RESEARCH I

WEEK 6

https://www.youtube.com/watch?v=ulk_JWckJ78

GUIDELINES FOR PARAMETRIC AND NON PARAMETRIC TEST

https://www.youtube.com/watch?v=pWEWHKnwg_0

P VALUE
https://www.youtube.com/watch?v=DAkJhY2zQ3c
https://www.youtube.com/watch?v=vemZtEM63GY

MEASURES OF POSITION & DISPERSION

MEASURE OF LOCATION

Measure of Position is a score that indicates the quantity of data below it when they are arranged
from highest to lowest. Example: P 20 reads percentile 20 indicates that 20% of the distribution lies below
it. In a test, it means that the lowest 20% of the takers are below it. A measure of position indicates the
ranks in the distribution.

Computing the Quartiles for Ungroup Data

Compute the quartiles of the following given:
75 75 81 83 85 86 87 90

It is easy to compute the quartiles because the given number of data is divisible by four, the sane
procedure is employed for any number data divisible by 4. This is getting the quartiles.
Q1 is (75+81)/2= 78.
Q2 is (83+85)/2= 84.
Q3 is (86+87)/2= 88.5.
 In distribution where the number of data is not divisible by 4, the quartiles approximating the
exact values. The procedure is as follows:
Get the indicator number to locate the value determined by the measure position. For example

75 76 78 80 80 83 83 85 86 87
Q1 Q2 Q3

The quartile 2 is 81.5. this is equal to the median.

To get the Q1, use the indicator value .25 (N+1). .25 (11) = 2.75
.25 (11)= 2.75
(2) is the 2nd score + .75 (d) d is the difference between the 2nd score and the next score

75 76 78 80 80 83 83 85 86 87

76 + .75 (78-76) = 77.5

For , Q3 it is .75 (11). Thus, the answer is 8.25
8th score + .25 d

75 76 78 80 80 83 83 85 86 87
Q3= 85+.25 (86-85)=85.25

For grouped Data in Frequency Distribution

The generic formula follows that of the median. Measures of position that are equivalent are

P 25=Q1 P 50= Q2 P 75= Q3 P 30=Decile 3 P 90=D 9

To illustrate:
P30= LL + i (.30N – cf)/f
P90= LL + i (.90N – cf)/f
P75= LL + i (.75N – cf)/f
P20= LL + i (.20N – cf)/f

Solve for the given measures of position using the following grouped table.
P30= LL + i (.30N – cf)/f

Compute first the indicator: .30N.

.30 (30)=9
Locate the cf that contains 9
Identify the other given
class f cf
70-76 1 30
63-69 3 29
56-62 13 26
49-55 7 f 13
42-48 2 6 cf
35-41 3 4
28-34 1 1
N= 30
P30= LL + i (.30N – cf)/f
P30=48.5 + 7 (9 – 6)/7
P30= 51.5

Solve for P 90:

P90= LL + i (.90N – cf)/f
indicator: .90N.
.90 (30)=27
Locate the cf that contains 27
Identify the other given
class F cf
70-76 1 30
63-69 3f 29
56-62 13 26 cf
49-55 7 13
42-48 2 6
35-41 3 4
28-34 1 1
N= 30

P90= LL + i (.90N – cf)/f

P90=62.5 + 7 (27 – 26)/3
P90= 64.83
To interpret this: Below 64.83 is 90% of the distribution.

Solve for P 75:

P 75= LL + i (.75N – cf)/f
indicator: .75N.
.75 (30)=22.5
Locate the cf that contains 22.5
Identify the other given

class F cf
70-76 1 30
63-69 3 29
56-62 13f 26
49-55 7 13 cf
42-48 2 6
35-41 3 4
28-34 1 1

P75= LL + i (.75N – cf)/f

P75=55.5 + 7 (22.5 –13)/13
P75= 60.62
Solve for P 85:
P 85= LL + i (.85N – cf)/f
indicator: .75N.
.85 (30)
Locate the cf that contains
Identify the other given

class F cf
70-76 1 30
63-69 3 29
56-62 13 26
49-55 7 13
42-48 2 6
35-41 3 4
28-34 1 1

The STANDARD DEVIATION

The standard deviation is the positive square root of the variance. In contrast to the range, it takes
into account all the data. The standard deviation measures the variation of the set by determining how far
from the mean are the data on the average. If there is a large amount of variation in the data, then on the
average, the data values will be far from the mean.

