2.1 measures of central tendency-ungrouped
DESCRIPTION
BIOSTATTRANSCRIPT
Measures of Central Tendencies
Numbers which, in some sense, give the central or middle values of the data
locates the center of the distribution of a set of data
the most typical value of a set of data
representative value of a given set of data
Mean
• arithmetic mean / average• the sum of the values divided by the number of values which were added.
Mean of ungrouped data
i1 2 3 nxx x x ... x
xn n
Where - sample mean xi - ith observation/item in the sample n - number of observations in the sample
x
i1 2 3 nxx x x ... x
xn n
Example 1: find the mean of the sample: 7, 11, 11, 8, 12, 7, 6, 6
7 11 11 8 12 7 6 6 68x 8.5
8 8
The sample mean is 8.5
Mean of ungrouped data
Example 2: find the mean of this sample:
18, 22, 25, 25, 26, 29, 45
Mean of ungrouped data
Weighted Mean
1 1 2 2 3 3 n nw
1 2 3 n
x (w ) x (w ) x (w ) ... x (w )x
w w w ... w
Where - weighted mean xi - ith observation/item in the sample wi – weight of the ith observation
wx
Examples of uses of weighted mean are in computing term GPA and in getting the mean responses for a Likert-type of questions.
It is used if the researcher wants
to know the feelings or
opinions of the respondents
regarding any topic or issues of
interest.
Likert-type questions
Choices are:5 – (SA) Strongly agree4 – (A) Agree3 – (N) Neutral2 – (D) Disagree1 – (SD) Strongly disagree
Check appropriate box 5 4 3 2 1
1 Student nurses serve as role models for their patients and the public.
2 Student nurses should set a good example by not smoking.
3 Patient's chances of quitting smoking are increased if a student nurses advises him or her to quit.
4 Smoking is harmful to your health.
5 Smoking other tobacco products is harmful to a person’s health.
Likert-type questions
54 3 2 1 Interpretati
on
1 7 11 2 0 0
2 9 10 1 0 0
3 2 8 8 2 0
4 16 4 0 0 0
5 17 3 0 0 0
x
Likert-type questions
4.25Strongly Agree4.40Strongly Agree
3.50 Agree
Likert-Type Mean Interpretation1.00 – 1.79 – Strongly Disagree1.80 – 2.59 – Disagree2.60 – 3.39 – Neutral3.40 – 4.19 – Agree4.20 – 5.00 – Strongly Agree
Strongly Agree4.80
4.85Strongly Agree
4.25 4.40 3.50 4.80 4.85Grand Mean 4.36
5
Strongly Agree4.36
Mean for Grouped Data
Where f – frequency of the class
X – Class Mark
n – sample size
nXf
x
Classes Freq. (f)
Class Mark(Xm)
fXm
12 - 22 23 - 33 34 - 44 45 - 55 56 – 66
47621
1728395061
Total 20
65932.95
20
Mean for Grouped Data
68
196
234100
61
659
nXf
x
Characteristics of the Mean
1. It can be calculated for any set of numerical data, so it always exist.
2. A set of numerical data has one and only one mean.
4. It is greatly affected by extreme or deviant values.
3. It is the most reliable since it takes into account every item in the set of data.
MedianThe median of a data is defined to be the ‘middle value’.
[(n 1) / 2]th term when n is odd
x (n / 2)th term + [(n/2)+1]th termwhen n is even
2
Note: it is important to arrange first the sample in ascending order before getting the median.
Thus, when n is odd, the median is the ‘center’ observation.
When n is even, it is the average of the two center observations.
Example 3: find the median of the this sample: 7, 11, 11, 8, 12, 7, 6, 6
The sample median is 7.5
(n / 2)th term + [(n/2)+1]th termx
2
Solution: Arrange the observations in ascending order.
6, 6, 7, 7, 8, 11, 11, 12
Since n = 8 (even), then
n/2 = 8/2 = 4 and (n/2) + 1=(8/2)+1=5 Thus,
4th term + 5th termx
2
7 + 8x 7.5
2
Median
Example 4: find the median of this sample: 18, 22, 25, 25, 26, 29, 45
The sample median is 25
Solution:The solution is already arranged in ascending order.
Since n = 7 (odd), then
(n + 1)/2 = (7 + 1)/2 = 4.
