LYCEUM-NORTHWESTERN UNIVERSITY

Tapuac District, Dagupan City

Institute of Graduate and Professional Studies

==================================================================

NAME OF STUDENT: BETHWEN C. DICHOSO

COURSE: MASTER IN EDUCATION major in FILIPINO

PROFESSOR: DR. RUFO DELA CRUZ

SUBJECTS: EDUCATIONAL STATISTICS

ACADEMIC YEAR: FIRST YEAR FIRST SEMESTER 2024

COMPILATION OF ACTIVITIES AND EXAMS

EXERCISE

1. Of what importance is statistics to you as a teacher/business person or engineer? (Choose

the role applicable to you)

Statistics are the collection, analysis, presentation, and interpretation of numerical data. They
can help people make informed decisions by providing accurate representations of data points
and trends. For the teachers, they can critically assess students, instruction, and the findings of
educational research by using statistics.

2. Why is the statistics considered as tool to quantitative research?

Statistics is a tool for quantitative research because it allows researchers to analyze numerical
data and draw meaningful conclusions. Some examples of quantitative data collection tools
include: Surveys, Questionnaires, Observation checklists, and Physical tests.
 EXERCISES (SET A)
Answer each of the following exercises by computing for the required/appropriate statistical
measure.
1. Given the following scores, determine the mean, median, and the mode.
a) Xi b) Xi
50 112
75 127
60 109
63 118
72 120
58 125
64 115
80 122
40 124
59 128
75

Answers:
A) MEAN = 50+75+60+63+72+58+65+80+40+60+75 =698
11 11
Arrange the the first set of scores inascending order to find the median:
40,50,58,60,60,63,65,72,75,75,80
MEDIAN isin the 6th scores = 63
MODE = 60 and 75

B) MEAN = 112+127+109+118+120+125+115+122+124+128 = 1200 =120

10 10
Arrange the set scores in ascending order to find the median:
109,112,115,118,120,122,.124,125,127,128
MEDIAn is the average of 5th and 6th = 120 and 122
MODE = each number appears only once so, there is no mode.
2. A grocery owner asked her cashier to report her average daily sales after one week. The
cashier came up with the daily sales record for one week. What amount should she report to
his boss?
Monday - P 4,206.50
Tuesday - P 3,945.00
Wednesday - P 3,796.25
Thursday - P 3,587.25
Friday - P 3,086
Saturday - P 3,229
Sunday - P 4,177
Answers:
4,206.50 + 3,945 + 3796.25 + 3587.25 + 3086 + 3229 + 4177 = P 26,026.00
26027 / 7 = P 3,718.00
5) Your brother, a second year criminology student is aiming for academics scholarship. If the
minimum grade-point average (GPA) required for academic scholarship is 1.75 with no grade
lower than 2.0, do you thin k he can avail himself of the scholarship privilege if his grades are as
follows?
Grade Units
1.5 3
2.0 2
1.75 3
2.0 5
1.25 3
1.75 3
1.75 5
Answers:
Calculate first the GPA
1.5 x 3 = 4.5
2.0 x 2 = 4.0
1.75 x 3 = 5.25
2.0 x 5 = 10.0
1.25 x 3 = 3.75
1.75 x 3 = 5.25
1.75 x 5 = 8.75
41.5
Total number of units 3 + 2+ 3 + 5+ 3 +3 +5 = 22
41.5 / 22 = 1.886
His GPA of 1.886 exceeds the minimum requirement of 1.75. however, he has a grade of 1.25,
which below the minimum grade requirement of 2.0
He does not meet the minimum grade requirement.

