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# Statistical analysis by iswar

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This ppt is prepared according to RGUHS syllabus, Modern Pharmaceutical analysis for 1st year M.pharm students

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### Statistical analysis by iswar

1. 1. Presented by:Iswar Hazarika1st yr M. Pharm PharmacologyThe Oxford college of Pharmacy
2. 2. 1. Statistics - Introduction2. Scope of statistic3. Normal Distribution4. Central Tendency1. Arithmetic mean2. Median3. Mode5. Dispersion1. Standard deviation (SD)6. Standard error of mean (SEM)7. Probability8. Test for significance1. Student ‘t’ test2. Chi square test
3. 3. “Statistics is a science which deals with thecollection, classification and tabulation ofnumerical facts as the basis for explanation,description and comparison of phenomena.”Here, the data are numbers which containinformation.
4. 4.  Industries Medical Science Agricultural biology Social Science Planning and economics Space research
5. 5.  When many independent random factorsact in an additive manner to createvariability, the data set follows a bellshaped distribution called as normaldistribution. Mathematicians De Moivre and Laplaceused this distribution in the 1700s. In the early 1800s, Germanmathematician and physicist Karl Gaussused it to analyze astronomical data, andknown as the Gaussian distribution.
6. 6.  When maximum frequency ofdistribution occurring at the centre of thecurve and the remaining evenlydistributed around it, it follows normaldistribution. Normal distribution is described by itsmean (µ) and standard deviation (σ).
7. 7.  Arithmetic mean Geometric mean Median Mode
8. 8.  It is defined as the sum of the all variatesof a variable divided by the total numberof item in a sample. It is expressed by the symbol Where, = Arithmetic meann = frequencyXi = all the varietes of Variable
9. 9.  Example:
10. 10.  It is defined as the nth root of the productof the n items in an ungrouped data. When percentage increase or decrease isexpressed over a period of time, the meanpercentage is find out by using geometricmean. If X1, X2, X3,…. Xn are the n variates of thevariable X then,Geometric Mean =
11. 11.  Example: Following administration of a drug in alaboratory mammal, the blood glucoselevel increased by 5% in the first hour, by8% in the second hour and 77% in thethird hour. What is the mean percentageincrease during the observation period? Here, we assume that the glucose level atthe beginning of every hour as 100mg%Then the level of blood sugar
12. 12.  At the end of 1 hour= 100+5 =105mg%At the end of 2 hour=100+8 = 108mg%At the end of 3 hour=100+77 = 177mg% So, geometric mean== 126.14 So the mean percentage increase= 126.14 – 100 = 26.14
13. 13.  It is the central value of all observationsarranged from the lowest to the highest. Example:(1) For Odd number of variatesWeight of frog in gram. n = 775, 66, 55, 68, 71, 78, 72. Data in ascending order of value:55, 66, 68, 71, 72, 75, 78. Here, Median is 71.
14. 14.  Example:(1) For Even number of variatesHeight of Students in cm, n = 8165, 175, 161, 155, 169, 171, 152, 166. Data in ascending order of value:152, 155, 161, 165, 166, 169, 171, 175. Here, Median is = 165.5
15. 15.  It is defined as the value which occursmost frequently in the sample. ExampleWeight of tablet in mg:52, 48, 50, 51, 50, 51, 50, 49. In the above data, 50 occurs 3 timesSo mode of above data = 50 mg
16. 16. Dispersion: Range Mean deviation Standard deviation Variance (σ2) Standard Error Mean (SEM)
17. 17.  It is defined as the square root of thearithmetic mean of the squareddeviations of the various items fromarithmetic mean. It is expressed as SD It is calculated by the following formula
18. 18.  Weight in gram of 6 Frogs.30, 90, 20, 10, 80, 70.For the above data: = 50.X weight in gram10 10 – 50 = - 40 +160020 20 – 50 = - 30 +90030 30 – 50 = - 20 +40070 70 – 50 = 20 +40080 80 – 50 = 30 +90090 90 – 50 = 40 +1600= 300 = 5800
19. 19.  SD ==== 34.05
20. 20. Text Book : Basic Concepts and Methodology forthe Health SciencesVariance: It measure dispersion relative to the scatter of the valuesabout there mean.a) Sample Variance ( ) : ,where is sample meanx2S1)(122nxxSnii
21. 21. Text Book : Basic Concepts and Methodology forthe Health Sciences b)Population Variance ( ) : where , is Population mean Example: slide no:20Varience=( )2= 11602NxNii122)(
22. 22.  In a small sample size the arithmeticmean would be an approximation of thetrue mean of the whole population, andtherefore subject to error. In such cases the error of the observedmean is calculated. The SE allows to find out the range inwhich the true mean would lie. It gives an estimate of the extent to whichthe mean will vary if the experiment isrepeated.
23. 23.  SE= SE of the previous example. SE== 13.05
24. 24.  The term probability means “chance” or“likelihood” of the occurrence of theevent. It is defined as the symbol ‘P’.Where, m= Number of favorable eventsN= Total number of events
