Statistics and Probability
Quarter 3 – Module 8:
Length of Confidence Interval and
Appropriate Sample Size
`
Statistics and Probability
Alternative Delivery Mode
Quarter 3 – Module 8: Length of Confidence Interval and Appropriate Sample Size
First Edition, 2020
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Statistics and
Probability
Quarter 3 – Module 8:
Length of Confidence Interval
and Appropriate Sample Size
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners,
can continue your studies and learn while at home. Activities, questions, directions,
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If you have any questions in using this SLM or any difficulty in answering the
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Thank you.
What I Need to Know
At the end of the lesson, the learner should be able to:
➢ Identify the length of a confidence interval. (M11/12SP-IIIj-1)
➢ Compute for the length of the confidence interval. (M11/12SP-IIIj-2)
➢ Compute for an appropriate sample size using the length of interval.
(M11/12SP-IIIj-3)
➢ Solve exercise involving sample size determination. (M11/12SP-IIIj-4)
.
What I Know
A. Find the length of the following confidence interval.
1. 0.0254 < p < 0.560
2. 0.385< p < 0.655
3. 0.363 < p < 0.655
4. Upper confidence limit = 0.632
Lower confidence limit = 0.475
5. Upper confidence limit = 0.686
Lower confidence limit = 0.415
B. Find the length of the confidence interval given the following data:
6. 𝑠 = 8.47, n = 300, confidence level= 95%
7. 𝑠 = 6.25, n = 150, confidence level= 95%
8. 𝑠 = 4.36, n = 350, confidence level= 99%
9. s = 5.25, n = 16, confidence interval = 95%
10. s = 7.15, n = 20, confidence interval= 99%
1
Lesson
1
Length of Confidence and
Appropriate Sample Size
How large should be the sample? Many researchers find this a problem. In
fact, many rely on old methods to determine the sample size they need in their
investigations. This problem must be addressed carefully because needlessly large
samples are a waste of important resources. On the other hand, insufficient sample
size may lead to poor results.
What’s In
A. Fill in the blank with the correct word or group of words to make a meaningful
statement. Write your answer on a separate sheet of paper.
1. Statistically, rounding up of 111.12 is____________________.
2. A confidence level is the confidence coefficient and is expressed as
__________________.
3. For a 95% confidence level, the confidence coefficient is __________________.
4. When a 99% confidence level is selected, a = _______________.
5. The standard deviation of a sampling distribution is called _____________.
6. A subset of the interest population is called ______________.
B. Answer the following:
1. 1 – 0.38
2. (0.35) (0.65)
3.
1.96(2)
0.06
4. (
5. (
(1.96)(5.3) 2
0.2
)
(2.58)(2.5) 2
0.10
)
2
Note to the Teacher
In this module, the learners will be able to know how to compute for
the length of the confidence interval and for an appropriate sample size
needed using the length of interval.
What’s New
Definition of Terms
1. Confidence interval - In statistics, this refers to the probability that
a Population parameter will fall between a set of values for a certain proportion of
times. Confidence intervals measure the degree of uncertainty or certainty in
a sampling method. They can take any number of probability limits, with the most
common being a 95% or 99% confidence level.
2. Narrowness of the interval – This pertains to a small width in relation to the
length of the confidence interval.
Length of Confidence Interval
This refers to the absolute difference between the upper confidence limit and
the lower confidence limit.
Formula:
LCI = |UCL – LCL|
= |LCL – UCL|
Or
LCI = UCL – LCL
Where:
LCI – length of confidence interval
UCL – upper confidence limit
LCL – lower confidence limit
3
Example 1. Find the length of the confidence interval 0. 275<p<0.360
Solution:
Steps
1. Determine the given.
Solution
UCL = 0.360
LCL = 0.275
LCl = UCL – LCL
2. Use the formula for Length of
Confidence Interval.
3. Substitute the given value, then
perform the indicated operation.
LCI = 0.360 - 0.275
LCI = 0.085
Thus, the length of confidence is 0.085.
Example 2. Find the length of the confidence interval
upper confidence limit = 0.805
lower confidence limit = 0.526
Solution:
Steps
1. Determine the given.
Solution
UCL = 0.805
LCL = 0.526
LCl = UCL – LCL
2. Use the formula for Length of
Confidence Interval.
3. Substitute the given value, then
perform the indicated operation.
