stat_ch11 by liuqingyan

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```									Chapter 11 Hypothesis Testing
 A statistical hypothesis is an assertion(主張)
or conjecture(推測) concerning one or more
populations.
 In hypothesis testing, we begin by making a
parameter. This tentative assumption is
called the null hypothesis, denote by H0.
 The opposite statement is called the
alternative hypothesis, denoted by Ha.
Acceptance and Rejection
 Acceptance of a hypothesis merely implies
that the data do not give sufficient evidence
to refute it.
 On the other hand, rejection implies that the
sample evidence refutes.
 i.e. rejection means that there is a small
probability of obtaining the sample
information observed when, in fact, the
hypothesis is true.
One- and Two-Tailed Tests
 One-tailed test:
– if the alternative is one-sided
H0:m  (=)m0         H0:m  (=) m0
Ha:m < m0             Ha:m > m0
 Two-tailed test:
– if the alternative is two-sided
H0:m = m0
Ha:m  m0
Testing a Statistical Hypothesis
 The value m0 is called the critical value.
 The region whose value satisfies the null
hypothesis is called the acceptance region.
 The region whose value satisfies the
alternative hypothesis is called the critical
region.
Example 11.1
 The manufacturer of a certain brand of
cigarettes claims that the average nicotine
content does not exceed 2.5 mg. State the
null and alternative hypotheses.
H0:m  2.5
Ha:m > 2.5
Example 11.2
 A real estate claims that 60% of all private
residences being built today are 3-bedroom
homes. To test this claim, a large sample of
new residences is inspected; the proportion
of these homes with 3 bedrooms is recorded
and used as our test statistic. State the null
and alternative hypotheses.
H0:p = 0.6
Ha:p  0.6
Testing Research Hypothesis
 The rejection of H0 supports the conclusion
and action being sought.
 The research hypothesis therefore should be
expressed as the alternative hypothesis.
Testing the Validity of a Claim
 The null hypothesis is generally based on
the assumption that the claim is true.
 The alternative hypothesis is then
formulated so that rejection of H0 will
provide statistical evidence that the stated
assumption is incorrect.
 Action to correct the claim should be
considered whenever H0 is rejected.
Testing in Decision-Making
 In previous tow cases, action is taken if H0
is rejected.
 In many instances, however, action must be
taken both when H0 cannot be rejected and
when H0 can be rejected.
Conclusion in Hypothesis Test
 The null and alternative hypotheses are
 Either the null hypothesis H0 is true or the
alternative hypothesis Ha is true, but not
both.
 Ideally, the hypothesis testing procedure
should lead to the acceptance of H0 when
H0 is true and the rejection of H0 when Ha
is true.
Type I and II Error
Type I Error
 Rejection of the null hypothesis when it is
true is called the type I error.
 The probability of committing a type I error
is called the level of significance, denoted
by a .
 We can control the probability of making
type I error by using level of significance.
 Commonly choices for a are 0.05 and 0.01.
Type II Error
 Acceptance of the null hypothesis when it is
false is called type II error.
 We do not control the probability of making
type II error.
 If we decide to accept H0 , we cannot
determine how confident we can be with
that decision.
 We use the statement “do not reject H0“,
p-value
 The p-value is the probability of obtaining a
sample result that is at least as unlikely as
what is observed.
 A small p-value indicates that the sample
results is unusual given the assumption that
H0 is true.
 As with the hypothesis tests, a small p-value
leads to the rejection of H0
p-value
 The p-value approach is designed to give
the user an alternative (in terms of
probability) to a mere “reject” or “do not
reject” conclusion.
 If p-value < a then we reject H0
– Two-tailed: P=2*P(z>Z when m = m0 )
– One-tailed: P=P(z>Z when m = m0 ) when Z>0
– One-tailed: P=P(z<Z when m = m0 ) when Z<0
Steps of Hypothesis Testing
 Determine the null and alternative
hypotheses.
 Select the test statistic that will be used.
 Specify the level of significance.
 Use the level of significance to develop the
rejection rule.
 Collect the sample data and compute the
value of the test statistic.
Steps of Hypothesis Testing
 Compare the value of test statistic to the
critical values specified in the rejection rule
to determine whether or not H0 should
rejected.
 Compute the p-value based on the test
statistic in Step 5. Use the p-value to
determine whether or not H0 should rejected.
One-Tailed Tests of m (n30)
H0:m  (=) m0
Ha:m < m0
 It is convenient to standardize  and
formally involve the standard normal
random variable Z, where
s known (n30)
 Under H0 , if m  m0 then Z has an N(0,1)
distribution, and hence the expression