Thus, = 25
x [(n 1) / 2]th term
x 4th term
Median
Characteristics
1. The score or class in a distribution, below which 50% of the score fall and above which another 50% lie.
2. Not affected by extreme or deviant values.3. Appropriate to use when there are extreme or deviant values.
Illustration:18, 22, 25, 25, 26, 29, 45
Compare the mean (example 2) and the median (example 4)of the above sample.
Mean = 27.1 Median = 25
Which is the better measure of central tendency in thisexample? Why?
Median
1. It is used when we want to find the value which occurs most often.
2. It is a quick approximation of the average.
3. It is an inspection average.
4. It is the most unreliable among the three measures of central tendency because its value is undefined in some observations.
Mode
Mode
Mode: VS
2. The ages of 5 students are: 17, 18, 23, 20, & 19No Mode
3. The grades of 5 students are: 4.0, 3.5, 4.0, 3.5, & 1.0
Mode: 4.0 & 3.54. The weight of 5 persons in pounds are: 117, 218, 233, 120, & 117
Mode: 117
1. The following are the descriptive evaluation of 5 teachers.
Teacher Evaluation A VS B S C VS D VS E S
Examples
ComparisonFactor Mean Median Mode
Type of data
Quantitative
Quantitative
Quantitative and
Qualitative
Extreme scoreproblem
Yes No No
Always measurable
Yes Yes No
Number of score
1 1 0,1,2…
Characteristics
All scores included in computatio
n
Middle value
Popular value
(Grouped)Approximating the Median from the Freq. Distribution Table
Steps:1.) Construct the < cumulative frequency distribution.2.) Starting from the top, locate the class with the <cf greater than or equal to n/2 for the first time. median class
3.) Approximate using the formula:
+i
where = the lower class boundary of the median class i= class width n=total no. of observations= less than cumulative freq of the class preceding the median class = frequency of the median class
Approximating the Mode from the Freq. Distribution Table
Steps:1.) Locate the modal class. The modal class is the class with the highest frequency
2.) Approximate using the formula:
+i
where = the lower class boundary of the modal class i= class width = frequency of the modal class = frequency of the class preceding the modal class = frequency of the class following the modal class
Find the mean, median and mode.FDT of Final grades of 100 Math 103 students
Class Freq42-48 449-55 1256-62 2263-69 2770-76 1777-83 984-90 891-97 1
Measures of Location or Fractiles
-values below which a specified fraction or percentage of the observations in a given set must fall.
Percentiles
Percentiles are values that divide a set of observations in an array into 100 equal parts.
Percentiles Sort all observations in ascending order Compute the position L = (K/100) * N, where N
is the total number of observations. If L is a whole number, the K-th percentile is
the value midway between the L-th value and the next one.
If L is not a whole number, change it by rounding up to the nearest integer. The value at that position is the K-th percentile.
Decilesvalues that divide the array into 10 equal parts
D1 - the value below which 10% of the values fallD2 - the value below which 20% of the values fall…
Deciles Sort all observations in ascending order Compute the position L = (K/10) * N, where N
is the total number of observations. If L is a whole number, the K-th decile is the
value midway between the L-th value and the next one.
If L is not a whole number, change it by rounding up to the nearest integer. The value at that position is the K-th decile.
Quartilesvalues that divide the array into 4 equal parts.
Q1 – the value below which 25% of the values fallQ2 – the value below which 50% of the values fallQ3 - the value below which 75% of the values fall
To compute for Lower Quartiles Sort all observations in ascending order Compute the position L1 = 0.25 * N, where N
is the total number of observations. If L1 is a whole number, the lower quartile is
midway between the L1-th value and the next one.
If L1 is not a whole number, change it by rounding up to the nearest integer. The value at that position is the lower quartile.
To compute for Upper Quartile Sort all observations in ascending order Compute the position L3 = 0.75 * N, where N
is the total number of observations. If L3 is a whole number, the upper quartile is
midway between the L3-th value and the next one.
If L3 is not a whole number, change it by rounding up to the nearest integer. The value at that position is the upper quartile.
Example
The surveyed weights (in kilograms) of the students in Stat 231 were the following: 69, 70, 75, 66, 83, 88, 66, 63, 61, 68, 73, 57, 52, 58, and 77. Compute1.) P23 5.) Q32.) P85 6.) Q13.) D304.) D90
Approximating the ith Percentile from the FDT
where =the lower class boundary of the Pith classc = the class size of the Pith classn = the total number of observations in the distribution=the <cf of the class preceding the Pith class
+c