LCI = 0.805 - 0.526
LCI = 0.279
Thus, the length of confidence interval is equal to 0.279.
Try this in your notebook.
A. Find the length of the confidence interval.
1. 0.355 < p < 0.470
2. 0.475 < p < 0.835
3. Upper confidence limit = 0.796
Lower confidence limit = 0.536
4. Upper confidence limit = 0.896
Lower confidence limit = 0.526
5. Upper confidence limit = 0.966
Lower confidence limit = 0.635
4
What is It
The formula for interval estimate of population mean (µ) with
known variance is:
̅ − 𝒛𝒂⁄𝟐
𝒙
𝝈
√𝒏
̅ − 𝒛𝒂⁄
<𝒖<𝒙
𝟐
𝝈
√𝒏
The formula to find the length of the confidence interval
= 𝟐𝒛𝒂⁄
𝟐
𝝈
√𝒏
Where:
𝒛𝒂⁄ – is the z value
𝟐
𝝈
– standard deviation
𝒏
– sample size
Example 3: Find the length of the confidence interval given the following data:
= 0.3, 𝑛=70, confidence level=95%
Solution:
Steps
1. Determine the given.
2. Determine the
confidence coefficient.
Solution
= 0.3, n =70, confidence level=95%
(1- 𝑎)100% = 95%
Find 𝑎 in equation (1𝑎)100% = 95% where in
95% is the given
confidence level.
(1- 𝑎) = 95%
100% of (1 - 𝑎) = (1 - 𝑎)
(1- 𝑎 )= 0.95
Charge 95% to decimal
number.
(1- 𝑎) - 1 = 0.95 -1
Subtraction property of
equality.
(-1)(-𝑎) = (-0.05)(-1)
Multiplication property of
𝒂 = 0.05
equality.
𝑎/2 = 0.05/2
𝒂/𝟐 = 0.025
5
Division property of
equality.
0.500 - 0.025 = 0.475
𝒛𝒂⁄ = 1.96
𝟐
3. Substitute the values
in the formula and
compute.
𝝈
LCl = 𝟐𝒛𝒂⁄ .
𝟐 √𝒏
𝒔
𝟐 √𝒏
LCl = 𝟐𝒛𝒂⁄ .
LCl = 𝟐(1.96) (
LCl =
0.3
)
√70
1.176
√70
Subtract 0.025 from
0.500 (which is half the
area of the Standard
Normal Curve).
Hence, Using the Area
under the Standard
Normal Curve Table,
as 𝑠ℎ𝑜𝑤𝑛 𝑏𝑒𝑙𝑜𝑤.
Use this formula to solve
for the length of
confidence interval.
Supplies the values,
then multiply
2(1.96)(0.3).
Divide 1.176 by √60.
LCI = 0.1406
Answer
Thus, the length of confidence interval is equal to 0.1406
Try this in your notebook.
B. Find the length of the confidence interval given the following data:
1.
= 0.3,
n = 45,
confidence level 98%
6
2.
= 0.5,
n = 50,
confidence level 95%
3.
= 1.5,
n = 70,
confidence level 99%
The formula for confidence interval of population mean, if n<30
𝝈
2 √𝑛
𝑥̅ − 𝑡𝑎⁄
< µ < 𝑥̅ − 𝑡𝑎/2
𝝈
√𝑛
The formula to find the length of the confidence interval.
𝝈
2 √𝑛
LCl = 2𝑡𝑎⁄
Where:
𝒕𝒂⁄ – is the t value
𝟐
𝝈
– standard deviation
𝒏
– sample size
Example 4: Find the length of the confidence interval, given the following data:
s = 6.5, n= 15, confidence level = 99%
Solution:
Steps
1. Determine the given.
2. Find the degree of
freedom df.
3. Find the 𝒂 in (1𝑎)100% confidence
level, then find 𝒕𝒂⁄
Solution
s = 6.5, n= 15,
confidence level = 99%
df = n-1
df = 15-1
df = 14
Thus, the degree of freedom is 14.
(1- 𝑎)100% = 99%
𝟐
Find 𝑎 in equation (1𝑎)100% = 99% where in
99% is the given
confidence level
(1- 𝑎) = 99%
Since, 100% of (1 – 𝑎) is
equal to (1 - 𝑎)
(1- 𝑎 )= 0.99
Charge 99% to decimal
number
(1- 𝑎) - 1 = 0.99 -1
Subtraction property of
equality.
