The document provides an overview of hypothesis testing, including: 1) The process of hypothesis testing involves deciding between a null and alternative hypothesis based on sample data. 2) The null hypothesis states there is no difference from the claimed population parameter, while the alternative hypothesis states there is a difference. 3) Hypothesis tests use critical values and rejection regions based on the level of significance to determine whether to reject or fail to reject the null hypothesis. 4) Examples are provided to demonstrate conducting hypothesis tests using z-tests and t-tests, and interpreting the results.

1. 1
Review!
• Hypothesis Testing - the process
of hypothesis testing involves
making a decision between two
opposing hypotheses.
• Null and Alternative Hypothesis
• Either reject H0 or fail to reject
H0.

2. 2
Review!
• The null hypothesis (denoted by H0)
is a statement saying that there is
NO significant difference between
population parameter and the value
that is being claimed.
 the starting point of the investigation.
The symbolic form of the null
hypothesis is the symbol =.

3. 3
Review!
 The alternative hypothesis (denoted by
H1 or Ha or HA) is the statement saying
that there is significant difference
between population parameter and the
value that is being claimed.
 This is a statement that will be true once
the null hypothesis is rejected.
 The symbolic form of the alternative
hypothesis must use one of these
symbols: , <, >.

4. 4
Decision
Accept Null Reject Null
R
E
A Null is true
L
I Null is false
T
Y
Type II Error
(β error)
Correct
decision
Type I Error
(α error)
Correct
decision
Possible outcomes of testing

5. 5
Lesson in Life!
There is NO
ERROR/MISTAKE
when we ACCEPT the
TRUTH and REJECT
what is FALSE.

6. 6
Identifies the
Rejection region
for a given
Level of significance

7. 7
Two types of test
Z- test
 The z- test is used to
predict the value the
population mean when
the variance (σ) is known,
or even when it is
unknown provided that
the sample size is large
based on the Central
Limit Theorem (CLT)
 n ≥ 30.
T-test
 When the population
variance (σ) is unknown
and the sample size is
limited, i.e., n < 30, then,
the t-test is the
appropriate test statistic.
 Different sample sizes
have different
distributions determined
by its degree of freedom
(df), df = n-1.

8. 8
Two types of test
Z- test T-test

9. 9
LEVEL OF SIGNIFICANCE
Level of
Significance
TYPES OF TEST
One-tailed test Two-tailed test
left-tailed (<) right-tailed (>) Both right and left
tailed (≠)
-1.645 1.645 ±1.960
-2.326 2.326 ±2.575
-1.282 1.282 ±1.645

10. 10
Level of Significance, a and the Rejection Region
H0: m  3
H1: m < 3
0
0
0
H0: m  3
H1: m > 3
H0: m = 3
H1: m  3
a
a
a/2
Critical
Value(s)
Rejection Regions
In the critical value approach, the computed statistic is
compared to the critical value of the test statistic. When the
absolute value of the computed statistic is greater than the
absolute critical value, the decision is to reject 𝐻𝑜.

11. 11
Example 1.
A new food supplement is claimed by its
manufacturer that the weight of woman
is 1.5 kilograms per month with a
standard deviation of 0.65 kg. 35
women chosen at random have reported
gaining weight an average of 1.65
kilograms within a month. Does this data
support the claim of the manufacturer at
0.05 level of significance?

12. 12
Solution

13. 13

14. 14
d. Since, the critical value -1.960 < the 𝑐𝑜𝑚𝑝
𝑢𝑡𝑒𝑑 z value 1.365 and the 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 z
value 1.365 < 1.960 which falls within the
acceptance region. Therefore, the null
hypothesis is accepted.
e. There is no significant difference between
the sample mean and the population
mean. Thus, the manufacturer is correct in
claiming the weight of women is 1.5 kg per
month in using the new food supplement.

15. 15
Example 2.
A sample of 8 measurements,
randomly selected from an
approximately normally distributed
population, resulted in the summary
statistics: 𝑋̅ =5.4, s= 1.3. Test the
null hypothesis that the mean of the
population is 6 against the
alternative hypothesis μ<6. Use
α=0.05.

16. 16
Solution
a. 𝐻𝑜: μ = 6
𝐻A: μ < 6
a. Type of test: one-tailed test
Test Statistic: t-test
Level of significance: α=0.05
degree of freedom = n-1 =8-1 = 7
Critical values: -1.860

17. 17

18. 18
d. Since, t 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 = -1.305 > -1.860
and falls within the acceptance
region. Therefore, the null
hypothesis is accepted
e. The sample does not provide enough
evidence to reject the null hypothesis.
Thus, there is no significant difference
between the means.

19. 19
Example 3
According to a study done last year, the average
monthly expenses for cell phone loads of high
school students in Manila was 350.00. A statistics
students believes that this amount has increases
since January of this year. Is there a reason to
believe that this amount has really increased if a
random sample of 60 students has an average
monthly expenses for cell phone loads of 380.00?
Use a 0.05 level of significance. Assume that the
population standard deviation is 77.00.

20. 20
Example 4
A History teacher claims that the average
height of Filipino males is 163 centimeters.
A student taking up statistics randomly
selects 20 Filipino males with a standard
deviation of 2.1cm and it is found out that
the mean is 165 cm. Use 0.05 level of
significance. Can it be concluded that the
average heights of Filipino males is different
from 163 cm?

21. 21
Group Work(Performance Task)
The students will show the steps in solving the
following problems through a video
presentation.
1. A recent survey stated that adults spend an
average of 8 hours a day playing mobile games. A
random sample of 50 adults is selected from a
normally distributed population of adults and noted
an average of 6 hours playing mobile games a day
with a standard deviation of 3 hours. Using the
0.05 level of significance, would you conclude that
the statement given in the survey is correct?

22. 22
How do you
relate the topic in
your life?