Chapter10.                                                  STATISTICS

 

10.1     Introduction:  A statistic is an algebraic expression that combines scores into a single number. Statistics basically serves two functions: they estimate parameters in population models and they describe the data. It is collection of methods for planning experiments, obtaining data, and then organizing, summarizing, presenting, analyzing, interpreting, and drawing conclusions.

 

10.2     Measure Of Central Tendency:  Central tendency is a representative score. The three measures of central tendency that will be discussed this semester are the mode, median, and mean.

 

10.2.1    Mean: The mean or the arithmetic mean is the most commonly used measure of central tendency. It is the sum of the numbers divided by the number of numbers. The symbol m is used for the mean of a population. The symbol M is used for the mean of a sample. The formula for m is ;

 

m =   ΣX

                                                                 N

 

Where, X is sum of all the numbers in the given set and N is the total number of numbers in the set.

 

 

Example:        Evaluate the mean of 1, 2, 5, 6, 4

 

Solution:          N = 5

                        ΣX = 1+2+5+6+4 = 18

            So,       m = 18/5

                             = 3.6

 

10.2.2  Median: The median is the middle of a distribution. Half the scores are above the median and half of them are below the median.

 

When there is an odd number of numbers in a given set, the median is simply the middle number.

 

For example: in a given set of 2, 8, and 9 the median is 8.

When there is an even number of numbers, the median is the mean of the two middle numbers.

 

For example: The median of the numbers 2, 4, 7, 12 is;

 

 (4+7)/2 = 5.5

 

10.2.3  Mode:  The mode in a list of numbers refers to the list of numbers that occur most frequently. It is important to note that there can be more than one mode and if no number occurs more than once in the set, then there is no mode for that set of numbers.

 

 

 

Example:        find the mode?

                        14,52,12,14,15,14,21,14,17,27,14

 

Solution: in the above given set of numbers, the most frequently occurring number is 14. 14 has occurred 5 times in the set. Hence,

                                    Mode = 14

 

A distribution may have more than one mode if the two most frequently occurring scores occur the same number of times. Such distributions are called Bimodal.

 

Example:        12,11,14,12,54,10,11,21,24,12,11

 

Solution:          12 and 11 both have occurred 3 times, so,

                       

                                    Mode = 12 and 11

 

 

 

10.3     Measures of Variability:  Variability refers to the spread or dispersion of scores. A distribution of scores is said to be highly variable if the scores differ widely from one another.

 

Consider the following two sets of scores:

Set A: 4, 5, 6, 7, 8

Set B: 2, 4, 6, 8, 10

We can see that the means for these two sets of scores are the same:

For Set A the mean is (4 + 5 + 6 + 7 + 8)/5 = 6

For Set B the mean is (2 + 4 + 6 + 8 + 10)/5 = 6

Although these two sets of scores have the same mean, they differ in how spread out the scores are. The scores in Set A vary over a smaller set of values (4 through 8) than does Set B which varies over score values from 2 through 10.

 

 

10.3.1  Range: The difference between the lowest and highest values.

In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9, so the range is 9-3 equals 6.

Range can also mean all the output values of a function.

 

Range = Highest Score - Lowest Score + 1

For Set A the Range = (8 - 4) + 1 = 4 + 1 = 5

For Set B the Range = (10 - 2) + 1 = 8 + 1 = 9

As we expected the set of scores with the greater spreadoutedness (Set B), has a larger range than the set with less variability (Set A).

 

 

10.3.2  Variance and Standard Deviation: The variance is the measure of variability about the mean and is symbolized by "σ²”

The square root of variance is called the standard deviation. It is represented by σ.

 

 

 

10.3.2.1           Methods of finding variance and standard deviation:

 

Case I :           when each term has frequency 1

 

Let x1 x2……xn are the n given observations and let x be their mean. Then the variance is given by;

 

                        σ² = (x₁ – x )² + (x₂ – x) + ….+ (x_n – x)

                                                     n

           

                             = Σ d_i²

                                    n

where,             

                        d_i = (x_i – x )

therefore,

                        σ =       Σ (x_i – x)²           =   Σ d_i²

                                           n                         n

 

 

 

 

 

 

 

 

 

Example:        Find the variance and standard deviation of the following onservations;

                                    10,11,8,15,16

 

 

 

 

Solution:          Mean = x  =  10+11+8+15+16 = 60/5 = 12

                                                       5

 

Variable             xi	Deviation from the mean                di = (xi – x )	di2
10           11            8           15           16	10 – 12 = -2         11 – 12 = -1           8 – 12 = -4         15 – 12 = 3         16 – 12 = 4	4           1         16           9         16
		Σdi2 = 46

 

Therefore,

                        Variance = σ² = Σd_i²  = 46/5  = 9.2

                                                   n

and standard deviation = σ = √9.2 = 3.303

 

 

 

 

Case II:             When the frequencies of the variable are given. The variance is given by

 

                                    σ² = Σ f_i d_i²   &  S.D = σ =   Σ f_i d_i²  

                                              Σf_i                                     Σf_i

 

 

Example: find the variance and standard deviation from the given distribution table.