7
(-1)(-𝑎) = (-0.01)(-1)
a = 0.01
Division property of
equality.
𝑎/2 = 0.01/2
𝒂/𝟐= 0.005
Hence, Using the t
distribution critical values
(t Table) at df = 14 and
𝑎⁄ = 0.005 , as 𝑠ℎ𝑜𝑤𝑛 𝑏𝑒𝑙𝑜𝑤.
2
𝒕𝒂⁄ = 2.977
𝟐
4. Substitute the values
in the formula and
compute.
LCl = 𝟐𝒕𝒂⁄ .
LCl = 𝟐𝒕𝒂⁄ .
Multiplication property of
equality.
𝒔
Use this formula to solve
for the length of
confidence interval.
𝟐 √𝒏
𝝈
𝟐 √𝒏
LCl = 𝟐(2.977) (
LCl =
38.701
√15
6.5
)
√15
Supplies the values,
then multiply 2(2.977)
(6.5).
Divide 38.701 by √15.
LCI = 9.99
Answer
Thus, the length of confidence interval is equal to 9.99
8
Try this:
C. Find the length of the confidence interval given the following data:
1. s = 4.5,
n = 20,
confidence interval = 98%
2. s = 3.5,
n = 25,
confidence interval = 95%
3. s = 2.5,
n = 28,
confidence interval = 99%
Deriving the Formula for a Sample Size
In deriving the formula for the sample size, there are two things to remember
when we decided on the quality of the sample size that we need: confidence and the
narrowness of the interval.
➢ Confidence interval describes the uncertainty inherent in this
estimate and describes a range of values within which we can be
reasonably sure that the true effect actually lies.
➢ Narrowness of the interval pertains to a small width in relation to the
length of the confidence interval.
If the confidence interval is relatively narrow (e.g., 0.70 to 0.80), the effect
size is known precisely. If the interval is wider (e.g., 0.60 to 0.93) the uncertainty is
greater, although there may still be enough precision to make decisions about the
utility of the intervention. Intervals that are very wide (e.g., 0.50 to 1.10) indicate
that we have little knowledge about the effect, and that further information is needed.
The computing formula in determining sample size is derived from the formula of the
margin of error E where:
𝐸=
•
•
𝜎
𝑧𝑎
⁄2 ( )
√𝑛
Study the derivation of the computing formula for determining minimum
sample size when estimating a population mean.
Discuss the elements of the formula.
To solve for n, which is a sample size needed in estimating a population mean,
do this. (Hint: Apply the appropriate properties of equality.)
𝐸=
𝑧𝑎/2 ⋅ 𝜎
Formula for margin of error.
√𝑛
𝑧𝑎 ⋅𝜎
(√𝑛)(𝐸) = (
(√𝑛)(𝐸)
=
𝐸
2
Multiply both sides of the equation by
√𝑛. Multiplication property of equality.
Divide both sides of the equation by E.
Division property of equality.
)(√𝑛)
√𝑛
𝑧𝑎 ⋅ 𝜎
2
𝐸
9
Square both sides.
2
𝑧𝑎
⁄2 ⋅ 𝜎
(√𝑛) = (
)
𝐸
2
𝑛=(
This is now the resulting formula for
sample size needed in estimating a
population mean.
𝑧𝑎⁄ ⋅𝜎 2
2
)
𝐸
Formula in Determining the Minimum Sample Size Needed when
Estimating the Population Mean
𝑛=(
2
𝑧𝑎
⁄2 ⋅ 𝜎
)
𝐸
Where:
𝒛𝒂⁄ – is the z value
𝟐
𝝈
– standard deviation
𝒏
– number of sample
E
– margin of error
Since the value of is usually unknown, it can be estimated by the standard
deviation ( ) from a prior sample. Alternatively, we may approximate the range R of
observations in the population and make a conservative estimate of ≈
R
4
. In any case,
round up the value of obtained to ensure the sample size will be sufficient to achieve
the specified reliability.
Example 1: Feeding Program
In a certain barangay, Mario wants to estimate the mean weight µ, in
kilograms, of all seven-year-old children to be included in a feeding program. He
wants to be 99% confident that the estimate of µ is accurate 0.06 kg. Suppose from
a previous study, the standard deviation of the weights of the target population was
0.5kg, what should the sample size be?
