Download Introduction to Hypothesis Testing

Introduction to Hypothesis Testing for One Population Mean
Research Question
Is the average body temperature of
healthy adults different from 98.6°F?
Introduction to
Hypothesis Testing
98.2 98.5 98.3 98.1
98.3 98.7 98.1 98.4
98.2 98.4 98.3 98.2
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Statistical Hypothesis
x  98.31
s  0.17
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Statistical Hypothesis
Alternative hypothesis (HA): [or H1 or Ha]
Null hypothesis (H0):
Hypothesis of no difference or no relation,
often has =, , or  notation when testing
value of parameters.
Example:
H0:  = 98.6°F
or
H0: Average body temperature is 98.6 HT - 3
Usually corresponds to research hypothesis
and opposite to null hypothesis,
often has >, < or  notation in testing mean.
Example:
HA:   98.6°F
or
HA: Average body temperature is not 98.6°F
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Hypotheses Statements Example
• A researcher is interested in finding out
whether average life time of male is higher
than 77 years.
One Sample Z-Test for Mean
(Large sample test)
H0:  = 77
( or   77 )
HA:  > 77
[Right-tailed test]
One-Sided Test
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Introduction to Hypothesis Testing for One Population Mean
I. Hypothesis
Evidence
What will be the key statistic (evidence) to use for
testing the hypothesis about population mean?
One wishes to test whether the average
body temperature for healthy adults is
less than 98.6°F.
Sample mean:
H0:  = 98.6°F v.s. HA:  < 98.6°F
x
A random sample of 36 subjects is chosen
and the sample mean is 98.46°F and
sample standard deviation is 0.3°F.
This is a one-sided test, left-side test.
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Sampling Distribution
II. Test Statistic
x  0 x  0
z

s
sx
-2.8 0
n
98.46  98.6  0.14


  2.8
0.3
0.05
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This implies that the statistic is 2.8 standard
deviations away from the mean 98.6 in H0 , and
is to the left of 98.6 (or less than 98.6)
If H0:  = 98.6°F is true,
sampling distribution of mean → Normal
with mean = 98.6
.3
standard deviation =
= 0.05.
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 x  0.05
X
98.6
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Level of Significance
III. Decision Rule
Critical value approach: Compare the test
statistic with the critical values defined by
significance level , usually  = 0.05.
We reject the null hypothesis, if the test statistic
z < –z = –z0.05 = –1.64.
Level of significance for the test ()
A probability level selected by the
researcher at the beginning of the
analysis that defines unlikely values of
sample statistic if null hypothesis is true.
Total tail area = 
c.v. = critical value
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Rejection
region
Left-sided Test
c.v.
0
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Not a common approach!
=0.05
–1.64
–2.8
Z
0
Critical values
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Introduction to Hypothesis Testing for One Population Mean
p-value
III. Decision Rule
 p-value 
The probability of obtaining a test
statistic that is as extreme or more
extreme than actual sample statistic
value given null hypothesis is true.
It is a probability that indicates the
extremeness of evidence against H0.
The smaller the p-value, the stronger
the evidence in supporting HA and
rejecting H0 .
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IV. Draw conclusion
p-value approach: Compare the probability of the
evidence or more extreme evidence to occur when
null hypothesis is true. If this probability is less
than the level of significance of the test, , then we
reject the null hypothesis.
p-value = P(z  2.8) = .003
 = .05
Left tail area .003
Left-sided Test
Z
0
–2.8
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Decision Rule
Since from either
critical value approach z = 2.8 < z= 1.64 or
p-value approach p-value = .003 <  = .05 ,
we reject null hypothesis.
Therefore we conclude that there is sufficient
evidence to support the alternative
hypothesis that the average body
temperature is less than 98.6°F.
p-value approach: Compute p-value,
if HA :   0 , p-value = 2·P( Z  | z |)
if HA :  > 0 , p-value = P( Z  z )
if HA :  < 0 , p-value = P( Z  z )
reject H0 if p-value < 
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Steps in Hypothesis Testing
Errors in Hypothesis Testing
1. State hypotheses: H0 and HA.
2. Choose a proper test statistic, collect
data, checking the assumption and
compute the value of the statistic.
3. Make decision rule based on level of
significance().
4. Draw conclusion.
(Reject null hypothesis if p-value < .)
Possible statistical errors:
• Type I error: The null hypothesis is true,
but we reject it.
• Type II error: The null hypothesis is false,
but we don’t reject it.
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“” is the probability of committing Type I Error.

0
Z
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Introduction to Hypothesis Testing for One Population Mean
Can we see data and then
make hypothesis?
1. Choose a test statistic, collect data,
checking the assumption and compute
the value of the statistic.
2. State hypotheses: H0 and HA.
3. Make decision rule based on level of
significance().
4. Draw conclusion.
One Sample t-Test for Mean
t
x  0
s
n
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One-sample Test with
Unknown Variance  2
I. State Hypothesis
In practice, population variance is unknown
most of the time. The sample standard
deviation s2 is used instead for 2. If the
random sample of size n is from a normal
distributed population and if the null hypothesis
is true, the test statistic (standardized sample
mean) will have a t-distribution with degrees of
freedom n1.
x
Test Statistic :
t
0
One-side test example:
If one wish to test whether the body
temperature is less than 98.6 or not.
H0:  = 98.6 v.s. HA:  < 98.6
(Left-sided Test)
s
n
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II. Test Statistic
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III. Decision Rule
If we have a random sample of size 16 from a
normal population that has a mean of 98.46°F,
and a sample standard deviation 0.2. The test
statistic will be a t-test statistic and the value will
be: (standardized score of sample mean)
Test Statistic : t 
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x  0 98.46  98.6  0.14


