BRYAN Q. RUEDAS
Doctorate in Science Education
ADVANCED EDUCATIONAL STATISTICS
Problem Set 1
A. Classify each of the following statements as descriptive statistics or inferential statistics.
1. Descriptive
2. Descriptive
3. Descriptive
4. Inferential
5. Inferential
B. Determine which of the following are qualitative and quantitative, discrete and continuous. Also, indicate the
level of measurement.
1. Age
2. Sex (male, female)
3. Student’s performance
(poor, fair, good, very
good)
4. Number of family
members
5. Grades of students
Qualitative or
Quantitative
Quantitative
Qualitative
Qualitative
Discrete or
Continuous
Continuous
Discrete
Discrete
Level of
Measurement
Ratio
Nominal
Ordinal
Quantitative
Continuous
Ratio
Quantitative
Continuous
Ordinal
C. Express the following using the summation notation.
8
1. ∑𝑖=3 𝑌 𝑖
25
2. ∑𝑖=1(𝑋𝑖
+ 𝑌𝑖 )
4
3. ∑𝑖=1(𝑋𝑖
− 8)
D. If 𝑿𝟏=𝟓, 𝑿𝟐=𝟔, 𝑿𝟑=9, 𝑿𝟒=𝟏𝟑, 𝑿𝟓=𝟏𝟒 𝒂𝒏𝒅 𝑿𝟔=𝟏𝟔, evaluate the following:
4.
∑3𝑖=1 2𝑋𝑖
= (2 • X1) + (2 • X2) + (2 • X3)
= (2 • 5) + (2 • 6) + (2 • 9)
= 10 + 12 + 18
= 40
5. ∑6𝑖=1(𝑋𝑖 − 4)
= (X1 – 4) + (X2 – 4) + (X3 – 4) + (X4 – 4) + (X5 – 4) + (X6 – 4)
= (5 – 4) + (6 – 4) + (9 – 4) + (13 – 4) + (14 – 4) + (16 – 4)
= 1 + 2 + 5 + 9 + 10 + 12
= 39
E. Define statistics.
Statistics deals with the collection, analysis, interpretation, presentation, and organization of numerical
data. Statistics also pertains to describing and summarizing data in a meaningful and informative way, such as
computing measures of central tendency (mean, median, mode) and measures of variability (range, standard
deviation). In order to improve the accuracy and reliability of the results, statistics are also used in research and
other sectors that rely on data. It interprets the data required in a certain field as well.
F. Direction: Identify the type of sampling used in each statement.
1. Random Sampling
2. Convenience Sampling
3. Stratified Random Sampling
4. Random Sampling
5. Systematic Sampling
G. Consider the following test scores in English of fifty students and construct a frequency distribution table.
Scores
(Classes)
f
TCB
CM
RF
20-24
25-29
30-34
35-39
40-44
45-49
50-54
2
6
9
10
12
4
7
19.5-24.5
24.5-29.5
29.5-34.5
34.5-39.5
39.5-44.5
44.5-49.5
49.5-54.5
22
27
32
37
42
47
52
4
12
18
20
24
8
14
Total
50
CF
<CF
2
8
17
27
39
43
50
RF
>CF
50
48
42
33
23
11
7
<RCF
4
16
34
54
78
86
100
100
From the data provided above the constructed frequency distribution table, do the following:
1. Histogram
>RCF
100
96
84
66
46
22
14
2. Frequency Polygon
3. Ogive
H. Given the following list of the employment status of 29 employees at the College of Teacher Education:
Employment Status of College of Teacher Education
Contractual
Temporary
Permanent
Substitute
Total
f
5
7
9
8
29
%
17
24
31
28
100
Bar Graph
Pie Graph
I. Find the area under the normal curve.
1. 𝑃 (𝑧= 1.58)
.4429 or 44.29%
2. 𝑃 (𝑧 > −2.41)
.4920
+ .5000
.9920 or 99.20%
3. To the left of 2.59
.4952
+ .5000
.9952 or 99.52%
4. 𝑃 ( −1.55 <𝑧> 1.55)
.4394
+ .4394
.8788 or 87.88%
5. a. between 5ft to 6.5 ft
𝑧=
=
𝑥− 𝜇
𝜎
5 𝑓𝑡− 5𝑓𝑡
0.75 𝑓𝑡
5−5
= 0.75
=0
𝑧=
=
𝑥− 𝜇
𝜎
6.5 𝑓𝑡− 5𝑓𝑡
0.75 𝑓𝑡
6.5−5
= 0.75
=2
.4772 or 47.72%
5. b. 𝑧 =
=
𝑥− 𝜇
𝜎
5 𝑓𝑡− 5𝑓𝑡
0.75 𝑓𝑡
5−5
= 0.75
=0
J. Provide for the necessary details below and test the hypothesis
1. Ho: The grades of the students in the diagnostic and summative are not significantly different from each other.
Ha: The grades of the student in summative are significantly different/higher than in the diagnostic test.
b. Level of significance: α = 0.05
c. Test statistics: T- Test
One-tailed or Two-tailed: Two tailed
Critical Value: 2.262
d. Computation
Student
A
B
C
D
E
F
G
H
I
J
Diagnostic Summative
Results
Results
79
81
81
80
83
80
84
85
84
80
86
87
86
86
87
88
88
86
90
91
D
D2
-2
1
3
-1
4
-1
0
-1
2
-1
4
1
9
1
16
1
0
1
4
1
∑𝑫 = 4
∑ 𝑫𝟐 = 38
̅=
𝐷
4
= 0.4
10
𝑛 ∑ 𝐷2 −(∑ 𝐷)2
𝑆𝐷 = √
𝑛(𝑛−1)
(10)(38)−(4)2
= √
10(10−1)
380−16
= √
90
= √3.82
= 1.95
1.95
𝑆𝐷̅ = 10
= 𝟎. 𝟔𝟐
√
𝑡=
0.4
0.62
= 𝟎. 𝟔𝟓
e. Decision:
Since the computed t-value 0.65 is less than the t-critical value of 2.262 at 0.05 with df = 9, accept Ho.
Therefore, the grades of the students in the diagnostic and summative are not significantly different from each
other.
2. Ho: The intelligence level of the students in the pilot section will have higher IQ scores.
Ha: The intelligence level of the students in the pilot section will have lower IQ scores.
b. Level of significance: α = 0.05
c. Test statistics: Z- Test
One-tailed or Two-tailed: One tailed
Critical Value: ±1.645
d. Computation
𝑍=
=
(𝑥̅ − 𝜇)√𝑛
𝜎
(102.75−100)√60
21.30
15
= 15
= 1.42
e. Decision:
Since the z computed value of 1.42 is less than the critical value of ±1.645 at 0.05 level of significance,
accept Ho. Therefore, the intelligence level of the students in the pilot section will have higher IQ scores.