Stat Q4 Mod 4 Week4
Stat Q4 Mod 4 Week4
Stat Q4 Mod 4 Week4
Population Mean
Learner's Module in Statistics and
Probability
Quarter 4 ● Module 4 ● Week 4
WINNIE C. MARTES
Developer
Published by:
DepEd Schools Division of Baguio City
Curriculum Implementation Division
Learning Resource Management and Development System
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2021
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What I Know
This pre-test will determine your prior knowledge on test statistics concerning
means. If you are able to answer all the test items correctly, then you may skip
studying this learning material and proceed to the next learning module.
Direction: Choose the letter of the correct answer and write it on each blank
provided. Use capital letters.
_____1. Which of the following test statistics is used when the population standard
deviation is known and ?
A. distribution B. distribution C. ̅ D.
______9. Given:
What conclusion can be drawn from the given information?
A. Reject the null hypothesis. C.
B. Do not reject the null hypothesis. D.
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For items 10-15, refer to this word problem: In one research, the average height
of Filipinos is 156.41 cm. Suppose a random sample of 25 Filipino citizens were
selected and it was determined that their mean height is 161.2 cm and the standard
deviation is 15 cm. Using 0.01 level of significance, can it be concluded that the
average height of Filipinos has increased? Assume normality over the population.
_____10. What is the null and alternative hypothesis in the problem above?
A. C.
B. D.
_____15. Based on the critical value and the computed value of the test statistic,
what conclusion can be drawn?
A. Reject . C. The critical value is low.
B. Do not reject . D.
What’s In
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_____5. A test statistic is a quantity calculated from the sample data and its value is
used to decide whether or not the null hypothesis should be rejected in a
hypothesis test.
What’s New
Consider this situation: The leader of the association of taxi drivers claimed that the
average daily take home pay of all taxi drivers in Baguio City is Php 400.00 due to
the current pandemic. A random sample of 100 taxi drivers were interviewed and it
was found out that their average daily take home pay is Php 425.00. Use a 0.05
level of significance to find out if the average daily take home pay of all taxi drivers in
Baguio City is different from Php 400.00. Assume that the population standard
deviation is Php 92.00. To test the claim of the leader of the association of taxi
drivers, what are the steps to be done?
The claim of the leader of the association of taxi drivers is considered a hypothesis.
So to help us determine whether the claim is acceptable or not, let us take a look at
the steps in hypothesis testing:
6 • Draw a conclusion.
What Is It
The - test and the - test are two statistical tools used in hypothesis testing
concerning means. In this module, we will be studying tests for a single population
mean and difference of means tests.
The -test, on the other hand, is the test statistic used when the sample standard
deviation is known and the sample size is less than 30 ( ). To find the -
value using one sample mean, the formula used is:
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̅
with
√
where ̅ sample mean,
population mean,
sample size,
sample standard deviation
degrees of freedom
After computing the test- statistic value, decision-making and stating the conclusion
are done. Generally, if the absolute value of the computed value of or is greater
than the absolute value of the critical value (or tabular value) of or , the null
hypothesis is rejected. That is,
reject if | | | | for the -test and
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Study the following examples:
Example 1: In the example given under What’s New, the leader of the association
of taxi drivers claimed that the average daily take home pay of all taxi
drivers in Baguio City is Php 400.00 due to the current pandemic. A
random sample of 100 taxi drivers was interviewed and it was found
out that their average daily take home pay is Php 425.00. Use a 0.05
level of significance to find out if the average daily take home pay of all
taxi drivers in Baguio City is different from Php 400.00. Assume that
the population standard deviation is Php 92.00.
Solution: Given: ̅
Step 3: Select the test statistic to be used. Since the population standard deviation
is known and , the appropriate test statistic to be used is the test:
̅
.
√
Use the Areas Under the Normal Curve Table. The area 0.4750 is under column
headed 0.06. Move along this row to the left until 1.9 under column headed is
reached. Therefore, At 5% level of significance, the critical values are
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Rejection Rejection
region region
𝛼 𝛼
Non-rejection
region
𝜇
-1.96 -1.96
Critical Value Critical Value
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Solution: Given: ̅
Step 3: Select the test statistic. Since and the population standard deviation
̅
is unknown, the appropriate test statistic is the test: with
√
.
b. Significance level:
Since the alternative hypothesis contains “ ”, we have a two-tailed test. This means
that the area of 0.05 will be divided into two tails. Use the Table of - Critical Values.
Locate 11 in the first column headed . Because the test is two-tailed with
, refer to the column headed 0.05 in two-tail. Therefore, At 5% level
of significance, the critical values are . We will reject if or
.
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Rejection Rejection
region region
𝛼 𝛼
Non-rejection
region
𝜇
-2.201 -2.201
Critical Value Critical Value
Step 6: Draw a conclusion. Since the computed test-statistic value is not greater
than or | | | | , do not reject the null hypothesis.
Conclude that there is no sufficient evidence to indicate that the national
average of coffee consumption of Filipinos is different from the sample
mean from the senior citizens.
Solution: Given: ̅
Step 3: Select the test statistic. Since population standard deviation is unknown, the
sample standard deviation is given, and which is less than 30, the
̅
appropriate test statistic is the test: with .
√
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Step 5: Determine the critical value.
a. Find the degrees of freedom.
b. Significance level:
The alternative hypothesis is directional. Hence, the one-tailed test shall be used.
Use the Table of - Critical Values. Locate 24 in the first column headed .
