2. What is a Hypothesis?
I assume the mean • an assumption about
GPA of this class the population
is 3.5!
parameter
• an educated guess
about the population
parameter
3. Hypotheses Testing: This is the process of
making an inference or generalization on
population parameters based on the results of
the study on samples.
Reject?
Accept?
Statistical Hypotheses: It is a guess or
prediction made by the researcher
regarding the possible outcome of the study.
5. Types of Statistical Hypotheses
Null Hypothesis (Ho): is
always hoped to be rejected
Always contains “=“ sign
Alternative Hypothesis (Ha):
•Challenges Ho
•Never contains “=“ sign
•Uses “< or > or ≠ “
•It generally represents the idea
which the researcher wants to prove.
6. The Null Hypothesis: Ho
Ex. Ho: The average GPA of this class is 3.5
µ = 3.5
H0:
The Alternative Hypothesis: Ha
Ha: The average GPA of this class is
a) higher than 3.5 (Ha: µ > 3.5)
b) lower than 3.5 (Ha: µ < 3.5)
c) not equal to 3.5 (Ha:
µ ≠ 3.5)
7. Types of Hypotheses Tests
1. One-tailed left directional test
– this is used if Ha uses < symbol
Critical value is
α = 0.05 obtained
Acceptance from the table
region
Area = 0.05
Rejection region
8. Types of Hypotheses Tests
2. One-tailed right directional test
– this is used if Ha uses > symbol
Critical value is
α = 0.05 Acceptance
obtained
from the table
region
Area= 0.05
Rejection region
9. Types of Hypotheses Tests
3. Two-tailed test: Non-directional
– this is used if Ha uses ≠ symbol
Critical value is
α = 0.05/2 obtained
Acceptance
from the table
region
Area=.025 Area=.025
Rejection region Rejection region
10. Level of Significance, α and the Rejection Region
α 0.05,
= means the probability of being right is 95% , and
the probability of being wrong is 5%. So what is α 0.01?
=
Rejection region
Acceptance Area is 0.05
Region
α = 0.05
.
11. Level of Significance, α and the Rejection Region
α= 0.01, means the researcher is taking a 1% risk of being
wrong and a 99% risk of being right. So, what is α = 0.05?
Rejection region
Area is 0.01 Acceptance
Region
α = 0.01
12. Level of Significance, α and the Rejection Region
α = 0.05 means the probability of committing Type I error is 5%.
α = 0.05, since it is 2-T, then α = 0.05/2= 0.025
Acceptance
region
Area=.025 Area=.025
Rejection region Rejection region
13. Level of Significance, α and the Rejection
Region
To summarize:
α
= 0.05, means the probability of being right is 95% and the
probability of being wrong is 5%. So what is α 0.01?
=
α= 0.01, means the researcher is taking a 1% risk of being
wrong and a 99% risk of being right. So, what is α = 0.05?
α = 0.05 means the probability of committing Type I error is
5%.
α
So what is = 0.01?
15. Errors in Hypothesis Testing
Type II (β error ) Errors
Accepting a false Ho: ERAP is not guilty
Ho!
If the court acquits
Errors in Decisions ERAP, when in fact
he is guilty, the
court commits
Type II error!
Errors in Conclusions
17. Testing the Significance of Difference Between Means
Z-test
σ is known
n ≥ 30
t-test
σ is unknown
n < 30
F-test
3 or more µ s
(ANOVA)
18. Testing the Significance of Difference Between Means
“n is large or when n ≥ 30 & σ is known.”
Z-test σ is known
n ≥ 30
• Hypothesized/population mean VS Sample mean
and population standard deviation is known.
Z=
(x − µ ) n x - is the sample mean
µ - is the population mean
σ n - is the sample size
σ - is the population std. dev.
Using PHStat: Go to…
“One-Sample Tests; Z-Test for the Mean:Sigma Known”
19. Testing the Significance of Difference Between Means
“n is large or when n ≥ 30 & σ is known.”
Z-test
σ is known
n ≥ 30
• Sample mean 1 VS Sample mean 2
and population standard deviation is known.
x1 − x 2 x1 - is the mean of sample 1
Z= x 2 - is the mean of sample 2
1 1
σ + n1 & n2 - are the sample
n1 n2 σ - is the population std. dev.
sizes
20. Testing the Significance of Difference Between Means
“n is large or when n ≥ 30 & σ is known.”