Note: The phrase “accurate to within 0.06 kg” indicates a narrowed width of the
confidence interval. Thus, the decide error E= 0.06 kg.
Population standard deviation
= 0.5 kg.
10
Solution:
Steps
1. Determine the given.
2. Determine the
confidence coefficient
Solution
99% confidence, E = 0.06 kg,
a = 100% - 99%
To determine the value of
a, we can simply subtract
the confidence level from
100%.
a = 1 - .99
a = .01
𝑎/2 = 0.01/2
𝒂/𝟐 = 0.005
0.500 - 0.005 = 0.495
𝒛𝒂⁄ = 1.96
𝟐
3. Substitute the values
in the formula and
compute.
= 0.5 kg
Division property of
equality.
Subtract 0.005 from
0.500.
Hence, Using the Area
under the Standard
Normal Curve Table.
2
𝑧𝑎
⁄2 ⋅ 𝜎
𝑛=(
)
𝐸
(2.58)(0.5) 2
𝑛=(
)
0.06
2
𝑧𝑎
⁄2 ⋅ 𝜎
𝑛=(
)
𝐸
1.29 2
𝑛=(
)
0.06
𝑛 = (21.5)2
𝒏 = 𝟒𝟔𝟐. 𝟐𝟓 or 463
4. Round up the resulting
value to the nearest Thus, Mario needs a sample size of 463 .
whole number.
Note: When Determining sample size, we always round up the resulting value to
the next whole number.
Rule: When the calculated sample size is not a whole number, it should
be rounded up to the next higher whole number.
Rule: Rounding up a sample size calculation for conservativeness ensures that
your sample size will always be the representative of the population.
Reference: https://www.isixsigma.com/tools-templates/sampling-data/rounding-and-round-rules
11
Example 1.
A sample size calculation determined that 2006.083 data points were
necessary to represent the population. In this case, 2007 data points
samples should be taken.
2. Suppose the calculated sample size is 409.14. What would be the
sample size?
409.14? ____________
Answer: 410
Try this:
D. Round up the following calculated sample size.
1. n = 207.39
2. n = 347.56
3. n = 976.09
Example 2. Replicating a study
Kristine wants to replicate a certain study, where the lowest observed value
is 10.4 while the highest is 12.8. She wants to estimate the population mean µ to
within an error of 0.05 of its true value. Using 98% confidence level find the sample
size n that she needs.
Solution:
Steps
1. Determine the given.
2. Determine the
confidence coefficient.
Solution
98% confidence, E = 0.06 kg, lowest observed value is
10.4 while the highest is 12.8.
a = 100% - 98%
a = 1 - .98
a = .02
𝑎/2 = 0.02/2
𝒂/𝟐 = 0.01
0.50 - 0.01 = 0.49
𝒛𝒂⁄ = 2.33
𝟐
12
To determine the value of
a, we can simply subtract
the confidence level from
100%.
Division property of
equality.
Subtract 0.005 from
0.500.
Hence, Using the Area
under the Standard
Normal Curve Table,
3. Determine the
standard deviation.
Since the range R = 12.8 – 10.4 = 2.4
Then, using the formula
= 𝑅/4 = 2.4/4 = .6
4. Substitute the values
in the formula and
compute.
𝒛𝒂
⁄ ⋅𝝈
𝒏=( 𝟐 )
𝑬
𝑛=
𝑧𝑎⁄ ⋅𝜎 2
( 2 )
𝐸
𝑛=(
𝟐
𝑛=(
(2.33)(0.6)
)
0.06
2
1.398 2
)
0.06
𝑛 = (23.3)2
5. Round up the resulting
value to the nearest whole
number.
𝒏 = 𝟓𝟒𝟐. 𝟖𝟗 or 543
Thus, Kristine needs a sample size of 543.
E. Solve this.
1. The school nurse of a certain school wants to conduct a survey about the average
number of students who buy snacks at the school canteen. If he plans to use 98%
confidence level, 3 as the margin of error, and a standard deviation of 15. How many
sample sizes does he need for the survey?