The standard deviation is the average amount of variability in your dataset. It tells you, on
average, how far each value lies from the mean.

A high standard deviation means that values are generally far from the mean, while a low
standard deviation indicates that values are clustered close to the mean.

What does standard deviation tell you?

Standard deviation is a useful measure of spread for normal distributions.

In normal distributions, data is symmetrically distributed with no skew. Most values cluster
around a central region, with values tapering off as they go further away from the center. The standard
deviation tells you how spread out from the center of the distribution your data is on average.

Many scientific variables follow normal distributions, including height, standardized test scores,
or job satisfaction ratings. When you have the standard deviations of different samples, you can compare
their distributions using statistical tests to make inferences about the larger populations they came from.

Example: Comparing different standard deviations

You collect data on job satisfaction ratings from three groups of employees using simple random
sampling.
The mean (M) ratings are the same for each group – it’s the value on the x-axis when the curve is
at its peak. However, their standard deviations (SD) differ from each other.
 The standard deviation reflects the dispersion of the distribution. The curve with the lowest
standard deviation has a high peak and a small spread, while the curve with the highest standard deviation
is more flat and widespread. A graph showing the distributions of three samples with different standard
deviations

THE EMPIRICAL RULE

The standard deviation and the mean together can tell you where most of the values in your
distribution lie if they follow a normal distribution.

The empirical rule, or the 68-95-99.7 rule, tells you where your values lie:

Around 68% of scores are within 2 standard deviations of the mean,

Around 95% of scores are within 4 standard deviations of the mean,
Around 99.7% of scores are within 6 standard deviations of the mean.
Example: Standard deviation in a normal distribution

You administer a memory recall test to a group of students. The data follows a normal
distribution with a mean score of 50 and a standard deviation of 10.
Following the empirical rule:

Around 68% of scores are between 40 and 60.

Around 95% of scores are between 30 and 70.
Around 99.7% of scores are between 20 and 80.
 A graph showing the empirical rule for normal distributions.
The empirical rule is a quick way to get an overview of your data and check for any outliers or
extreme values that don’t follow this pattern.

For non-normal distributions, the standard deviation is a less reliable measure of variability and
should be used in combination with other measures like the range or interquartile range.

Example: The scores of a 5-player team in the quiz bee are as follows: 5,6,7,7, and 9

X X2
5 25
6 36
7 49
7 49
9 81
∑X= 34 ∑X2= 240

The population Standard Devetion

S= √ ∑X2/N
The sample Standard Deviation:
s= √ ∑X2/N-1
where:
∑X 2= ∑X2 – (∑X)2/N

∑X2 = 240 – (34)2/5

=8.8

S= √8.8/5
S= 1.3266; the population standard deviation

The sample standard deviation

s= √8.8/5-1
s= 1.483; the sample standard deviation

Find the standard deviation of the following

Class limit f Class midpoint fX fX2

X
15-19 5 17 85 1445 (17x17x5)
10-14 15 12 180 2160 (12x12x13)
5-9 13 7 91 637 (7x7x13)
0-4 7 2 14 28 (2x2x7)
∑ 40 370 4270

∑X2= ∑fX2 – (∑fX)2/N

∑X2= 4270-(370) 2/N

= 847.5

S= ∑X2/N
= √847.5/40
= 4.60

s=∑X2/N-1
s= √847.5/39
=4.66
 ACTIVITY FOR WEEK 6

1. A survey on the study habits of school children reveals the number of hours they spent studying per
week. The data are as follows:

Hours per Week f cf

35-39 10 404
30-34 69 394
25-29 130 325
20-24 85 195
15-19 80 f 110
10-14 21 30 cf
5-9 9 9
N= 404

a. P20
b. P30
c. P60
d. P78
e. P90

2. The scores of Grade 9 STE students on the Second Quarter Examination are distributed as follows:

Scores f Midpoint (x) fx fx2

90-94 1
85-89 2
80-84 3
75-79 10
70-74 10
65-69 28
60-64 10
55-59 9 57
50-54 5 52
45-49 4 47
40-44 2 42 84 (42x42x2)=3528
35-39 1 37 37 1369

Compute for standard deviation for Population (S) and sample (s).