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Hypothesis Testing

Applied to population parameters by specifying H0 that
contains a null value for the population parameter—a
value that would indicate a baseline, or that nothing of
interest is happening: ―old news‖, ―no difference‖, etc.

Based on a point estimate (sample statistic), and
assessing how unlikely to obtain this sample statistic if the
null parameter value were correct.
Example

According to MA 115 B1 Intro Survey, 52 out of
108 students reported feeling stressed at the
beginning of the semester.

What is the approximate probability to obtain the
sample proportion of 48% or lower, if the true
proportion of all students who feel stressed during
the semester is 85%?
Hypothesis Testing

Achieving statistical significance is equivalent to
rejecting the idea that the observed results are
plausible if the null value is correct, i.e., rejecting the
null hypothesis (H0) in a favor of alternative
hypothesis (Ha).

Ha does not specify any specific value for the
true population parameter.

Ha gives an open interval that may contain
possible values of the true parameter, but never
contains the null value.
Hypothesis Testing
H0: μ=μ0(p=p0)
Ha: μ>μ0(p>p0)
upper-sided
Ha: μ≠μ0(p≠p0)
two-sided
Ha: μ<μ0(p<p0)
lower-sided
5 Basic Steps in Any Hypothesis Test
Step 1: Determine hypotheses (H0 and Ha).
Step 2: Verify necessary data conditions, and if met,
summarize the data into an appropriate test statistic.
Step 3: Assuming the null (H0) hypothesis is true, find either
Rejection Rule (region or the p-value).
Step 4: Decide whether or not the result is statistically
significant based on Rejection Rule.
Step 5: Report the conclusion in the context of the problem
(question of interest).
One Sample Hypothesis
Test for Population Mean
Test
Scenario
Data
Populati 1
on Mean Sample
Population
Sample
Parameter
Statistics
Response
Explan
atory
Variable
Numerical
(Age, Price)
____
Can we claim that the average GPA of all BU graduates is higher than
3.0?
Example: Testing Mean
Systolic Blood Pressure

A large, national study conducted in 2003 reported
that the mean systolic blood pressure for males
aged 50 was 130 with std =15.

2004, an investigator hypothesized that due to
increased stress in the work-place, faster-paced
lifestyles, and poorer nutritional habits, average
systolic blood pressure have increased.
Example: Testing Mean
Systolic Blood Pressure

In 2003, systolic blood pressure for males aged 50
had mean=130 with std =15.

Have average systolic blood pressure increased in
2004, on average?
Step 1: H0: μ = 130 (―no change‖)
Ha:μ>130 (i.e., mean blood pressure increased)
Example: Testing Mean
Systolic Blood Pressure

Have average systolic blood pressure increased in
2004, on average? (2003, mean=130 with std =15)
Step 2:

Select a random sample from population of interest (n
= 108 males aged 50 in 2004)

Record the systolic blood pressure on each male

Generate a point estimate for the population mean μ

Compute an appropriate test statistic
Example: Testing Mean
Systolic Blood Pressure

Have average systolic blood pressure increased in
2004, on average? (n=108, mean=130 , std =15)
Step 2: Consider the following cases: (H0:μ =130)
1.
X
2.
X 150 130
3.
X 135 130
130
P( X
135 | H 0 _ true)
P( X
135 |
130)
P( X
150 | H 0 _ true)
P( X
150 |
150)
Critical Value

If the sample mean is less than the critical
value, we will conclude that H0 is true (e.g.,
μ = 130)

If sample mean is greater than the critical
value, we will conclude that Ha is true (e.g.,
μ >130)
Test Statistic

Instead of determining critical values for sample mean
(specific to each application), we use the CLT to
standardize and produce a z-score:

Assuming H0 is true Z ~ N(0,1):

Z is close to zero then H0 is most likely true.