What’s More
Activity 2
A. Round up the following calculated sample size.
1. n = 507.49
2. n = 247.51
3. n = 1, 276.08
13
4. n = 644.o41
5. n = 932.63
B. Find the length of the following confidence interval.
1. 0.241 < p < 0.653
2. 0.355 < p < 0.570
3. 0.475 < p < 0.735
4. Upper confidence limit = 0.996
Lower confidence limit = 0.436
5. Upper confidence limit = 0.886
Lower confidence limit = 0.245
C. Find the length of the confidence interval given the following data:(s – sample
standard deviation)
1. 𝑠 = 2.36, n = 350,
confidence level: 99%
2. 𝑠 = 3.35,
n = 250,
confidence level: 99%
3. 𝑠 = 3,
n = 275,
confidence level: 95%
4. 𝑠 = 6,
n = 425,
confidence level: 98%
5. 𝑠 = 9,
n = 501,
confidence level: 99%
D. Determine the sample size, given the following data: (s – sample standard
deviation)
1. 𝑠 = 5,
E = 2.45,
confidence level: 95%
2. 𝑠 = 7,
E = 3.65,
confidence level: 98%
3. 𝑠 = 4,
E = 2.76,
confidence level: 99%
4. 𝑠 = 8,
E = 2.22,
confidence level: 95%
5. 𝑠 = 3.3,
E = 1.03,
confidence level: 99%
What I Have Learned
I. Fill in the blanks. Write your answer on your answer sheet.
A confidence interval, in statistics, refers to the probability that
a _____(1)_______ parameter will fall between a set of values for a certain proportion
of times. Confidence intervals measure the degree of uncertainty or certainty in
14
a ______(2)______ method. They can take any number of probability limits, with the
most common being a 95% or 99% confidence level.
_____(3)______ is the absolute difference between the upper confidence limit
and the lower confidence limit.
There are two things to remember when we decided on the quality of the
sample size we need: ____(4)______ and the _____(5)_____ of the interval.
What I Can Do
Solve the following problems. Write your answer on your answer sheet.
1. You want to estimate the mean gasoline price within your town to the margin of
error of 5 centavos. Local newspaper reports the standard deviation for gas price in
the area is 25 centavos. What sample size is needed to estimate the mean gas prices
at 95% confidence level?
2. Carlos wants to replicate a study where the highest lowest observed value is 13.8
while the lowest is 13.4. He wants to estimate the population mean µ to the margin
of error of 0.025 of its true value. Using 99% confidence level, find the sample size n
that he needs.
Assessment
A. Find the length of the following confidence interval.
1. 0.242 < p < 0.653
2. 0.345 < p < 0.570
3. 0.275 < p < 0.463
15
4. Upper confidence limit = 0.820
Lower confidence limit = 0.490
5. Upper confidence limit = 0.715
Lower confidence limit = 0.350
B. Find the length of the confidence interval given the following data:
6. s = 5.36, n = 350, confidence level: 99%
7. s = 2.35, n = 250, confidence level: 99%
8. s = 1.20, n = 200, confidence level: 95%
9. s = 8.15, n = 29, confidence interval: 99%
10. s = 3.25, n = 17, confidence interval: 95%
C. Solve the following problems.
11. In a group presentation, the average geometric reasoning of Grade 10 students
in a Mathematics camp was observed to be 80 with s standard deviation of 4. A
researcher wants to replicate the study to estimate the true population mean µ to
within 0.5 maximum error. If the 98% level of confidence is adopted, how many
respondents are needed?
12. Teacher Carol, wants to conduct a survey about the average number of students
in a certain school who wants online class instead of distance modular learning. If
she plans to use 99% confidence level, 0.5 as the margin of error, and a standard
deviation of 5. How many sample sizes does she need for the survey?
16
Additional Activities
A. Determine the sample size, given the following data:
1. 𝑠 = 6,
E = 0.5,
confidence level= 95%
2. 𝑠 = 5,
E =0.04,
confidence level=98%
3. 𝑠 = 8,
E = 3,
confidence level=99%
4. 𝑠 = 10,
E = 5,
confidence level=95%
5. 𝑠 = 7,
E = 2,
confidence level= 99%
B. Solve the following problems.
1. A researcher wants to estimate the average number of children with congenital
heart disease who are between the ages of 1–5 years old. How many children should
be enrolled in this study, if the researcher plans on using a 95% confidence level and
wants a margin of error of 0.5 and standard deviation 4?
