1. 1 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x Example 1 : Raw Data Method A Mean

2. 2 Example 1 : Raw Data 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

3. 3 Example 1 : Raw Data 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

4. 4 Example 1 : Raw Data 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

5. 5 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 Example 1 : Raw Data Method A Standard deviation