Z is large, then H1 is most likely true
Test Statistic

Have average systolic blood pressure increased in
2004, on average? (n=108, mean=130 , std =15)
Step 2:

Compute an appropriate test statistic
Appropriate Test Statistics
Decision/Rejection Rule
Step 3: Assuming H0 is true, how likely is

Using CLT:

Similarly:
Rejection Rule
Step 3:
 Is set up a rule reject H0 if Z> 1,
α = P(Type I error) =0.1587
 Is set up a rule reject H0 if Z> 2,
α = P(Type I error) =0.0228
Better to fix α in advance!
Rejection Rule (Table 5.4)
Rejection Rule
Say α=0.05, then we have to
find such point A, such that
P(Z≥Z1-α) =0.05, therefore
Z1-α =1.645.
Rejection Rule is given by:
 Reject Ho if Z≥ 1.645,
 and fail to reject H0 if Z< 1.645.
Rejection Rule
H0: μ=μ0(p=p0)
Ha: μ>μ0(p>p0)
upper-sided
Reject H0
if Z≥ Z1-α
(t≥ t1-α, df)
Ha: μ≠μ0(p≠p0)
two-sided
Reject H0
if Z≥ Z1-α/2 (t≥ t1-α/2,df)
or Z≤- Z1-α/2 (t≤- t1-α/2,df )
Ha: μ<μ0(p<p0)
lower-sided
Reject H0
if Z≤- Z1-α
(t≤- t1-α,df)
Decision

Step 4: The final step in the test of
hypothesis is to compare the test statistic to
the decision rule to draw a conclusion.
The test statistic falls in the rejection region
and therefore we reject H0 because
Test statistic (3.46) > critical value (1.645)

Conclusion


Step 5: Conclusion:
Based on the sample of n=108 male, there
is significant evidence, at level α = 0.05, to
conclude that the mean systolic blood
pressure for males aged 50 in 2004 has
increased from 130.
P-value
Step 4: We rejected H0 because
Test statistic (3.46) > critical value (1.645)
 Option 2: Compute p-value


p-value is the probability of getting a test
statistic as extreme or more extreme (in the
direction of Ha) than the observed value of the
test statistic, assuming the null hypothesis(H0)
is true.
P-value
p-value = P(test stat is more extreme|H0 is true)

From Step 1: H0: μ = 130 (―no change‖)
Ha:μ>130 (i.e., mean blood pressure increased)


From Step 3: Test Statistic
p-value = P(Z≥ 3.46| H0 is true=> Z~N(0,1))=
= 0.0001
P-value


If the p-value ≤ α , then the result IS statistically
significant, the decision is to reject H0.
If the p-value >α , then the result IS NOT
statistically significant, the decision is to fail to
reject H0.

p-value = P(Z≥ 3.46| H0 is true=> Z~N(0,1))=
= 0.0001<0.05 , so reject H0
Example (Student Sleep Deprivation)

A National Sleep Foundation survey found that
college/university-aged students get an average of 6.8
hours of sleep each night.

Sleep deprivation is common in college freshmen.

At the beginning of the fall semester, based on a sample of
n=108 students in MA 115, we estimated mean =7.21
hours with the std=1.12 hours.

Based on the data, is there significant evidence (at α =5%)
to conclude that students tend to sleep more at the
beginning of school year?
Example (Student Sleep Deprivation)

A National Sleep Foundation survey found that
college/university-aged students get an average of 6.8
hours of sleep each night.

Based on the data (108 MA 115 students), is there
significant evidence (at α =5%) to conclude that students
tend to sleep more at the beginning of school year?

Step 1:

Parameter: μ = _____________________________

H0:
Ha:
Significance level α = ______
Example (Student Sleep Deprivation)

A National Sleep Foundation survey found that
college/university-aged students get an average of 6.8
hours of sleep each night.

Based on the data (108 MA 115 students), is there
significant evidence (at α =5%) to conclude that students
tend to sleep more at the beginning of school year?

