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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
  tentative assumption about a population
  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
  competing statements about the population.
 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“,
  instead of “accept 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.
Test about Population Proportion
 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.
Test about Population Variance
 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
  return to the supplier of poor quality.
  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
 statements about the shipment:
  – 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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