 

Variable (xi)	2	4	6	10	12
Frequency (fi)	4	4	5	5	6

           

 

 

Solution:

           

Variable       xi	Frequency         fi	fixi	di = (xi – x )	di2	fi di2
2       4       6       8      10      12      14      16	4        4        5      15        8        5        4        5	8       16       30     120       80       60       56       80	-7 -5 -3 -1   1   3   5   7	49 25 9 1 1 9 25 49	196 100 45 15 8 45 100 245
	Σ fi = 50	Σfixi = 450			Σfi di2 = 754

 

Therefore,

Variance = σ² = Σ f_i d_i²     = 754 / 50 = 15.08

                            Σf_i    

 

Standard deviation = σ = √15.08 = 3.88

 

 

 

Case III:         When the mean is a decimal fraction.

 

In this case we will use the formula;                ₂

                                    σ² = Σ f_i d_i²  -     Σf_id_i

                                              Σf_i                Σf_i

 

 

 

Example:        The scores of 10 students in an examination, in which maximum marks were 50 is given. Find the variance?

 

Variable     xi	Frequency        fi	di = xi – A	di2	fidi2	fidi
19 22 27 28 34 35 36 41 48	1 1 1 2 1 1 1 1 1	-15 -12 -7 -6 0 1 2 7 14	225 144 49 36 0 1 4 49 196	225 144 49 72 0 1 4 49 196	-15 -12 -7 -12 0 1 2 7 14
	Σfi = 10			Σ fidi2 = 740	Σfidi = -22

 

Using the formula                                            ₂

                                    σ² = Σ f_i d_i²  -     Σf_id_i

                                              Σf_i                Σf_i

                                          

     = (740/10) – (-23/10)²

     = 74 – 4.84

     = 69.16

 

 

 

10.3.3  Variance of grouped data:

 

Step Deviation method:  When data are grouped into a frequency distribution having class intervals of equal size h, the formula used is;

                                                                             2

                                    σ² = h²    Σ f_i u_i² –   Σf_i u_i

                                                     Σf_i            Σf_i

 

Where u_i = x_i – A    , A being assumed as the mean.

                      h

 

 

Example:        Calculate the mean and standard deviation for the following distribution:

 

 

Marks	Number of students
20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90	3 6 13 15 14 5 4

 

 

Solution: firstly let’s find the mid values of the class intervals and consider a suitable assumed mean, A

 

Class interval	Frequency      fi	Mid-value        xi	ui = xi – A           10	fi ui	fiui2
20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90	3 6 13 15 14 5 4	25 35 45 55 65 75 85	-3 -2 -1 0 1 2 3	-9 -12 -13 0 14 10 12	27 24 13 0 14 20 36
	Σ fi  = 60			Σfi ui = 2	Σfiui2 = 134

 

 

Therefore,

                        Mean = x = A +      Σf_i u_i  x  h

                                                     Σ f_i

                                 

          =    55 + ( 2/60) x10

          =   55+0.33

          = 55.33

 

Variance =                                                              2

                                    σ² = h²    Σ f_i u_i² –   Σf_i u_i

                             Σf_i            Σf_i

 

                                      = 100 [(134/60) –(2/60)²]

                                      =  222.9

 

Hence, standard deviation = σ = √222.9

                                                 = 14.94

 

 

 

 

10.4      Mean Deviation about the Median:  The mean of absolute deviations of values of various observations from their median is called the mean deviation about the median.

Thus, if x₁,x₂,….x_n be the n observations and M be their median then;

 

 

                                                   _n

                        Mean deviation = Σ | x_i – M |

                                      i ⁼¹

                                                        n

 

 

10.4.1    Methods of finding Mean deviation about Median:

 

Case I:            For discrete series:

 

i)                    Let n₁ be the number of those x_i’^s for which x_i ≥ M and let n₂ be the number x_i’^s for which x_i < M. then n₁ +n₂ = n.

ii)                  Let s₁ and s₂ denote the sum of n₁ and n₂ oservations respectively. Then,

 

                        Mean deviation = (s₁ – s₂) – (n­₁ – n₂)M

                                                            (n₁ + n₂)

 

 

 

 

 

Case II:           For Grouped Data:

 

i)                    Let n₁ be the sum of frequencies f_i’^s of those x_i’^s for which x_i < M and let s₁ = Σ f_i x_i for these f_i’^s.

ii)                  Let n₂ be the sum of frequencies f_i’^s of those x_i’^s for which x_i < M and let s₂ = Σ f_i x_i for these f_i’^s.

 

Mean Deviation = (s₁ – s₂) – (n­₁ – n₂)M

                                    (n₁ + n₂)