Step 1:

Parameter: μ = number of hours of sleep for students

H0: μ = 6.8 Ha: μ > 6.8 Significance level α = 0.05
Example (Student Sleep Deprivation)
H0: μ = 6.8 Ha: μ > 6.8 Significance level α = 0.05
At the beginning of the fall semester, based on a sample of
n=108 students in MA 115, we estimated sample mean
=7.21 hours with the s=1.12 hours.
Step 2: Compute appropriate test statistic
Appropriate Test Statistics
Example (Student Sleep Deprivation)
H0: μ = 6.8 Ha: μ > 6.8 Significance level α = 0.05
At the beginning of the fall semester, based on a sample of
n=108 students in MA 115, we estimated sample mean
=7.21 hours with the s=1.12 hours.
Step 2: Compute appropriate test statistic
Example (Student Sleep Deprivation)

Step 3: Assuming the null (H0) hypothesis is true, define
decision rule (or rejection region)

The decision rule for any hypothesis testing application
depends on three factors:
(1) Whether the test is an upper-, lower-, or two-tailed test
(2) The level of significance:
(3) The form of the test statistic:
Rejection Rule
H0: μ = 6.8 Ha: μ > 6.8
Significance level α = 0.05
H0: μ=μ0(p=p0)
Ha: μ>μ0(p>p0)
upper-sided
Reject H0
if Z≥ Z1-α
(t≥ t1-α, df)
Ha: μ≠μ0(p≠p0)
two-sided
Reject H0
if Z≥ Z1-α/2 (t≥ t1-α/2,df)
or Z≤- Z1-α/2 (t≤- t1-α/2,df )
Ha: μ<μ0(p<p0)
lower-sided
Reject H0
if Z≤- Z1-α
(t≤- t1-α,df)
Example (Student Sleep Deprivation)

Step 3: Assuming the null (H0) hypothesis is true, define
decision rule (or rejection region)

The decision rule for any hypothesis testing application
depends on three factors:
(1) Whether the test is an upper-, lower-, or two-tailed test
(2) The level of significance:
α =0.05
(3) The form of the test statistic: Z statistic.

The decision rule:
Reject H0 if Z ≥ 1.645 Do not
reject Ho if Z < 1.645
Example (Student Sleep Deprivation)



A National Sleep Foundation survey found that
college/university-aged students get an average of 6.8
hours of sleep each night.
Based on the data, is there significant evidence (at α =5%)
to conclude that students tend to sleep more at the
beginning of school year?
Step 4: We reject H0 (μ = 6.8 ) because 3.79 > 1.645.
Example (Student Sleep Deprivation)



A National Sleep Foundation survey found that
college/university-aged students get an average of 6.8
hours of sleep each night.
Based on the data, is there significant evidence (at α =5%)
to conclude that students tend to sleep more at the
beginning of school year?
Step 5: Based on the sample of n=108 students, there is
significant evidence, at level α = 0.05, to conclude that
students tend to sleep more at the beginning of school year,
on average.
Facts:

In 1997 the University of Minnesota found that students
who went to school at 7:15 a.m. got higher grades than
those who went to school at 8:40 a.m.

Randy Gardner holds the scientifically documented record
for not sleeping for 264 hours ~eleven days without using
stimulants of any kind

Never scientifically verified: Thai Ngoc, born 1942,
claimed in 2006 to have been awake for 33 years or 11,700
nights.
Facts:

Caffeine is often used over short periods to
increase alertness and counteract the effects of
sleep deprivation; however, caffeine is less
effective if taken routinely.

Other strategies recommended by the American
Academy of Sleep Medicine include prophylactic
sleep, daytime naps, increase in night sleep time.
Types of Errors
Truth
H0 True
Ha True
Decision
Type of Error
Probability
Reject H0
TYPE I
Error
P(TYPE I)=α
Fail to
Reject H0
NO Error
1-α
Reject H0
NO Error
power=1-β
Fail to
Reject H0
TYPE II
Error
P(TYPE II)=β