can be used to write an approximate
acceptance region.
s known (n30)
 The critical region is designed to control a,
the probability of type I error.
 Given a computed value        , the formal test
involves:
– if the computed test statistic Z<- Za , reject H0.
– If Z> -Za, do not reject H0.
One-Tailed Tests of m (n30)
H0:m  (=) m0
Ha:m > m0
 Under H0 , if m  m0 the expression

can be used to write an approximate
acceptance region.
s known (n30)
 Given a computed value           , the formal test
involves:
– if the computed test statistic Z> Za , reject H0.
– If Z <Za, do not reject H0.
Example 11.3
 A random sample of 100 recorded deaths in
the United States during the past year
showed an average life span of 71.8 years.
Assuming a population standard deviation
of 8.9 years, does this seem to indicate that
average life span today is greater than 70
years? Use a 0.05 level of significance.
Solution
H0:m = 70 years Ha:m >70 years
a=0.05
Critical region: Z > 1.645, where

Conclusion: reject H0 and conclude that the
average life span today is greater than 70
years.
Solution
 p-value:
Use the table, we have
P=P(z>2.02)=0.0217 < 0.05
As a result, the evidence in favor of Ha is
even stronger than that by a 0.05 level of
significance.
s unknown (n30)
 It is convenient to standardize  and
formally involve the standard normal
random variable Z, where
One-Tailed Tests (n<30)
 If the population has a normal distribution,
it is convenient to standardize     .
– Use the student-t random variable T, when s is
unknown.

– Or use the standard normal random variable Z,
when s is known.
s unknown (n<30)
 Under H0 , if m  (=) m0 then T has an t-
distribution, and hence the expression

can be used to write an approximate
acceptance region.
s unknown (n<30)
 Given a computed value           , the formal test
involves:
– if the computed test statistic T<- ta , reject H0.
– If T>- ta , do not reject H0.
s unknown (n<30)
 Under H0 , if m  (=) m0 then T has an t-
distribution, and hence the expression

can be used to write an approximate
acceptance region.
s unknown (n<30)
 Given a computed value           , the formal test
involves:
– if the computed test statistic T> ta , reject H0.
– If T< ta , do not reject H0.
Example 11.4
 The Edison Electric Institute has claimed
that a vacuum cleaners expends an average
of 46 kwh/ year. If a random sample of 12
homes included in a planned studied
indicates that vacuum cleaners expend an
average of 42 kwh/year with standard
deviation of 11.9 kwh, does this suggest at
the 0.05 level of significance that vacuum
cleaners expend, on average, less than 46
kwh/year? Assume normal population.
Solution
H0:m = 46 kwh Ha:m <46 kwh         a=0.05
Critical region: t < -1.796, where

Conclusion:Do not reject H0 and conclude
that the average number of kw expanded
annually by vacuum cleaners is not
significantly less than 46.
Solution
 p-value:
Use the table, we have
P=P(T<-1.16)=0.135 > 0.05
Conclusion: Do not reject H0
Two-Tailed Tests of m (n30)
H0:m = m0
Ha:m  m0
 It is convenient to standardize  and
formally involve the standard normal
random variable Z, where
s known (n30)
 Under H0 , if m = m0 then Z has an N(0,1)
distribution, and hence the expression

can be used to write an approximate
acceptance region.
s known (n30)
 The critical region is designed to control a,
the probability of type I error.
 Given a computed value        , the formal test
involves:
– if the computed test statistic Z>Za/2 or Z<- Za/2 ,
reject H0.
– If -Za/2 <Z< Za/2 , do not reject H0.
Example 11.5
 A manufacturer of sports equipment has
developed a new synthetic fishing line that
he claims has a mean breaking strength of 8
kg with a standard deviation of 0.5 kg. Test
the hypothesis that m=8 kg against the
alternative that m8 kg if a random sample
of 50 lines is tested and found to have a
mean breaking strength of 7.8 kg. Use a
0.01 level of significance.
Solution
H0:m = 8 kg      Ha:m  8 kg a=0.01
Critical region: Z<-2.575 and Z > 2.575,
where

Conclusion: reject H0 and conclude that the
average breaking strength is not equal to 8
kg, in fact, is less than 8 kg.
Solution
 p-value:
Use the table, we have
P=P(|z|>2.83)=0.0046 < 0.01
reject the null hypothesis that m=8 kg at a
level of significance smaller than 0.01.
s unknown(n<30)
 If the population has a normal distribution,
it is convenient to standardize     .
– Use the student-t random variable T, when s is
unknown.

– Or use the standard normal random variable Z,
when s is known.
s unknown (n<30)
 Under H0 , if m = m0 then T has an t-
distribution, and hence the expression

can be used to write an approximate
acceptance region.
s unknown (n<30)
 The critical region is designed to control a,
the probability of type I error.
 Given a computed value        , the formal test
involves:
– if the computed test statistic T>ta/2 or T<- ta/2 ,
reject H0.
– If -ta/2 <T< ta/2 , do not reject H0.
s known (n<30)
 Given a computed value           , the formal test
involves:
– if the computed test statistic Z>Za/2 or Z<- Za/2 ,
reject H0.
– If -Za/2 <Z< Za/2 , do not reject H0.
Relation to Confidence Interval
 (1-a)100% confidence interval:

when s is known. And

when s is unknown.
A Confidence Interval Approach
H0:m = m0
Ha:m  m0
1 Compute the confidence interval m0 .
2 If the confidence interval contains the
hypothesized value m0 ,do not reject H0.
3 Otherwise, reject H0.
 One-tailed test:
– if the alternative is one-sided
H0: p  (=)p0        H0: p  (=)p0
Ha : p < p 0          Ha : p > p 0
 Two-tailed test:
– if the alternative is two-sided
H0 : p = p 0
Ha : p  p 0
Test Statistic
One-Tailed Test of P
H0: p  (=)p0
Ha : p > p 0
 Given a computed value Z , the formal test
involves:
– if the computed test statistic Z>Za reject H0.
– If Z< Za , do not reject H0.
One-Tailed Test of P
H0: p  (=)p0
Ha : p < p 0
 Given a computed value Z, the formal test
involves:
– if the computed test statistic Z<- Za , reject H0.
– If -Za <Z , do not reject H0.
Two-Tailed Test of P
H0: p = p0
Ha : p  p 0
 Given a computed value Z, the formal test
involves:
– if the computed test statistic Z>Za/2 or Z<- Za/2 ,
reject H0.
– If -Za/2 <Z< Za/2 , do not reject H0.
p-value
 If p-value < a then we reject H0
– Two-tailed: P=2*P(z>Z when p = p0 )
or P=(|z| >Z when p=p0 )
– One-tailed: P=P(z>Z when p = p0 ) when Z>0
or P=(z < Z when p=p0 ) when Z<0
p-value
H0: p  (=)p0
Ha : p < p 0
P=P(z<Z when p=p0)
H0: p  (=)p0
Ha : p > p 0
P=P(z > Z when p=p0)
H0 : p = p 0
Ha : p  p 0
P=(|z| > Z when p=p0)
Example 11.6
 A builder claims that heat pumps are
installed in 70% of all homes being
constructed today in the city of Richmond.
Would you agree with this claim if a
random survey of new homes in this city
shows that 8 out of 15 had heat pumps
installed? Use a 0.10 level of significance.
Solution
H0: p = 0.7
Ha: p  0.7
a=0.10 Z0.05=1.645 - Z0.05=-1.645

z > - Z0.05  Do not reject H0
Solution
 p-value:
Use the table, we have
P=2*P(z  -1.4124)=0.1586 > 0.10
 Do not reject H0
 Conclusion: there is insufficient reason to
doubt the builder’s claim.
Binomial Distribution Approach
Binomial variable X with p=0.7 and n=15
x=8 and np0=15*0.7=10.5
p-value:

 Do not reject H0
 Conclusion: there is insufficient reason to
doubt the builder’s claim.
 One-tailed test:
– if the alternative is one-sided
H0: s2  (=) s20 H0: s2  (=) s20
Ha: s2 < s20          Ha: s2 > s20
 Two-tailed test:
– if the alternative is two-sided
H0: s2 = s20
Ha: s2  s20
Test Statistic
One-Tailed Test of             s 2

H0: s2  (=) s20
Ha: s2 > s20
 Given a computed value x2 , the formal test
involves:
– if the computed test statistic x2 > x2a reject H0.
– If x2 < x2a , do not reject H0.
One-Tailed Test of              s 2

H0: s2  (=) s20
Ha: s2 < s20
 Given a computed value x2, the formal test
involves:
– if the computed test statistic x2 < x21-a , reject
H0 .
– If x2 > x21-a , do not reject H0.
Two-Tailed Test of                s 2

H0: s2 = s20
Ha: s2  s20
 Given a computed value x2, the formal test
involves:
– if the computed test statistic x2 > x2a/2 or x2 <
x21-a/2 , reject H0.
– If x21-a/2 < x2 < x2a/2 , do not reject H0.
p-value
 If p-value < a then we reject H0
– Two-tailed: P=2*P(X2> x2 when s2 = s20 )
or P=(| X2 | > x2 when s2 = s20 )
– One-tailed: P=P(X2 > x2 when s2 = s20 )
or P=(X2 < x2 when s2 = s20 )
Calculating Type II Error
H0:m  (=)m0
Ha:m < m0
Calculating Type II Error

Denoting the probability of making type II
error as b .
Example 11.7
 A quality manager must decide to accept a
shipment of batteries from a supplier or
Assume that the required battery life is at
120 hours. To evaluate the quality of an
incoming shipment, a sample of 36 batteries
are sampled. The standard deviation of the
population is 12. What is the type II error if
the actual mean life is 112 and the manager
accept this shipment?
Solution
H0:m  120
Ha:m < 120
a= 0.05, s =12, n =36
The test statistic is
Solution
 If m=112 is really true, what is the
probability of accepting H0: m120 and
hence committing type II error?
Type II Error

 The probability of correctly rejecting H0
when it is false is called power, denoted by
1-b .
Determine the Sample Size
H0:m  (=)m0
Ha:m < m0

Notation:
Za = z-value for type I error.
Zb = z-value for type II error
m0 = population mean in Ho
ma = population mean used for type II error
Determine the Sample Size
Determine the Sample Size
Example 11.8 (followed 11.7)
 Assume the manager make the following
– Type I error: If the mean life of the batteries in
the shipment is 120, I am willing to risk an a =
0.05 probability to reject this shipment.
– Type II error: If the mean life of batteries in the
shipment is 115, I am willing to risk a b = 0.10
probability to accept this shipment.
How many samples should he take?
Solution

The recommended sample size is 50 to
satisfy the manager’s allowance.

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