Lesson05_Static11

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  • 1. Statistics for Management Fundamentals of Hypothesis Testing
  • 2. Lesson Topics 1. What is a Hypothesis? Hypothesis Testing Methodology Hypothesis Testing Process Level of Significance, a Errors in Making Decisions 2. Hypothesis Testing: Steps Z Test for the Mean (s Known) Connection to Confidence Interval Estimation Hypothesis Testing Methodology
  • 3.
    • A hypothesis is an assumption about the population parameter.
      • A parameter is a Population mean or proportion
      • The parameter must be identified before analysis.
    I assume the mean GPA of this class is 3.5! © 1984-1994 T/Maker Co. 1. What is a Hypothesis?
  • 4.
    • States the Assumption (numerical) to be tested
    • e.g. The average # TV sets in US homes is at least 3 ( H 0 :  3 )
    • Begin with the assumption that the null hypothesis is TRUE.
        • (Similar to the notion of innocent until proven guilty )
    The Null Hypothesis, H 0
    • Always contains the ‘ = ‘ sign
    • The Null Hypothesis may or may not be rejected.
  • 5.
    • Is the opposite of the null hypothesis e.g. The average # TV sets in US homes is less than 3 ( H 1 :  < 3 )
    • Never contains the ‘=‘ sign
    • The Alternative Hypothesis may or may not be accepted
    The Alternative Hypothesis, H 1
  • 6.
    • Steps:
      • State the Null Hypothesis ( H 0 :   3)
      • State its opposite, the Alternative Hypothesis ( H 1 :  < 3)
        • Hypotheses are mutually exclusive & exhaustive
        • Sometimes it is easier to form the alternative hypothesis first.
    Identify the Problem
  • 7. Population Assume the population mean age is 50. (Null Hypothesis) REJECT The Sample Mean Is 20 Sample Null Hypothesis Hypothesis Testing Process No, not likely!
  • 8. Sample Mean  = 50 Sampling Distribution It is unlikely that we would get a sample mean of this value ... ... if in fact this were the population mean. ... Therefore, we reject the null hypothesis that  = 50. 20 H 0 Reason for Rejecting H 0
  • 9.
    • Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True
      • Called Rejection Region of Sampling
      • Distribution
    • Designated   (alpha)
      • Typical values are 0.01, 0.05, 0.10
    • Selected by the Researcher at the Start
    • Provides the Critical Value(s) of the Test
    Level of Significance, 
  • 10. Level of Significance,  and the Rejection Region H 0 :   3 H 1 :  < 3 0 0 0 H 0 :   3 H 1 :  > 3 H 0 :   3 H 1 :   3    /2 Critical Value(s) Rejection Regions
  • 11.
    • Type I Error
      • Reject True Null Hypothesis
      • Has Serious Consequences
      • Probability of Type I Error Is 
        • Called Level of Significance
    • Type II Error
      • Do Not Reject False Null Hypothesis
      • Probability of Type II Error Is  (Beta)
    Errors in Making Decisions
  • 12. H 0 : Innocent Jury Trial Hypothesis Test Actual Situation Actual Situation Verdict Innocent Guilty Decision H 0 True H 0 False Innocent Correct Error Do Not Reject H 0 1 -  Type II Error (  ) Guilty Error Correct Reject H 0 Type I Error (  ) Power (1 -  ) Result Possibilities
  • 13.   Reduce probability of one error and the other one goes up.  &   Have an Inverse Relationship
  • 14.
    • True Value of Population Parameter
      • Increases When Difference Between Hypothesized Parameter & True Value Decreases
    • Significance Level 
      • Increases When  Decreases
    • Population Standard Deviation 
      • Increases When   Increases
    • Sample Size n
      • Increases When n Decreases
    Factors Affecting Type II Error,       n
  • 15.
    • Convert Sample Statistic (e.g., ) to Standardized Z Variable
    • Compare to Critical Z Value(s)
      • If Z test Statistic falls in Critical Region, Reject H 0 ; Otherwise Do Not Reject H 0
    Z-Test Statistics (  Known) Test Statistic X
  • 16.
    • 1. State H 0 H 0 :  3
    • 2. State H 1 H 1 : 
    • 3. Choose   = .05
    • 4. Choose n n = 100
    • 5. Choose Test : Z Test (or p Value)
    2. Hypothesis Testing: Steps Test the Assumption that the true mean # of TV sets in US homes is at least 3.
  • 17.
    • 6. Set Up Critical Value(s) Z = -1.645
    • 7. Collect Data 100 households surveyed
    • 8. Compute Test Statistic Computed Test Stat.= -2
    • 9. Make Statistical Decision Reject Null Hypothesis
    • 10. Express Decision The true mean # of TV set is less than 3 in the US households.
    Hypothesis Testing: Steps Test the Assumption that the average # of TV sets in US homes is at least 3. (continued)
  • 18.
    • Assumptions
      • Population Is Normally Distributed
      • If Not Normal , use large samples
      • Null Hypothesis Has  or   Sign Only
    • Z Test Statistic:
    3. One-Tail Z Test for Mean (  Known)
  • 19. Z 0  Reject H 0 Z 0 Reject H 0  H 0 :  H 1 :  < 0 H 0 :  0 H 1 :  > 0 Must Be Significantly Below  = 0 Small values don’t contradict H 0 Don’t Reject H 0 ! Rejection Region
  • 20.
    • Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed X = 372.5 . The company has specified   to be 15 grams. Test at the  0.05 level.
    368 gm. Example: One Tail Test H 0 :  368 H 1 :  > 368 _
  • 21. Z .04 .06 1.6 . 5495 . 5505 .5515 1.7 .5591 .5599 .5608 1.8 .5671 .5678 .5686 .5738 .5750 Z 0  Z = 1 1.645 .50 -. 05 .45 . 05 1.9 .5744 Standardized Normal Probability Table (Portion) What Is Z Given   = 0.05 ?  = .05 Finding Critical Values: One Tail Critical Value = 1.645
  • 22.
    •  = 0.05
    • n = 25
    • Critical Value: 1.645
    Test Statistic: Decision: Conclusion: Do Not Reject at  = .05 No Evidence True Mean Is More than 368 Z 0 1.645 .05 Reject Example Solution: One Tail H 0 :  368 H 1 :  > 368
  • 23.
    • Does an average box of cereal contains 368 grams of cereal? A random sample of 25 boxes showed X = 372.5 . The company has specified   to be 15 grams. Test at the  0.05 level.
    368 gm. Example: Two Tail Test H 0 :  368 H 1 :  368
  • 24.
    •  = 0.05
    • n = 25
    • Critical Value: ±1.96
    Test Statistic: Decision: Conclusion: Do Not Reject at  = .05 No Evidence that True Mean Is Not 368 Z 0 1.96 .025 Reject Example Solution: Two Tail -1.96 .025 H 0 :  386 H 1 :  386
  • 25. Connection to Confidence Intervals
      • For X = 372.5oz,  = 15 and n = 25,
    • The 95% Confidence Interval is:
    • 372.5 - ( 1.96 ) 15 / 25 to 372.5 + ( 1.96 ) 15 / 25
    • or
    • 366.62    378.38
    • If this interval contains the Hypothesized mean ( 368 ), we do not reject the null hypothesis.
    • It does. Do not reject.
    _
  • 26.
    • Assumptions
      • Population is normally distributed
      • If not normal, only slightly skewed & a large sample taken
    • Parametric test procedure
    • t test statistic
    t-Test:  Unknown
  • 27. Example: One Tail t-Test Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5 , and  S= 15 . Test at the  0.01 level. 368 gm. H 0 :   368 H 1 :  368  is not given,
  • 28.
    •  = 0.01
    • n = 36, df = 35
    • Critical Value: 2.4377
    Test Statistic: Decision: Conclusion: Do Not Reject at  = .01 No Evidence that True Mean Is More than 368 Z 0 2.4377 .01 Reject Example Solution: One Tail H 0 :  368 H 1 :  368
  • 29.
    • Involves categorical variables
    • Fraction or % of population in a category
    • If two categorical outcomes , binomial
    • distribution
      • Either possesses or doesn’t possess the characteristic
    • Sample proportion ( p s )
    4. Proportions
  • 30. Example:Z Test for Proportion
    • Problem: A marketing company claims that it receives 4% responses from its Mailing.
    • Approach: To test this claim, a random sample of 500 were surveyed with 25 responses.
    • Solution: Test at the  = .05 significance level.
  • 31.
    •  = .05
    • n = 500
    Do not reject at  = .05 Z Test for Proportion: Solution H 0 : p  .04 H 1 : p  .04 Critical Values:  1.96 Test Statistic: Decision: Conclusion: We do not have sufficient evidence to reject the company’s claim of 4% response rate. Z  p - p p (1 - p) n s = .04 -.05 .04 (1 - .04) 500 = -1.14 Z 0 Reject Reject .025 .025
  • 32. 5. Comparing two independent samples
    • Comparing Two Means:
      • Z Test for the Difference in Two Means
      • (Variances Known)
      • t Test for Difference in Two Means
      • (Variances Unknown)
    • Comparing two proportions
    • Z Test for Differences in Two Proportions
  • 33.
    • Different Data Sources:
      • Unrelated
      • Independent
    • Sample selected from one population has no effect or bearing on the sample selected from the other population.
    • Use Difference Between the 2 Sample
    • Means
    • Use Pooled Variance t Test
    Independent Samples
  • 34.
    • Assumptions:
      • Samples are Randomly and Independently
      • drawn
      • Data Collected are Numerical
      • Population Variances Are Known
      • Samples drawn are Large
    • Test Statistic:
    Z Test for Differences in Two Means (Variances Known)
  • 35.
    • Assumptions:
      • Both Populations Are Normally Distributed
      • Or, If Not Normal, Can Be Approximated by
      • Normal Distribution
      • Samples are Randomly and Independently
      • drawn
      • Population Variances Are Unknown But
      • Assumed Equal
    t Test for Differences in Two Means (Variances Unknown)
  • 36. Developing the Pooled-Variance t Test (Part 1)
    • