MMW TOPIC

LESSON 7-8 STATS

1ST SEMESTER
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Data Management
Determine the value of k = 1 + 3 log n where n = 60.
FREQUENCY DISTRIBUTION log60 = 1.7782=15125
k = 1 + 3 (1.7781512)
 is a grouping of data into categories showing the number of k = 1 + 5.3344536
observations in each of the non-overlapping classes. k = 6. therefore, 6 is the estimated number of classes.
 is the organization of data in a tabular form, using mutually r = 29- 10 = 19
exclusive classes showing the number of observations in Class size = 17/6 = 3.16 or 3.
each. Interval is 3.

RAW DATA MEASURES OF CENTRAL TENDENCY

 is the data collected in original form.  Central tendency determines a numerical value in the central
region of a distribution. It refers to the center of distribution of
RANGE observations. There are three measures of central tendency:
 is the difference of the highest value and lowest value in a - Mean
distribution - Median
- Mode
CLASS LIMITS  Any data set can be characterized by measuring central
 is the highest and lowest values describing a class. tendency. A measure of central tendency, commonly referred
to as an average, is a single value that represents a data set.
CLASS BOUNDARIES Its purpose is to locate the center of a data set.
 is the upper and lower values of a class for group frequency
distribution whose values has additional decimal place more
MEAN
than the class limits and end with the digit 5.
 The mean is also called arithmetic mean or average. It can
INTERVAL be affected by extreme score. It is stable, and varies less
 is the distance between class lover boundary and the class from sample to sample. It is used if the most reliable measure
upper boundary and it is denoted by the symbol i. is desired and when there are few very high values and few
very low values. The mean is the balance point of a score
FREQUENCY distribution.
 is the number of values in a specific class of a frequency
distribution. PROPERTIES OF MEAN
PERCENTAGE 1. A set of data has only one mean.
 is obtained by multiplying the relative frequency by 100%. 2. Mean can be applied for interval or ratio.
3. All values in the data set are included in computing the mean.
CUMULATIVE FREQUENCY 4. The mean is very useful in comparing two or more data sets.
 is the sum of the frequencies accumulated up to the upper 5. Mean is affected by the extreme small or large values on a
boundary of a class in a frequency distribution. data set.
6. Mean is most appropriate in symmetrical data.
MIDPOINT
 is the point halfway between the kinits of each class and is
representative of the data within that class Mean=
∑ of all values
Number of values
Determining class Interval

k Sample mean: x =
∑x
2 2 to the k rule”:
n
Range HV −LV ∑x
Suggested Class Interval= =
Number of classes k Population mean: μ=
N
Where: HV = Highest value in a data set
LV = Lowest value in a data set where: x=¿sample mean (it is read as “x bar”).
k = number of classes μ = population mean (it is read as “mu”).
I = suggested class interval x = the value of any particular observation or
k
2 ≥n measurement.
∑ x = sum of all x’s.
n = total number of values in the sample.
Range N = total number of values in the population.
Suggested Class Interval=
1+3 log n
MEDIAN
 The median is the midpoint of the data array. When the data
set is ordered, whether ascending or descending, it is called
data array. Median is an opportunity measure of central

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 tendency for data that are ordinal or above, but is more dispersed from average. Dispersion is the distance between
valuable in an ordinal type of data. the actual value and the average value.

PROPERTIES OF MEDIAN RANGE

1. The median is unique, there is only one median for a data set. Probably the simplest and easiest way to determine measure of
2. The median is found by arranging the set of data from lowest dispersion is the range. The range is the difference of the highest value
or highest (or highest to lowest) and getting the value of the and the lowest value in the data set.
middle observation.
3. Median is not affected by the extreme small or large values. Advantages:
4. Median can be applied for ordinal, interval and ratio data. - it is easy to compute
5. Median is most appropriate in a skewed data. -it is easy to understand
Disadvantages:
- it can be distorted by a single extreme value (outlier)
• If n is odd, the median is the middle ranked. - only two values are used in the calculation.
• If n is even, then the median is the average of two middle
ranked values.
n+1 How to compute for the mean?
middle (Rank Value) = (Ungrouped data)
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Note that n is the population/sample size. The mean is the balance point of a distribution.

MODE
Mean =
∑ of values
 The mode is the value in a data set that appears most the number of values
frequently. Like the median and unlike the mean, extreme
values in a data set do not affect the mode. A data may not Example 1
contain any mode if none of the values is “most typical”. A
data set that has only one value that occurs the greatest Jeffrey has bee working on programming and uploading a Web site for
frequency is said to be unimodal. If the data has two values his company for the past 24 months. The following numbers represent
with the same greatest frequency both values are considered the number of hours Jeffrey has worked on this Web site for each of the
the mode and the data set is bimodal. If the data set has past 7 months: 24, 25, 33, 50, 53, 66, 78. What is the mean (average)
more than two more modes, then the data set is said to be number of hours that h Jeffrey worked on his Web site each month.
multimodal. There are some cases when a data set value has
the same amount of frequency. When this occurs, the data Solution:
set is said to be no mode.
Step 1. Add the numbers to determine the total number of hours he
PROPERTIES OF MODE worked.
1. The mode is found by locating the most frequent occurring 24 + 25 + 33 + 50+ 53 + 66 + 78 = 329
value.
2. The mode is the easiest average to compute. Step 2. Divide the total by the number of months.
3. There can be more than one mode or even no mode in any
given data set.
Mean =
∑ of values =
329
= 47
4. Mode is not affected by the extreme small or large values. 7
5. Mode can be applied for nominal, ordinal. Interval and ratio
the number of values
data. 47 was the average number of hours that Jeffrey worked on this Web
site each month.
WEIGHTED MEAN
Weighted Mean
 The weighted mean is particularly useful when various
classes or groups contribute differently to the total. The
weighted mean is found by multiplying each value by its WM =
∑ fX
corresponding weight and dividing by the sum of the weights. N
Where: WM = weighted mean
f = frequency
x w + x w + x w +…+ xn w n
x w =¿ 1 1 2 2 3 3 X = score
w1 +w 2+ w3 +…+ wn ∑ fX = sum of the product of frequency and score
N = total frequency
Where: x=weighted mean .
w i = corresponding weight. Grouped Data: MEAN
x i = the value of any particular observation or measurement
M=
∑ fX
MEASURES OF DISPERSION N
 Another important characteristic of a data set is how it is
distributed, or how far each element is from some measure of Where:
central tendency (average). There are several ways to
measure the variability of the data. Although the most M = mean
common and most important is the standard deviation, which f = frequency
provide an average distance for each element from the mean. X = class mark

STANDARD DEVIATION
∑ fX = sum of the product of frequencies and class marks
 Standard deviation is a statistical term that provides a good N = total frequency
indication of volatility. It measures how widely values are
VARIANCE & STANDARD DEVIATION (sample)

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VARIANCE (Ungrouped: Population)

VARIANCE (Ungrouped: Sample)

Ex.1: The daily salaries of a sample of eight employees at GMS, Inc.

are Php550, Php420, Php560, Php500, Php700, Php670, Php860,
Php480. Find the mean daily rate of employees.

x=
∑x = x1 + x 2 + x 3+ x 4+ x5 + x 6 + x 7 + x 8
n n
VARIANCE (Grouped: Population)
550+420+560+ 500+700+670+ 860+480
x= =
8
4,740
=592.50
8

The sample mean daily salary of employees is Php592.50.\

Subtract the mean from each value in the data set.

VARIANCE (Grouped: Sample)

Variance and standard deviation

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s=
∑ ( x−x )2
n−1
142,950
¿
8−1
s ¿ 20,421.43
2

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 s=
√ ∑ ( x−x )2
n−1
MEASURES OF RELATIVE POSITION: DECILE

Dk =
k ( N +1 )
¿
√ 142,950
8−1
¿ √ 20,421.43
Where:
Dk = Decile
10

s=142.90 N=Population
k =Decile
location

( )
N
−cf b
10
Dn= X LB + i
fm
Where:
Dn=the score corresponding ¿ the ithquartile rank .
X LB =lower limit of the quartileinterval class .
MEASURES OF RELATIVE POSITION: QUARTILE
f m=frequency of the quartile class .
k ( N +1 ) cf b=cumulative frequenc of theinterval before the quartile cla
Qk = i=class ¿
4
Where: 10 = stands for the class size division
N = total frequency
Qk =Quartile
N=Population
MEASURES OF RELATIVE POSITION: PERCENTILE
k =Quartile
location k ( N +1 )
Pk =
100
Where:

( )
N Pk =Percentile
−cf b
4 N=Population
Qn= X LB + i k =Percentile
fm
Where:
location
Qn=the score corresponding ¿ theith quartile rank .

( )
N
X LB =lower limit of the quartileinterval class . −cf b
100
f m=frequency of the quartile class . Pn= X LB + i
fm
cf b=cumulative frequenc of theinterval before the quartile clas
Where:
i=class ¿ Pn=the score corresponding ¿ theith quartile rank .
4 = stands for the class size division
N = total frequency X LB =lower limit of the quartileinterval class .
Find the 1st, 2nd, 3rd quartiles of the ages of 9 middle-management
f m=frequency of the quartile class .
employees of a certain company. cf b=cumulative frequenc of theinterval before the quartile cla
The ages are 53, 45, 59, 48, 54, 46, 51, 58 and 55.
i=class ¿
1.Arrange the data in order. 100 = stands for the class size division
45, 46, 48, 51, 53, 54, 55, 58, 59 N = total frequency
2. Select the 1st, 2nd, 3rd quartiles value using the formula.
1 ( 9+1 ) (10)
Q 1= = =2.5
4 4
2 ( 9+1 ) (20)
Q 2= = =5
4 4
3 ( 9+1 ) (30)
Q 3= = =7.5
4 4
3. Identify the 1st, 2nd, 3rd quartiles values in the data set.
45, 46, 48, 51, 53, 54, 55, 58, 59

Q1 = 47, Q2 = 53, Q3 = 56.5

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