Because the test is one-tailed with , refer to the column headed 0.01 in
one-tail. The critical value is -2.492. There is a negative sign because of the
direction of the alternative hypothesis which is “<”.
Non-rejection
region
𝛼
𝜇
-2.492
Critical Value
In comparing two means, there are assumptions and conditions that need to be met
or checked. The first one is the independence assumption. The data in each group
must be drawn independently and at random from a homogeneous population, or
generated by a randomized comparative experiment. Second is the normal
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population assumption. We should check the assumption that the underlying
populations are each Normally distributed for both groups. And finally, the
independent groups assumption. The two groups we are comparing must be
independent of each other. Since no statistical test can verify this assumption, we
have to think of how the data were collected.
̅ ̅
The formula for test using two-sample means is
√
̅ ̅
The formula for test using two-sample means is
√ ( )
Neighborhood
Sample A Sample B
The bank wishes to test the null hypothesis that the two neighborhoods have equal
mean income. What should the bank conclude? Test the hypothesis using .
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Solution: For purposes of wider options and comparative discussions, solutions for
both one-tailed and two-tailed test are presented below.
Using the Areas Under the Normal Curve Table, the area 0.4500 is between 0.4495
and 0.4505. Therefore, At 5% level of
significance, the critical value is .
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2. Two-tailed Test Using Two-Sample Means
Given: Sample A Sample B
Using the Areas Under the Normal Curve Table, the area 0.4750 is
Therefore, at 5% level of significance, the critical values are
Example 2: The weights of twelve babies who were given two types of diets were
recorded as follows:
Diet A Diet B
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At 0.01 level of significance, test whether the average weights of babies under diet
A is significantly higher than those under diet B. Assume that the date are normally
distributed.
Solution:
Given:
Diet A Diet B
c. Determine the test statistic to be used. Since the variances and are
unknown and and are both equal to 12 ( 30), use the - test for two-
sample mean:
̅ ̅
with degrees of freedom.
√
√ √
( )
The alternative hypothesis is directional so the one-tailed test shall be used. Using
the Table of - Critical Values. Locate 22 in the first column headed . Because the
test is one-tailed with , refer to the column headed 0.01 in one-tail. The
critical value is +2.508 because of the direction of the alternative hypothesis which
is “>”.
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What’s More
1. Given: ̅
2. Given: ̅
1. _________
2. _________
3. _________
4. _________
In a certain high school, male and female students were rated on their tolerance level
to school-related stress. The result are as follows:
At 0.01 level of significance, test whether the tolerance level to school-related stress
of the male students is significantly higher than the female students. Assume that the
date is normally distributed.
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What I Have Learned
To check your understanding on the test on population mean, complete the following
statements:
What I Can Do
In a long bond paper, create an infographic chart about hypothesis testing concerning
means. Be creative!
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Assessment
This assessment aims to measure how much you have learned from this module.
Direction: Read and understand each statement before choosing the correct answer.
Write your answer on the space provided.
_____1. Which of the following test statistics is used when the population standard
deviation is known and ?
A. distribution B. distribution C. ̅ D.
______9. Given:
What conclusion can be drawn from the given information?
A. Reject the null hypothesis. C.
B. Do not reject the null hypothesis. D.
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For items 10-15, refer to this word problem: In one research, the average height
of Filipinos is 156.41 cm. Suppose a random sample of 25 Filipino citizens were
selected and it was determined that their mean height is 161.2 cm and the standard
deviation is 15 cm. Using 0.01 level of significance, can it be concluded that the
average height of Filipinos has increased? Assume normality over the population.
_____10. What is the null and alternative hypothesis in the problem above?
A. C.
B. D.
_____15. Based on the critical value and the computed value of the test statistic,
what conclusion can be drawn?
A. Reject . C. The critical value is low.
B. Do not reject . D.
Additional Activity
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WHAT I KNOW/ ASSESSMENT WHAT I HAVE LEARNED: Answers may vary
1. B 11. C
2. B 12. A WHAT I CAN DO: Answers vary
3. C 13. D
4. B 14. B
5. D 15. B
6. A
7. C
8. A
9. B
10. D
WHAT’S MORE ADDITIONAL ACTIVITY: Answers vary
Activity 1:
1. 𝑧 𝑡𝑒𝑠𝑡; 𝑧
2. 𝑡 𝑡𝑒𝑠𝑡 𝑡
Activity 2: Activity 3:
1. Reject 𝐻𝑜 a. 𝐻 𝜇 𝜇 𝐻𝑎 𝜇 𝜇
2. Do not reject 𝐻𝑜 b. 𝑡 𝑡𝑒𝑠𝑡 𝑡
3. Do not reject 𝐻𝑜 c. critical value:
4. Reject 𝐻𝑜 d. Reject 𝐻𝑜
ANSWER KEY
REFERENCES
Books:
Bock, David et al. 2007. Stats Modeling the World, 473-565. Teacher’s Edition
Mercado, Jesus and Orines, Fernando. 2016. Statistics and Probability for Senior High
School, 25-41. Phoenix Publishing House, Inc.
Shio, Christian Paul and Reyes, Maria Angeli. 2017. Statistics and Probability for Senior High
School, 223-232. C & E Publishing, Inc.
Zorilla, Roland et al. 2016. Statistics and Probability for Senior High School, 84-97. Mutya
Publishing House, Inc.
Lubrica, Maria Azucena B. Statistics for the Social Sciences, 38-49. Department of Math-
Physics-Statistics, College of Arts and Sciences, Benguet State University
Online sources:
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
https://onlinelibrary.wiley.com/doi/pdf/10.1002/0471308889.app2
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