Statistical distributions

Different Statistical Distributions in Biomedical Research

1. STATISTICAL DISTRIBUTIONS Dr. TANVEER REHMAN MODERATOR : Dr. VENKATACHALAM

2. OUTLINE Introduction Frequency distributions Central tendency Variability Z score Theoretical distributions Application 28-08-2017 2 Dr Tanveer Rehman PSM JIPMER

3. Figure 1. “Distribution” as a lens Introduction 28-08-2017 3 Dr Tanveer Rehman PSM JIPMER

4. Figure 2. Empirical vs Theoretical distribution Introduction 28-08-2017 4 Dr Tanveer Rehman PSM JIPMER

5. Figure 3. Examples of different shapes for distributions Frequency Distributions Symmetrical Skewed 28-08-2017 5 Dr Tanveer Rehman PSM JIPMER

6. Figure 4. Measures of central tendency for three symmetrical distributions: normal, bimodal, and rectangular. Central Tendency 28-08-2017 6 Dr Tanveer Rehman PSM JIPMER

7. When To Use The Median Or Mode Figure 5. Measures of central tendency for skewed distributions 28-08-2017 7 Dr Tanveer Rehman PSM JIPMER

8. Figure 6. Population distributions of adult heights and adult weights. Variability Distribution Meaning 28-08-2017 8 Dr Tanveer Rehman PSM JIPMER

9. Figure 6. the scores in the sample are less variable (spread out) than the scores in the population Measures of Variability Range Variance Standard deviation 28-08-2017 9 Dr Tanveer Rehman PSM JIPMER

10. Figure 7.Following a z- score transformation, the X -axis is relabelled in z -score units. Z or Standard Score Purpose Equation 28-08-2017 10 Dr Tanveer Rehman PSM JIPMER

11. Figure 8. The distribution of original score, a z -score, and a new, standardized score. Z-score Distributions Mean S.D. Shape Standardized distribution Disadvantage 28-08-2017 11 Dr Tanveer Rehman PSM JIPMER

12. Distributions 28-08-2017 12 Dr Tanveer Rehman PSM JIPMER

13. Distributions 28-08-2017 13 Dr Tanveer Rehman PSM JIPMER

14. Figure 9. The normal distribution following a z -score transformation. Normal Distribution Discoverer Shape Equation Proportions 28-08-2017 14 Dr Tanveer Rehman PSM JIPMER

15. Figure 10. A portion of the unit normal table. 28-08-2017 15 Dr Tanveer Rehman PSM JIPMER

16. Application Of Normal Distribution Q1. The population distribution of health expenditures is normal with a mean of Rs . 5000 and a standard deviation of Rs . 1000. What is the probability of randomly selecting an individual from this population whose health expenditures is greater than Rs . 7000? 28-08-2017 16 Dr Tanveer Rehman PSM JIPMER

17. Key to Q1 28-08-2017 17 Dr Tanveer Rehman PSM JIPMER

18. First, the probability question is translated into a proportion question: Out of all possible scores, what proportion is greater than 700? The mean is μ 500, so the score X 700 is to the right of the mean. Because we are interested in all scores greater than 700, we shade in the area to the right of 700. This area represents the proportion we are trying to determine. Identify the exact position of X > 700 by computing a z -score. For this example, That is, a score of X 700 is exactly 2 standard deviations above the mean and corresponds to a z -score of z 2.00.. The proportion we are trying to determine may now be expressed in terms of its z -score: p ( z > 2.00) ? all normal distributions regardless of the values for μ and , have 2.28% of the scores in the tail beyond z> 2.00. Thus, for the population of scores, p ( X > 700) p ( z = 2.00) 2.28% 28-08-2017 18 Dr Tanveer Rehman PSM JIPMER

19. Figure 11 depicts the typical distribution of sample means Central Limit Theorem Marquis de Laplace The distribution of sample means Any of two conditions to satisfy 28-08-2017 19 Dr Tanveer Rehman PSM JIPMER