25. 25. Test of SignificanceIn scientific research, a sample investigationproduces results which are helpful inmaking decisions about a populationWe are interested in comparing thecharacteristics of two or more groups.The two samples drawn from the samepopulation will show some differenceDifference can be controlled by “Test ofsignificance”
26. 26. Procedure for Testof Significance1. Laying down Hypothesis:a) Null hypothesis: Hypothesis which is to be actually tested foracceptance.b) Alternative hypothesis: Hypothesis which is complementary tothe to the null hypothesis.Eg. avg of gene length is 170 kbpHo:µ=170H1:µ=170i.e, µ>170 or µ<1702. Two types of error in testing of hypothesisa) Type I error: Rejection of null hypothesis which is trueb) Type II error: Acceptance of null hypothesis which is false
27. 27. 3. Level of significance Minimize Type I & II error Level of significance is denoted by α α is conventionally chosen as 0.05 (moderate precision) or 0.01(high precision) In most biostatistical test α is fixed at 5%, means probability ofaccepting a true hypothesis is 95%4. One & two tailed tests of hypothesis In a test the area under probability curve is divided into Acceptance region Critical/ rejection region
28. 28. Types of test ofSignificance Two types of test used in interpretation ofresults.(1)Parametric test:- It involves normal distribution. It includes: Student’s t-testAnalysis of variance(ANOVA)RegressionCorrelationZ- test
29. 29. Test of Significance(2)Non-Parametric test:-It involves when the sample data doesnot follow normal distribution. It includes: Chi-squared testWilcoxon Signed-rank testKruskal-Wallis test
30. 30. Student ‘t’ test:This test is applied to assess the statistical significanceof difference between two independently drawn samplemeans obtained from two series of data with anassumption that the two mean are from normaldistribution population, with no significant variationt= (difference of means of two samples)/(std error ofdifference)Standard error of difference(Sd) = √{(S12/n1)+(S22/n2)}t= |X1 – X2|/ √{(S12/n1)+(S22/n2)}Degrees of freedom = (n1+n2-2)
31. 31. Ex. Following data related to disintegration time(DT) ofChloroquine tablets using diluent, Lactose monohydrate(LM),dibasic calcium phosphate (DCP).Determine whether the twomeans are significantly different.Lactose Monohydrate DCPn 3o 35mean 32 38variance 9.62 14.23Null hypothesis: Ho: There is no significant differencebetween the mean DT in choroquine tablets betweenLM & DCPSd = √{(S12/n1)+(S22/n2)} =√(9.62/30)+(14.23/35)=√0.73 = 0.85
32. 32. Difference between mean = 38-32 = 6t= |X1 – X2|/ √{(S12/n1)+(S22/n2)}= |32-38|/ √{(9.62/30)+(14.23/35)}= 6/√o.73 = 7.06Degrees of freedom = (n1+n2-2)= (30+35-2)=63Conclusion:Calculated value of t(7.06)> tabulated value of t for 63(at1%=2.66)So the two mean are very much differentSo the null hypothesis is rejected at p=0.01The difference between the two sample means is a realdifference because the level of significance is very high
33. 33. Chi-square test:- In biological research apart from quantitative characters one hasto deal with qualitative data like flower color or seed color Results of breeding experiments and genetical analysis comesunder chi-square test The quantity x2 describes the magnitude of difference betweenthe observed & the expected frequencyx2 = ∑(fo - fe)2/fe fo – observed frequency fe – effective frequency
34. 34. Determination of value of x21. Calculate the expected frequency(fe)2. Find out the difference between the observedfrequency(fo) and expected frequency(fe)3. Square the value of (fo-fe) i.e (fo-fe)24. Divide each value of fe & obtain the total ∑(fo - fe)2/fevalue5. The calculated value of x2 is compared with the tablevalue for the given degrees of freedom(d.f)d.f= (r-1) (c-1)where, r- no. of rows in tablec- no. of columns in table
35. 35. Examples of x2 test In F2 generation, Mendel obtained 621 tall plants & 187dwarf plants out of the total of 808. test whether thesetwo types of plants are in accordance with theMendelian monohybrid ratio of 3:1 or they deviatefrom ratioSolution:Tall plants Dwarf plants TotalObserved frequency(fo) 621 187 808Expected frequency(fe) 606 202 808Deviation(fo-fe) 15 -15
36. 36.  Formula applied x2 = ∑(fo - fe)2/fe=(15)2/606+(-15)2/202= 225/606+ 225/202= 0.3713+ 1.1139= 1.4852Tabulated value is 3.84 at 5% level of probabilityfor d.f= 2-1 =1Therefore the difference between the observed & expectedfrequencies is not significantHence the null hypothesis is true
37. 37. Application of x2 test1. To test the goodness of fit2. To test the independence of attributes3. To test the homogeneity of independent estimates ofthe population varience4. To test the detection of linkage
38. 38. References Khan IA, Khatum A. Fundamentals of Biostatistics.3rd revised edition. Ukazz publication, Hyderabad Brahmankar DM, Jaiswal SB. Biopharmaceutics &Pharmacokinetics. Kulkarni SK. Textbook of Experimental pharmacology. Khan IA, Khatum A. Biostatistics in Pharmacy. 3rd edition.Ukazz publikation, Hydrabad Jeffery GH, Bassett J,Mendham J, Denney RC. Textbook ofquantitative chemical analysis. Fifth edition. Vogel’spublication.