6) Given the following scores, construct a frequency distribution:

60 43 51 20 48
46 83 44 25 38
37 12 52 55 39
27 64 60 45 55
75 35 40 30 42
78 50 50 32 57
80 36 65 50 45
68 62 65 60 70
Answer:
List the scores in ascending order:
12,20,25,27,30,32,35,36,37,38,39,40,42,43,44,45,46,48,50,50,50,51,52,55,55,57,59,60,60,62,64
,65,68,70,75,78,80,83
83 - 12 = 71
frequency scores within each interval: 10-19:1
20-29:3
30-39:8
40-49:8
50-59:9
60-69:8
70-79:3
80-89:2
Frequency distribution table:

Class interval Frequency

10-19 1

20-29 3

30-39 8

40-49 8

50-59 9

60-69 8

70-79 3

80-89 2
 CHAPTER 3

1. The following distributions represents the distributions of point ratings obtained by 60 teacher-
applicants. Using the distributions, compute for

A. Quartile Deviation

B. Standard Deviation

C. Variance

c.i. f

90-94 2

85-90 8

80-84 8

75-79 18

70-74 10

65-69 9

60-64 5

N=60

2. A group of 50 junior high school students were tested on two aspects--mental ability and endurance
test. With the following results, in which test were they more varied?

Mental Ability Test Endurance Test

M -115 M - 37 min.

SD - 7.25 SD - 5.5 min.

ANSWERS:
 CHAPTER 4
Using the given distribution, answer the following questions:
1. Determine P₉₅.
2. Under what point/value lies the lower 55% of the distribution?
3. Above what point/value do we find the upper 45% of the cases?
4. The middle 38% of the distribution can be found between which two points/values?
5. What is the PR of the value 107?

c.i. f
120-122 2
117-119 1
114-116 3
111-113 4
108-110 5
105-107 8
102-104 7
99-101 4
 96-98 3
93-95 2
90-92 1
N =40
Answers:
c.i. f cf
120-122 2 40
117-119 1 38
114-116 3 37
111-113 4 34
108-110 5 30
105-107 8 25
102-104 7 17
99-101 4 10
96-98 3 6
93-95 2 3
90-92 1 1
N =40
1. P₉₅ = 95% of 40 is 38
(117-119)
P₉₅ = 118

2. 55% of 40 is 22
(105-107)
55% is 106

3. 45% of 40 is 18
 (105-107)
45% is 106

4. 38th of 40 is 15.2
It lies between 102-104 and 105-107, therefore 103 and 106

5. PR of 107
105-107 is 25
PR = 25/40 x 100%
= 62.5%

CHAPTER 7
Determine the relationship between the two variables X and Y using the three methods -
Spearman’s formula, Product-Moment Method (Actual Mean Deviation), Product-Moment
Method (Assumed Mean Deviation)

Spearman’s Formula
X Y x y D D²
Ranking
1.50 125 9.5 9.5 0 0
1.45 120 11 11 0 0
1.52 130 8 7.5 .5 25
1.60 150 3.33 1.5 1.83 3.35
1.55 125 6 9.5 -3.5 12.25
1.75 140 1 4.5 -3.5 12.25
1.65 140 2 4.5 -2.5 6.25
160 145 3.33 3 .33 .11
1.54 135 7 6 1 1
1.50 130 9.5 7.5 2 4
1.40 115 12 12 0 0
1.60 150 3.33 1.5 1.83 3.35
ED² = 57.66

1 −6 ∑ D ²
N(N²-1)
P=

1- 6(57.66) = 1-.21
12(12²-1)
1- 345.96 P = .79 (high relationship)
1716

Product-Moment Method using Actual Mean Deviation

X Y x y x² y² xy
1.50 125 -.06 -8.75 .0036 76.56 .53
1.45 120 -.11 -13.75 .012 189.06 1.51
1.52 130 -.04 -3.75 .0016 14.06 .15
1.60 150 .04 16.25 .0016 254.06 .65
1.55 125 -.01 -8.75 .0001 76.56 .088
1.75 140 .19 6.25 .036 39.06 1.19
1.65 140 .09 6.25 .0081 39.06 .56
1.60 145 .04 11.25 .0016 126.56 .45
1.54 135 -.02 1.25 .0004 1.56 .025
1.50 130 -.06 -3.75 .0036 14.06 .23
1.40 115 -.16 -18.75 .026 351.56 3
1.60 150 .04 16.25 .0016 264.06 .65
18.66 1605 .096 1425.22 9.033
Mx = 18.66 = 1.56
12
My = 1605 = 133.75
12

