Mean and Standard Deviation
Major topics covered in this chapter
Repeated measurements in analytical experiments in order to reveal the presence of random errors.
Table 2.1 Results of 50 determinations of nitrate ion concentration, in μg ml-1
The distribution of the results can most easily be appreciated by drawing a histogram
standard deviation gives a measure of the spread of a set of results about the mean value, it does not indicate the shape of the distribution
The results can be summarised in a frequency table. The distribution of the results appreciated by drawing a histogram. This shows that the distribution of the measurements is roughly symmetrical about the mean, with the measurements clustered towards the centre.
Although it cannot be proved that replicate values of a single analytical quantity are always normally distributed, there is considerable evidence that this assumption is generally at least approximately true. Moreover we shall see when we come to look at sample means that any departure of a population from normality is not usually important in the context of the statistical tests most frequently used.
Sample in its statistical sense of a group of objects selected from the population of all such objects, for example a sample of 50 measurements of nitrate ion concentration from the (infinite) population of all such possible measurements, or a sample of healthy human adults chosen from the whole population in order to measure the concentration of serum albumin for each one.
For 95% confidence limits, z 1.96 For 99% confidence limits, z 2.58 For 99.7% confidence limits, z 2.97
Example 2.6.1 Calculate the 95% and 99% confidence limits of the mean for the nitrate ion concentration measurements in Table 2.1. From previous examples we have found that $\mu$ = 0.500 ,$\sigma$ = 0.0165 and n = 50. Using Eq. (2.6.3) gives the 95% confidence limits as: $$x +- 1.96 * s/ = 0.500 + 1.96 * 0.0165>250 = 0.500 ; 0.005 mg ml-1$$ and the 99% confidence limits as: x ; 2.58s>2n = 0.500 ; 2.58 * 0.01651>250 = 0.500 ; 0.006 mg ml-1
In a new method for determining selenourea in water the following values were obtained for tap water samples spiked with 50 ng ml1 of selenourea: 50.4, 50.7, 49.1, 49.0, 51.1 ng ml1
The quantum yield of fluorescence, , of a material in solution is calculated from the expression: where the quantities involved are defined below, with an estimate of their relative standard deviations in brackets:
$$ f = {I_f}$$