1. Lecture 4 Basic Statistics Dr. A.K.M. Shafiqul Islam School of Bioprocess Engineering University Malaysia Perlis 21.09.2011

2. PROPAGATION OF ERRORS Addition and subtraction: — errors add as the square root of the squares of the absolute values of the uncertainties 3.067 ± 0.040 4.02 ± 0.01 - 2.9846 ± 0.3308 4.1024 ± 0.333

3. PROPAGATION OF ERRORS Multiplication and division: ─The relative uncertainties are additive, and the most probable error is represented by the square root of the sum of the relative variances. ─ i.e., the relative variance of the answer is the sum of the individual relative variances.

4. PROPAGATION OF ERRORS Multiplication and division: — relative errors add as the square root of the squares of the relative uncertainties

5. Find out the relative uncertainty of the calculation PROPAGATION OF ERRORS

6. =356.0 = ± 0.002566 = ± 356.0 X ± 2.6 X 10 3 = ± 2.6 X 10 3 = ± 0.93 So the answer is 356.0 ± 0.9.

7. A quality control chart is a time plot of a measured quantity that is assumed to be constant (with a Gaussian distribution) for the purpose of ascertaining that the measurement remains within a statistically acceptable range. The control chart consists of a central line representing the known or assumed value of the control and either one or two pairs of limit lines, the inner and outer control limits. CONTROL CHARTS

8. Calculation of the standard deviation for a set of data provides an indication of the precision inherent in a particular procedure or analysis. But unless there is a large number of data, it does not by itself give any information about how close the experimentally determined mean x might be to the true mean value . Statistical theory allows us to estimate the range within which the true value might fall, within a given probability, defined by the experimental mean and the standard deviation. CONFIDENCE LIMIT

9. This range is called the confidence interval, and the limits of this range are called the confidence limit. The likelihood that the true value falls within the range is called the probability, or confidence level, expressed as a percent. The confidence limit is given by where t is a statistical factor that depends on the number of degrees of freedom and the confidence level desired. CONFIDENCE LIMIT

11. A soda ash sample is analyzed in the analytical chemistry laboratory by titration with standard hydrochloric acid. The analysis is performed in triplicate with the following results: 93.50, 93.58, and 93.43% Na 2 CO 3. Within what range are you 95% confident that the true value lies? So you are 95% confident that, in the absence of a determinate error, the true value falls within 93.31 to 93.69%. CONFIDENCE LIMIT

13. Why we do testing to our experimental data? 1.to compare data among friends with the intention of gaining some confidence that the data observed could be accepted or rejected. 2.to decide whether there is a difference between the results obtained using two different methods. All these can be confirmed by doing some significant tests. SIGNIFICANT TESTING

14. Is used to determine if 2 sets of measurements are statistically different. The comparison between 2 set of measurements which made by 2 method, one will be test method and the other one will be accepted method. By using T test, we can determine whether these two method are significant difference. T TEST

15. This is a test designed to indicate whether there is a significant difference between two methods based on their standard deviations. F is defined in terms of the variances of the two methods, where the variance is the square of the standard deviation: F TEST

16. The F test evaluates differences between the spread of results, while the t test looks at differences between means. F TEST

17. You are developing a new colorimetric procedure for determining the glucose content of blood serum. You have chosen the standard Folin-Wu procedure with which to compare your results. From the following two sets of replicate analyses on the same sample, determine whether the variance of your method differs significantly from that of the standard method. F TEST

18. Solution The F value is > 1. F TEST = 1.73

19. F TEST

20. The tabulated F value for v 1 = 6 and V 2 = 5 is 4.95. Since the calculated value is less than this, we conclude that there is no significant difference in the precision of the two methods, i.e., the standard deviations are from random error alone and don't depend on the sample. F TEST