance The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n – 1 : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf x2fx2f 10900 2124 323570140 431957171 54832128 6531575 7621272 Total 21100210610 = 2.1 n = f Variance= = = 1.71 (to 3 s.f.) x 2 f

31
Example 2 : Method B Standard deviation The number of children per family, x, for a random selection of 100 families, is given by the following table: Divide the sum of squares S xx by n – 1 and take the square root : No. of children, x 0123456>6 Frequency, f 92435198320 xfxf x2fx2f 10900 2124 323570140 431957171 54832128 6531575 7621272 Total 21100210610 = 2.1 n = f Standard deviation s= = = 1.31 (to 3 s.f.) x 2 f RETURN

32
S xx Definition Alternative version Raw data Frequency distribution Various forms of sum of squares S xx Variance Standard deviation s Mean squareRoot mean square deviationdeviation (rmsd) RETURN

33
Notes on using the presentation The presentation covers calculations using Raw data (Example 1) or a Frequency distribution (Example 2). The sum of squares S xx is evaluated using the Definition formula (Method A) or the Alternative formula (Method B). For each example and each method the sum of squares S xx is used to calculate the mean square deviation and root mean square deviation or the variance and standard deviation Use the links in the presentation to choose the appropriate example and method, together with the desired calculations. RETURN

Similar presentations

Presentation is loading. Please wait....

OK

Before Between After.

Before Between After.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on total internal reflection diagrams Apple vs samsung market share ppt online Ppt on the digestive system of human Ppt on anticancer therapy Ppt on nutrition in animals class 7 Ppt on save tigers in india download Ppt on service oriented architecture ppt Ppt on statistics in maths Ppt on kinetic energy and potential energy Ppt on principles of object-oriented programming encapsulation