20. 28-08-2017 20 Dr Tanveer Rehman PSM JIPMER

21. Figure 12 shows that the size of the standard error decreases as the sample size increases. Standard Error of M Serves the same two purposes The overall mean is equal to μ SD is the “starting point” 28-08-2017 21 Dr Tanveer Rehman PSM JIPMER

22. Application Of Normal Distribution Q2. A positively skewed distribution has μ = 60 and σ = 8. What is the probability of obtaining a sample mean greater than M = 62 for a sample of n = 4 scores? What is the probability of obtaining a sample mean greater than M = 62 for a sample of n = 64 scores? 28-08-2017 22 Dr Tanveer Rehman PSM JIPMER

23. Key to Q2. a. The distribution of sample means does not satisfy either of the criteria for being normal. Therefore, you cannot use the unit normal table, and it is impossible to find the probability. b. With n = 64, the distribution of sample means is nearly normal. The standard error is 8/ square root 64 = 1, the z-score is = 2.00, and the probability is 0.0228 28-08-2017 23 Dr Tanveer Rehman PSM JIPMER

24. Figure 13. The critical region (very unlikely outcomes) for α = 0.05. Hypothesis testing State the hypothesis Set the criteria for a decision Collect data and compute sample statistics Make a decision 28-08-2017 24 Dr Tanveer Rehman PSM JIPMER

25. Application Of Normal Distribution Q3. A health care project begins with a known population of all PHC staff—in this case, scores on a standardized test that are normally distributed with µ = 65 and σ = 15 . The Project officer suspects that special training in reading skills will produce a change in the scores for the staff. Because it is not feasible to administer the treatment (the special training) to everyone in the population, a sample of n = 25 individuals is selected, and the treatment is given to this sample. Following treatment, the average score for this sample is M = 70 . Is there evidence that the training has an effect on test scores? 28-08-2017 25 Dr Tanveer Rehman PSM JIPMER

26. Key to Q3 State the hypothesis and select an alpha level. The null hypothesis states that the special training has no effect. In symbols, H0: µ=65 (After special training, the mean is still 65.) The alternative hypothesis states that the treatment does have an effect. H1: µ is not equal to 65 (After training, the mean is different from 65.) At this time you also select the alpha level. For this demonstration, we will use α =0.05. Thus, there is a 5% risk of committing a Type I error if we reject H0. Locate the critical region. With alpha .05, the critical region consists of sample means that correspond to z-scores beyond the critical boundaries of z=1.96. Obtain the sample data, and compute the test statistic. For this example, the distribution of sample means, according to the null hypothesis, is normal with an expected value of mean 65 and a standard error of 3. In this distribution, our sample mean of M=70 corresponds to a z-score of 1.67 Make a decision about H0, and state the conclusion. The z-score we obtained is not in the critical region. This indicates that our sample mean of M=70 is not an extreme or unusual value to be obtained from a population with mean = 65. Therefore, our statistical decision is to fail to reject H0. Our conclusion for the study is that the data do not provide sufficient evidence that the special training changes test scores. 28-08-2017 26 Dr Tanveer Rehman PSM JIPMER

27. Figure 14 shows t distributions have more variability, indicated by the flatter and more spread-out shape. t Distribution W. S. Gossett Use t statistic formula Every sample like z-score Approximation depends on df Shape 28-08-2017 27 Dr Tanveer Rehman PSM JIPMER

28. Application Of t Distribution Q4. A research project by Central Government would like to determine whether there is a relationship between depression and aging. It is known that the general population averages μ = 40 on a standardized depression test. The project officer obtains a sample of n = 9 individuals who are all more than 70 years old. The depression scores for this sample are as follows: 37, 50, 43, 41, 39, 45, 49, 44, 48. On the basis of this sample, is depression for elderly people significantly different from depression in the general population? 28-08-2017 28 Dr Tanveer Rehman PSM JIPMER