2. Allan, a Grade 12 senior high school student, wants to estimate the average
number of students who will pursue collage degree in a certain school. How many
sample sizes does he need, if he plans to use 98% confidence, 0.5 as the margin of
error, and a standard deviation of 5.
C. Find other methods of determining sample size, then compare these with the
formula proposed in this module. If you are to select a method, which will it be?
Explain your idea.
17
What I Have Learn
1. Population
2. Sampling
3. Length of confidence
interval
4. confidence
5. narrowness
What I Can Do
1. n = 97
2. n = 107
What’s New
A.
1. 0.115
2. 0.36
3. 0.26
4. 0.37
5. 0.331
18
Additional Activities
A.
1. 554
2. 849
3. 48
4. 16
5. 82
B.
1. 246
2. 543
C.
Students’ answers may vary.
Assessment
1. 0.411
2. 0.215
3. 0.178
4. 0.33
5. 0.365
6.1.4783
7. 0.7669
8. 0.3326
9. 8.3631
10. 3.2327
11. n = 543
12. n = 666
What I Know
1. 0.5345
2. 0.27
3. 0.302
4. 0.157
5. 0.271
6. 1.9169
7. 2.0004
8. 1.2025
9. 5.5939
10. 9.1483
What’s In
A.1. 112
2. percentage
3. 1.96
4. 0.01
5. standard error
6. Sample size
B. 1. 0.62
2. 0.2275
3. 65.33
4. 2,615.81
5. 4,160.25
What Is It
B.
1. 0.2084
2. 0.2771
3. 0.9251
C.
1. 5.1096
2. 2.8896
3. 2.6183
D.
1. 208
2. 348
3. 977
E.
1. 136
What’s More
Activity 2
A.
C.
1.508
2.248
3.1,277
4.645
5. 933
B.
D.
1. 0.412
2. 0.215
3. 0.26
4. 0.56
5. 0.641
1. 0.6509
2. 1.0933
3. 0.7092
4. 1.5018
5. 2.0748
1. 16
2. 20
3. 14
4. 50
5. 69
Answer Key
References
Books
Alonzo, George A. (2017). Statistics and Probability For Senior High School, Salinlahi
Publishing House, Inc., 1206 Cardonia St. Barangay Poblacion, Makati City,
Philippines.
Belecina, Rene R. et. al. (2016). Statistics and Probability. 1st ed. Rex Bookstore, 856
Nicanor Reyes Sr. St. Sampaloc, Manila.
Websites
Cole, Neal. “Z Score – definition and How to Use – Conversion Uplift”.
https://www.google.com/search?q=z+values+table&tbm=isch&ved=2ahUKE
wiZkOnikrfuAhUbI6YKHSBlDBEQ2-cCegQIABAA&oq=z+values&gs_lcp=CgN
pbWcQARgAMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgIIAD
ICCAAyAggAMgIIADoGCAAQBxAeOggIABAIEAcQHjoKCAAQsQMQgwEQQ1D
8oxBY6v8QYOCREWgAcAB4AIABgwGIAfYEkgEDNy4xmAEAoAEBqgELZ3dzL
Xdpei1pbWfAAQE&sclient=img&ei=x8EOYNmQLpvGmAWgyrGIAQ&bih=912
&biw=1920&rlz=1C1CHBD_enPH913PH913#imgrc=F0JP0o8NizNdrM
Confidence
Intervals.
https://handbook-51.cochrane.org/chapter_12/12_4_1_confidence_intervals.htm
iSixSigma-Editorial. “Sample Size Round-off”. https://www.isixsigma.com/toolstemplates/sampling-data/rounding-and-round-rules
Mathematics
Stack
Exchange.
“Calculate
Critical
Value”.
https://www.google.com/search?q=t+table+critical+values&rlz=1C1CHBD_e
nPH913PH913&sxsrf=ALeKk00PCvAoqB3o1ENz-ywgil0pIj0PyQ:1611579845
695&tbm=isch&source=iu&ictx=1&fir=Ak3E8SGWtJZSvM%252C3IfNW_1KN
-XacM%252C_&vet=1&usg=AI4_-kQMLXnmB3gxGk_tBC4xSwP6EoE1oA
19
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Schools Division of Bataan - Curriculum Implementation Division
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Telefax: (047) 237-2102
Email Address: bataan@deped.gov.ph