1. Community medicine \3rd stage Practical November 30th ,2016
By Dr. Muslim N.Saeed
Community medicine \3rd stage
Practical November 30th ,2016

2. Mathematical ((Numerical)) representation of data

3. 1- Measurements of location ((Measurement of central tendency)).
a) Arithmetic Mean (Average)  The most commonly used measure of central tendency. This is the sum (Σ) of all observations divided by the number of observation. We have:  Population Mean (μ) = Σx / N (N= No. of population)  Sample Mean (x) = Σx /n (n= No. of sample)

4. Advantages.  Uniqueness, only Arithmetic Mean for a giving set of data.  Simplicity, easy to compute & understand.  Take in consideration all values, not ignore any one (takes into account the magnitude of each and every observation in the data).

5. EX: Suppose data are: 90, 80, 95, 85, and 65 (n=5) Sample mean x = = 83 The mean can be thought of as a “balancing point”, “center of gravity”

7. Disadvantages.  Affected by the extreme values, which in some cases so distorted that it become undesirable as a measure of central tendency. As an EX. Of how the extreme values may affect the mean: Suppose five physicians receive the following charges, $95, $95, $85, and $280. The charge for the five physicians is found to be $118, a value that is not very representative to their charges.

8. The mean is the most commonly used measure of central tendency
The mean is the most commonly used measure of central tendency. However, if the data has extreme values (very high or very low values produce skewness in the data), such values can pull the mean towards one side and make it less usefull. The median is less influenced by extreme values and this makes it useful in certain epidemiological studies where skewed distributions are studied.

10. Ex: Measurement results from the Forced Expiratory Volume (FEV) test in one second (in litres): 2.30, 2.15,3.50,3.60,3.50, 2.82, 4.05, 2.25,2.68, 3, 2.68, 2.85, 3.38, The mean is calculated as: ∑ X\n = 2.93 liters .

11. b) Median (50th percentile, 5th deciles)  It is the value that divides the data into two equal parts (50% greater and 50% lesser than it)) giving that the values are arranged in o ordered of magnitude.  If the number of values is odd, then the position of the median = n+1/ 2. If the number of values is even, then the median is taken to be the mean of (n/2) and (n/2 +1) observations.

12.  Advantages. 1) Uniqueness, only 1 median for a giving data
 Advantages. 1) Uniqueness, only 1 median for a giving data. 2) Simplicity, easy to calculate. 3) Not affected by the extreme values.  Disadvantages. Didn't take in consideration all values, but only the median one.

13. EX: give the median of these values: 1st set of data: 5, 15, -7, 20, 25, 3, -1, 0 & -3. 2nd set of data: 7, 9, 16, -5, -9, 3, -4, & 6. Solution: (1st set )ordered array: -7, -3, -1, 0, 3, 5, 15, 20, 25 2nd set of data : -9, -5, -4, 3, 6,7,9,16 Median position= (9+1)/2=5, i.e. the 5th reading, median = 3 in 1st set 2nd set median order= (8/2) & (8/2) + 1 =4 & 5 so position 3 & 6, (3+6)/2= 4.5 in 2nd set. The Median is a Better Description (than the Mean) of the Majority When the distribution is skewed (not affected by the extreme values).

14. Ex: Data are: 14, 89, 93, 95, 96 • Skewness is reflected in the outlying low value of 14 • The sample mean is 77.4 • The median is 93  When do we use mean & median? If there are extreme values in a set of data we use the median, so as not to be misled by extreme value when we use the mean.

16. c) Mode.  The value which occurs most frequently
c) Mode.  The value which occurs most frequently. Mode is not unique  If all values are different → No mode.  If one value which occur most frequently → Uni-modal.  If two values which occur most frequently → Bi-modal.  If 3 values which occur most frequently → Trimodal…. etc.

17.  Advantages. 1) Simplicity, easy to calculate
 Advantages. 1) Simplicity, easy to calculate. 2) Unlike mean & median, the mode can be used for qualitative data (e.g., modal diagnosis as HT, DM, IHD …..etc.).

18. Important to know:
If the data is homogenous, we use the mean, but if it is not, we use the median.  For uni-modal frequency distribution (normal distribution) which is symmetric: The mean = The median= The mode.  Knowing the values of the mean and median of a distribution can provide some information on the skewness of a distribution.

19. If the mean is greater than the median then: the distribution is positively skewed; the long tail is on the high (usually right) side of the graph. If the mean is equal to the median then: the distribution is symmetric. If the mean is less than the median then: the distribution is negatively skewed; the long tail is on the low (usually left) side of the graph.

20. Measures of Central Tendency; The Weighted Mean:
The individual values in the set are weighted by their respective frequencies. _ ∑ (n . X) X w = N

21. Examples (24+25+29+29+30+25+131)/6 =44.7 days
Although the mean is often an excellent summary measure of a set of data, the data must be approximately normally distributed, because the mean is quite sensitive to extreme values that skew a distribution.
Example
In an outbreak of hepatitis A, 6 persons became ill with clinical symptoms. The incubation periods for the affected persons (xi) were 29, 31, 24, 29,30, and 25 days.
_ ∑ X
X = = 168/6 = 28 days n
If the largest value of the six listed incubation periods were 131 instead of 31, the mean would change from 28.0 to ?
( )/6 =44.7 days

22. Group (1) - Exercise
Consider the data taken from a study that examines the response to ozone and sulfur dioxide among adolescents suffering from asthma. The following are the measurements of forced expiratory volume (liters) for 10 subjects: 3.5, 2.6, 2.8, 4.0, 2.3, 2.7, 3.0, 4.0, 2.9, 3.0 Calculate the measures of central tendency.

23. Group (2) - Exercise 2
For the following haemoglobin values (gm/dl), find the mean, mode, and median. 12, 14, 16, 15, 8, 10, 10, 13, 11, 14, 15, 10, 10, 17, 14

24. Group (3) - Exercise 2
A sample of 15 patients making visits to a health center traveled these distances in miles, calculate measures of central tendency: 5, 9, 11, 3, 12, 13, 12, 6, 13, 7, 3, 15, 12, 15, 5

25. Group (4) - Exercise 2
Arterial blood gas analysis performed on a sample of 15 physically active adult males yielded the following resting PaO2 values: 75, 80, 80, 74, 84, 78, 89, 72, 83, 76, 75, 87, 78, 79, 88 Calculate the measures of central tendency.

26. Group 5; Exercise 1 Mean age (months) No. of children Village 58 44 1
The mean age in months of preschool children in five villages are presented down; calculate the weighted mean of preschool children in these villages.
Mean age
(months)
No. of
children
Village 58 44 1 45 78 2 62 48 3 60 4 59 47 5