29. Key to Q4 28-08-2017 29 Dr Tanveer Rehman PSM JIPMER

30. H0: μ = 40. With df = 8 the critical values are t = +-2.306. For these data, M = 44, SS = 162, s2 = 20.25, the standard error is 1.50, and t = 2.67. Reject H0 and conclude that depression for the elderly is significantly different from depression for the general population. 28-08-2017 30 Dr Tanveer Rehman PSM JIPMER

31. Figure 15 The binomial distribution is always a discrete histogram, and the normal distribution is a continuous, smooth curve Binomial distributions Two categories Criteria for normality Mean and SD Z score 28-08-2017 31 Dr Tanveer Rehman PSM JIPMER

32. Application Of Binomial Distribution Q5. A national health organization predicts that 50% of residents of a particular area will get the flu this season. If a sample of 40 residents is selected from the population, what is the probability that at least 26 of the people will be diagnosed with the flu? 28-08-2017 32 Dr Tanveer Rehman PSM JIPMER

33. Key to Q5 28-08-2017 33 Dr Tanveer Rehman PSM JIPMER

34. Probability of flu = 0.50=p q= 1- p= 0.5; n= 40 pn = 20 & qn =20, both above 10; so it’s a binomial distribution Mean = pn = 20; SD= square root of npq = square root of 10 = 3.16 We are looking for 26 or more to get flu; so in a continuous data – real lower limit is 25.5 Z score is 25.5-20 divided by 3.16= 1.74 In unit normal table, it comes to 4.09% 28-08-2017 34 Dr Tanveer Rehman PSM JIPMER

35. Figure 16 Chi-square distributions are positively skewed. The critical region is placed in the extreme tail, which reflects large chi-square values. Chi-square Distribution Formula Chi-square value Characteristics 28-08-2017 35 Dr Tanveer Rehman PSM JIPMER

36. Application of Chi-square Distribution Q6. Consider the following results from an influenza vaccine trial carried out during an epidemic. Of 460 adults who took part, 240 received influenza vaccination and 220 placebo vaccination. Overall 100 people contracted influenza, of whom 20 were in the vaccine group and 80 in the placebo group. We now wish to assess the strength of the evidence that vaccination affected the probability of contracting influenza. 28-08-2017 36 Dr Tanveer Rehman PSM JIPMER

37. Key to Q6 28-08-2017 37 Dr Tanveer Rehman PSM JIPMER

38. 28-08-2017 38 Dr Tanveer Rehman PSM JIPMER

39. Poisson Distribution The number of discrete occurrences of an event during a period of time Independent and random The occurrences in each interval can range from zero to infinity Depends upon just one parameter: the mean number of occurrences in periods of the same length 28-08-2017 39 P ( x ) = e -µ µ x x ! Dr Tanveer Rehman PSM JIPMER

40. Figure 17. The distribution is very skewed for small means, when there is a sizeable probability that zero events will be observed. It is symmetrical for large means and is adequately approximated by the normal distribution for values of λ =10 or more. 28-08-2017 40 Dr Tanveer Rehman PSM JIPMER

41. Application of Poisson distribution Q7. A district health authority which plans to close the smaller of two maternity units is assessing the extra demand this will place on the remaining unit. At present the larger unit averages 4.2 admissions per day and can cope with a maximum of 10 admissions per day. This results in the unit’s capacity being exceeded only on about one day per year. After the closure of the smaller unit the average number of admissions is expected to increase to 6.1 per day . Estimate the proportion of days on which the unit’s capacity is then likely to be exceeded. 28-08-2017 41 Dr Tanveer Rehman PSM JIPMER

42. Key to Q7 28-08-2017 42 Dr Tanveer Rehman PSM JIPMER

43. Figure 18 shows of all the values in the distribution, only 5% are larger than F= 3.88, and only 1% are larger than F = 6.93. F - Distribution ANOVA Advantage over t tests F-ratio Characteristics Exact shape depends on df 28-08-2017 43 Dr Tanveer Rehman PSM JIPMER

44. Table 1. A portion of the F distribution table. 28-08-2017 44 Dr Tanveer Rehman PSM JIPMER