  2.8
s
0.2
0.05
n
16
Under null hypothesis, this t-statistic has a tdistribution with degrees of freedom n – 1, that
is, 15 = 16  1.
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Critical Value Approach:
To test the hypothesis at  level 0.05,
the critical value is –t = –t0.05 = –1.753.
Rejection
Region
–1.753
–2.8
0
t
Descion Rule: Reject null hypothesis if t < –1.753
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Introduction to Hypothesis Testing for One Population Mean
III. Decision Rule
Decision Rule: Reject null hypothesis if
p-value < .
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p-value Calculation
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IV. Conclusion
p-value corresponding the test statistic:
For t test, unless computer program is used, pvalue can only be approximated with a range
because of the limitation of t-table.
p-value = P(T<2.8)
P(T<-2.8) =<? P(T<2.602) = 0.01
Since the area to the left of –2.602 is .01, the
area to the left of –2.8 is definitely less than 0.01.
Area to the left of
–2.602 is 0.01
Decision Rule:
If t < 1.753, we reject the null hypothesis, or
if p-value < 0.05, we reject the null hypothesis.
Conclusion: Since t = 2.8 < 1.753, or say
p-value < 0.01 < 0.05, we reject the null
hypothesis. There is sufficient evidence to
support the research hypothesis that the
average body temperature is less than 98.6°F.
t
Smaller than 0.01 –2.8 –2.602
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What if we wish to test whether
the average body temperature is
different from 98.6°F using t-test
with the same data?
The p-value is equal to twice the p-value of
the left-sided test which will be less than .02.
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Decision Rule
p-value approach: Compute p-value,
if HA :   0 , p-value = 2·P( T  | t |)
if HA :  > 0 , p-value = P( T  t )
if HA :  < 0 , p-value = P( T  t )
reject H0 if p-value < 
–2.8
0
2.8
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Introduction to Hypothesis Testing for One Population Mean
When do we use this t-test for
testing the mean of a population?
Remarks
• If the sample size is relatively large (>30)
both z and t tests can be used for testing
hypothesis.
• “t-test” is robust against normality.
• If the sample size is small and the sample is
from a very skewed or other non-normal
distribution, we can use nonparametric
alternatives Signed-Rank Test.
• Many commercial packages only provide t-test
since standard deviation of the population is
often unknown.
• A random sample from normally distributed
population with unknown variance.
or
• When sample size is relatively large the tscore is approximately equal to z-score
therefore t-test will be almost the same as ztest.
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Sampling Distribution
Average Weight for Female Ten
Year Children In US
S.E. =
Info. from a random sample:
x = 80 lb,
s = 18.05 lb.
Is average weight greater than 78 lb
at  = 0.05 level?
H0 : Mean is 78 lb
Ha: Mean is greater than 78 lb
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18.05
 5.71
10
n = 10
X
78
80
t
S.E. =
80  78
S .E.
18.05
 0.90
400
n = 400
X
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Practical Significance?
78
80
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Type I & Type II Error
• Type I Error: reject the null hypothesis
when it is true. The probability of a Type
I Error is denoted by .
• Type II Error: accept the null hypothesis
when it is false and the alternative
hypothesis is true. The probability of a
Type II Error is denoted by .
Power and Sample Size in
Testing One Mean
35
1 –  : Power of the test
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Introduction to Hypothesis Testing for One Population Mean
Power of the Test
Null hypothesis
The power of the statistical test is the ability
of the study as designed to distinguish between
the hypothesized value and some specific
alternative value. That is, the power is the
probability of rejecting the null hypothesis if the
null hypothesis is false.
Decision
True
False
Reject H0
Type I Error
Correct
Decision

1
Correct
Decision
1
Type II Error
Do not reject H0
Power = P(reject H0 | H0 is false)
Power = 1  

The power is usually calculated given a simple alternative
hypothesis: Ha:  = 211 mg/100ml
Power = P(reject H0 | Ha is true)
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For a two-sided test, the formula is
Formula: The estimated sample size for a onesided test with level of significance  and a power
of 1   to detect a difference of a – 0 is,
 ( z  z  )   
n

 a  0 
 ( z / 2  z  )   
n

 a  0 
2
(z  z ) 
n    /2  



2
2
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Example:
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 ( z / 2  z  )   
n

 a  0 
For testing hypothesis that
H0:  = 0 = 70 v.s.
(Two-sided Test)
2
Or, if  is the effect size, in terms of
number of standard deviations, to be
detected, the sample size would be
Or, if  is the effect size, in terms of number of
standard deviations, to be detected, the sample size
 ( z  z ) 
n 




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Ha:   70
 = .05
1   = .90
  = .10
with a level of significance of  = .05. Find
the sample size so that one can have a power
of 1   = .90 to reject the null hypothesis if the
actual mean is 5 unit different from 0. The
standard deviation, , is approximately equal
15.
2
z/2 = z.025 = 1.96
z = z.1 = 1.28

 (1.96  1.28) 15 
n
  94.48  95
5

2
The sample size needed is 95.
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