Z-test
σ is unknown
n ≥ 30
• Sample mean 1 VS Sample mean 2
and 2 sample standard deviations are known.
x1 - is the mean of sample 1
x1 − x 2
Z= x 2 - is the mean of sample 2
s12 s2 2 n1 & n2 - are the sample
+
n1 n2 s1 & s2 - are the sample std. devs.
sizes
Using Microsoft Excel: Go to…“Z-Test: Two-Sample For Means”
21. The Critical Value Approach in
Testing the Significance of Difference
Between Means
The 5-step solution
Step 1. Formulate Ho and Ha
Step 2. Set the level of significance α , usually
it is given in the problem.
Step 3. Formulate the decision rule (when to reject Ho);
Find the critical value/P-value.
Step 4. Make your decision.
Step 5. Formulate your conclusion.
22. Approaches in
Hypothesis Testing
Critical value p- value
approach approach
Computed vs. Critical p-valueα vs.α
α ≤
5-step solution 5-step solution
1.Ho: ___________ 1.Ho : ___________
Ha: ___________ Ha : ___________
2.α = ___; Cri-value= ______ 2. α = ___; p- value=________
α
3. Decision rule: Reject Ho 3. Decision rule: Reject Ho if
if Comp − value ≥ Cri − value p- value ≤ α
≤α
4. Decision: α 4. Decision:
5. Conclusion: 5. Conclusion:
23. Finding Critical Values: One-Tailed
What Is Z Given α = 0.05?
.45
.05
Α= α = .05
Z 4 5
Z=1.65
Critical value 1.5 .4382 .4394
1.6 .4495 .4505
24. Critical Values: Z - Table
α .01 0.05
Type
One-T ± 2.33 ± 1.65
Two-T ± 2.58 ± 1.96
You will refer to this table to get the critical value of Z
or the Z tabular .
25. CRITERION:
1. One-tailed test (right directional)
“Reject H0 if Zc ≥ Zt “
2. One-tailed test (left directional)
“Reject H0 if Zc ≤ Zt
3. Two-tailed test (Zc = +)
“Reject H0 if Zc ≥ Zt “
4. Two-tailed test (Zc = -)
‘Reject H0 if Zc ≤ Zt “
26. EXERCISES:
1. Past records showed that the average
final examination grade of students in
Statistics was 70 with standard deviation
of 8.0. A random sample of 100 students
was taken and found to have a mean final
examination grade of 71.8. Is this an
indication that the sample grade is better
than the rest of the students? Test at 0.05
level of significance.
27. 2. A certain type of battery is known to have
a mean life of 60 hours. In random sample
of 40 batteries, the mean life was found to
be 58 hours with a standard deviation of
4.5 hours. Does it indicate that the mean
lifetime of such battery has been reduced?
Test at 0.01 level of significance.
28. 3. The manager of a rent-a-car business
wants to know whether the true average
numbers of cars rented a day is 25 with a
standard deviation of 6.9 rentals. A
random sample of 30 days was taken and
found to have an average of 22.8 rentals.
Is there a significance between the mean
and the sample mean? Test at 0.05 level
of significance.
29. 4. Advertisements claim that the average
nicotine content of a certain kind of
cigarette is 0.30 milligram. Suspecting that
this figure is too low, a consumer
protection service takes a random sample
of 50 of these cigarette from different
production slots and find that their nicotine
content has a mean of 0.33 milligram with
a standard deviation of 0.18 milligram.
Use the 0.05 level of significance to test
the null hypothesis µ = 0.30 against the
alternative hypothesis µ < 0.30.
30. 5. An experiment was planned to compare the mean time
(in days) required to recover from common cold for
person given a daily doze of 4 mgs. of vitamin C versus
those who were not given a vitamin supplement.
Suppose that 35 adults were randomly selected for each
treatment category and that the mean recovery times
and standard deviations for the 2 groups were as
follows:
n X δ
W/ vit. C 35 5.8 1.2
W/o vit. C 35 6.9 5.8
Suppose your research objective is to show that the use
of vit. C increases the mean time required to recover
from common cold. Test using α = 0.05.
