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= = = (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)= = = (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= = = = 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= = = (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 = = – = 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= = = (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)= = = (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= = = = 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= = = (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 >6 Frequency, f xfxf Total 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 >6 Frequency, f xfxf Total = 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 >6 Frequency, f xfxf Total = 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 >6 Frequency, f xfxf Total = 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 >6 Frequency, f xfxf Total 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 >6 Frequency, f xfxf Total 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 >6 Frequency, f xfxf Total 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 >6 Frequency, f xfxf Total 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 >6 Frequency, f xfxf Total 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 >6 Frequency, f n = f xfxf x2fx2f Total

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 >6 Frequency, f xfxf x2fx2f Total S xx = = 610 – = 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 >6 Frequency, f xfxf x2fx2f Total = 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 >6 Frequency, f xfxf x2fx2f Total = 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 >6 Frequency, f xfxf x2fx2f Total = 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 >6 Frequency, f xfxf x2fx2f Total = 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

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google