Rxy = Exy
√(Ex)(Ey)

= 9.033
√(.096)(1456.22)

= 9.033
√139.80

= 9.033
11.82

Rxy = .76 (high relationship)

Product Moment Method using Assumed Mean Deviation

X Y xʾ yʾ xʾ² yʾ² xʾyʾ

1.50 125 -.1 -15 .01 225 1.5
1.45 120 -.15 -20 .023 400 3
1.52 130 -.08 -10 .0064 100 .8
1.60 150 0 10 0 100 0
1.55 125 -.05 -15 .0025 225 .75
1.75 140 .15 0 .023 0 0
1.65 140 .05 0 .0025 0 0
1.60 145 0 5 0 25 0
1.54 135 -.06 -5 .0036 25 .3
1.50 130 -.1 -10 .01 100 1
1.40 115 -.2 -25 .04 625 5
1.60 150 0 10 0 100 0
18.66 1605 Ex²=.328 1925 12.354

Mx = 1.56 Cx = .078 r = Exʾ² - cxcy

My = 133.75 Cy = 10.47 N
AMx = 1.60 C²x = .0061 σxσy
AMy = 140 C²y = 109.62
= .328 - .87
R= Exʾ² - cxcy 12
N .14 x 7.13
Σxσy
= -.84
σx = Exʾ² - c²x = .328 - .0061 = .14 .100
N 12
r = 8.4
Σy = Eyʾ² - c²y = 1925 - 109.62 = 7.13
N 12
 LYCEUM NORTHWESTERN UNIVERSITY
Institute of Graduate and Professional Studies
Dagupan City, Pangasinan

PRELIMINARY EXAMINATION

Name: __Dichoso, Bethwen C., MEd-Filipino

I. Directions: Choose the letter of the correct answer.
__B__1. Standard Deviation refers to:
a. The distance between the highest score and the lowest score c. Compare the
distances of scores
b. Most common measure of the spread of scores d. To spread out the
score
__ A_2. What is the range of the set of scores 11, 51, 51, 91?
a. 80 b. 81 c. 82 d. 83
__B__ 3. In a P.E. Class, there are 25 freshmen, 30 sophomores and 10 juniors. If the freshmen averaged
89 in P.E test, the sophies averaged 85, and the juniors averaged 82, What is the mean grade of
the entire class?
a. 86.08 b. 85.08 c. 84.04 d. 83.02
For items 4-5. Calculate the mean deviation of 2,4,6,8,10
___C_ 4.What is the value of the Mean?
a. 3 b. 4 c. 5 d. 6
__C__ 5. What is the value of the Mean Deviation?
a. 2.4 b.2.5 c.2.6 d. 2.7
For items 6-7. Find the variance of the following 5,6,2,3,1,7,4,8.
__D__ 6.What is the value of the Mean?
a. 3 b. 3.5 c. 4 d. 4.5
__B__ 7.What is the value of the Variance?
a. 4 b. 5 c. 6 d. 7
For items 8-9. Calculate the standard deviation of the following scores 5, 4, 3 , 6, 2
__ C__ 8. What is the value of the Mean?
a. 3 b. 4 c. 5 d. 6
__A__ 9. What is the value of the standard deviation?
a. 1.58 b. 2.58 c. 3.58 d. 4.58
__A__ 10. What is the mean given the following data: 5, 46, 31, 31, 43, 20, 19, 5?
a. 25 b. 25.5 c. 31 d. 40

Complete the following table. (10pts).

X f x fX x- x (x- x )2 f(x- x )2
30-34 4 32 128 8.75 76.56 306.24
25-29 5 27 135 3.75 14.06 70.4
20-24 6 22 132 -1.25 1.56 9.36
15-19 2 17 34 -6.25 39.06 78.12
10-14 3 12 36 -11.25 126.56 379.68
N= 20 ∑ fx=¿465 ∑ f(x- x )2 =843.8

Solve the following:

1. x =
∑ fx = 465/20 = 23.25
N
2. s2 = ∑
2
f ( x−x)
= 843.8/20-1 = 824.8
n− 1

3. s =
√∑ f (x − x )2
n −1
= 843.8/20-1 = 824.8 = 28.72

Prepared by:

RUFO F. DELA CRUZ, Ed.D

Instructor
 CHAPTER 8
The mean of 16 independent observations of a certain magnitude is 100 and the SD is 24. (a) at
the .05 confidence level, what are the fiduciary limits of a true mean? (b) taking the .99
confidence interval as our standard, we may be assured that the true mean is at least as large
as what value?

Answers:
Calculate the standard error of the mean (SEM). SEM = SD / √n = 24 / √16 =6

a) For a 95% confidence interval (.05 significance level), the critical z-value is 1.96
Margin of error = z
SEM = 1.96
6 = 11.76
Lower limit = mean-margin of error = 100 - 11.76 = 88.24
Upper limit = mean + margin of error = 100 +11.76 = 111.76

b) For a 99% confidence interval, the critical z-value is 2.576.

 Margin of error = 2.576
6 = 15.456

Lower limit = 100 - 15.456=84.544

CHAPTER 9
Some 35 incoming high school freshen who graduated from three type of elementary schools
took the entrance test of high school X. Their scores are given below. Do their mean
performance differ significantly? Use the .05 level of significance in testing the null hypothesis.

Group I Group II Group III

(Public) (private sectarian) (Private non-Sectarian)

45 60 62
51 68 66
42 64 72
47 70 65
38 59 58
40 73 74
55 78 75
50 66 68
 46 57 54
44 61 59
75 70
63 60
64

ANSWERS:
 CHAPTER 11
Product researchers have designated two microchips that have increased processing speed of
the computers. In a test of the speed quality of these microchips, an experimental group of 20
chips were fed to serve identical except for brand name. Here are the speeds (in khz) after 20
trials.

Experimental Control
375 401 447 361 356 366 321 380

426 393 403 434 462 402 349 283

407 467 318 406 399 329 410 356

392 477 420 427 272 316 384 350

326 410 339 430 431 360 455 345

Question: Is there a significant difference between the set of the chips? Prove your answer. Test
your hypothesis at the .05 level of significance.

ANSWERS:
 FINAL EXAMINATION
Directions: interpret and justify the following problems using statistical tools.
1. A teacher wants to determine the relative effect of the 30 minutes daily
coaching in reading. She has two comparative groups: one was given extra
coaching daily while the other was not coached. After the one month
experiment, she obtained the following data. It is the mean difference
between the two groups significant? Test the null hypothesis at the .05 level
of significance (1.96).
Group I Group II
M₁ = 4.28 M₂ = 44.3
SD = 4.9 SD = 4.7
N₁ = 40 N₂ = 40

2. A researcher conducted a survey where she can involve 40 college

students. She looked into their performance in English and classified their
grades in terms of the dominant language they spoke at home. Is there
significant difference in their mean performance? Test your hypothesis at the
point .095 level of significance.

A {Filipino} 2.0, 2.75, 2.25, 1.50, 1.25, 1.5, 1.25, 1.75, 1.25
B {English} 1.5, 1.75, 1.5, 2.00, 1.25, 1.5, 1.25
C {Pangasinenese} 2.25, 2.5, 2.00, 2.75, 3.00, 2.50, 3.00, 2.50, 2.75, 2.75,
2.50
D {Combination} 2.50, 2.75, 2.00, 2.75, 1.75, 2.25, 3.0, 2.75, 2.50, 2.50,
2.75, 2.75
ANSWERS: