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Dr Hidayathulla Shaikh. Objectives At the end of the lecture student should be able to – Discuss normal curve Classify parametric and non parametric tests.

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Presentation on theme: "Dr Hidayathulla Shaikh. Objectives At the end of the lecture student should be able to – Discuss normal curve Classify parametric and non parametric tests."— Presentation transcript:

1 Dr Hidayathulla Shaikh

2 Objectives At the end of the lecture student should be able to – Discuss normal curve Classify parametric and non parametric tests Mention assumptions in parametric tests. Explain some of parametric and non parametric test.

3 Introduction When we take different samples from same population the values may differ. The difference in values is called sampling variability. Hence while dealing with values from 2 or more samples, one is interested to know whether the difference in values between groups are due to sampling variation or not. So these tests of significance deals with techniques to know how far the difference in values of different samples is due to sampling variation.

4 Normal distribution When we take large number of observations of any variable such as height, blood pressure, pulse rate etc. And then we draw a frequency distribution curve or a graph, we get normal distribution or normal curve.

5 Characteristics of normal curve Bell shaped Smooth curve Perfectly symmetrical Does not touch the baseline Mean, median & mode = 0 Standard deviation =1. Total area = 1 Has Two inflections

6 Symmetrical and Skewed Distributions Mean Median Mode Median Mean SymmetricalSkewed When data is normally distributed When data is not normally distributed

7 Tests of significance Parametric tests: Used for quantitative & normally distributed data Nonparametric tests or distribution free tests: Used for qualitative data & data which are not normally distributed

8 Samples are randomly selected from larger populations. Samples are independent. Observations within each sample were obtained independently. Data from each population is normally distributed. Population variances in each groups have the same value. Assumptions in Parametric tests:

9 Parametric tests : Large sample tests: Z-test Small sample test : t-test a) Unpaired t-test b) Paired t-test F-test or ANOVA – Analysis Of Variance

10 Non parametric tests : Chi Square test Wilcoxon signed rank test Mann Whitney U-test Mc Nemar’s test Kruskall Wallis test

11 Z-test [Standard normal test]: Most frequently used in research studies. Used when the sample size is greater than 30 Given by formula Difference in means (X – X ̅̅ ) Z = Standard Deviation (SD) Examples 1) Plaque scores after using chlorhexidine & Listerine mouth washes in 100 patients. 2) DMFT scores after using NaF & SnF 2 mouth rinses in 100 patients.

12 t-test [student’s t test]:  Done w hen the sample size is smaller than 30.  The t test assesses whether the means of two groups are statistically different from each other.  t = ratio of observed differences between two Means to standard error (SE).  Its is two types a) Paired t test b) Unpaired t test.

13 Paired t-test :  Used when each individual gives a pair of observations.  Here the observations are from one sample only. t = X / S.E  Where X is Mean of different observation and SE is Standard Error  Standard Error – when we take random sample from population (n) & similar samples over and over again then we will find that every sample will have different Mean  And the (SD) Standard Deviation of Means is called Standard Error (SE) Examples - 1) Plaque scores before & after the use of chlorhexidine mouth wash. 1) DMFT scores before & after the use of fluoridated tooth pastes.

14 Unpaired t-test : It is done when data is obtained from two different or separate groups or samples drawn from two populations. X 1 – X 2 S.E E.g. 1) Radiation exposure from OPG & occlusal radiographs of 28 patients. 1) Lap top prices of two different companies for various models. t =

15 ANOVA (Analysis Of Variance) It is done in situations which involves collecting data on three or more groups of individuals. The objective is to determine whether any true differences in mean exist. Usually done in experimental study designs. For example 1) Various therapeutic approaches to a specific problem 2) Various dosage levels of particular drug under comparison. ANOVA is of different types a) ONE WAY ANALYSIS OF VARIANCE b) TWO WAY ANALYSIS OF VARIANCE c) MULTIPLE ANALYSIS OF VARIANCE [MANOVA]

16 Nonparametric tests : χ 2 test for qualitative data: Developed by Karl Pearson and pronounced as kye square test Used to test the association between two events [To test a given hypothesis] and when there are more than two groups to be compared. E.g. 1) Association between smoking & periodontitis. 2) Association between tongue thrusting & malocclusion.

17 χ 2 test How to calculate χ 2 test Step 1 – Test null hypothesis If we want to check the association between oral hygiene instructions and occurrence of new dental caries Then null hypothesis is there is no association between oral hygiene instruction and occurrence of new dental caries Step 2 - Observed frequency - Expected frequency is calculated χ 2 Step 3 - Applying the values to the formula χ 2 = ∑ (O - E) 2 E Step 4 – Finding degree of freedom (d.f)  It depends upon number of columns and rows in table  df = (column – 1)(row – 1) χ 2 Step 5 – Comparing the χ 2 values in probability table

18 Lets explain this with an example We want to check the association between oral hygiene instructions and occurrence of new dental caries /cavities. We conducted our study on 90 people and the data we have is

19 Step 1 testing null hypothesis - it says there is no association between oral hygiene instruction and occurrence of new dental caries Step 2 - Observed frequency - Expected frequency is calculated Row total x Column total Expected frequency [E] = Grand total χ 2 Step 3 Applying the values to the formula χ 2 = ∑ (O - E) 2 E We get 7.76+6.88+9.27+8.22 = 32.13

20 Step 4 – Finding degree of freedom (d.f)  It depends upon number of columns and rows in table  df = (column – 1)(row – 1) (2-1)(2-1) = 1 χ 2 Step 5 – Comparing the χ 2 values in probability table χ 2 In the probability table with the degree of freedom 1, χ 2 the value for probability of 0.05 is 3.84 Our value 32.13 is much higher than 3.84 Hence we can conclude null hypothesis is false and there is difference in caries occurrence in two groups with caries being lower in those who received oral hygiene instructions.

21 Values of chi-square are compared in Probability table with degrees of freedom df 0.10 0.05 0.02 0.01 0.001 1 2.706 3.841 5.024 6.635 10.828 2 4.605 5.991 7.378 9.210 13.816 3 6.251 7.815 9.348 11.345 16.266 4 7.779 9.488 11.143 13.277 18.467 5 9.236 11.070 12.833 15.086 20.515 6 10.645 12.592 14.449 16.812 22.458 7 12.017 14.067 16.013 18.475 24.322 8 13.362 15.507 17.535 20.090 26.125 9 14.684 16.919 19.023 21.666 27.877 10 15.987 18.307 20.483 23.209 29.588


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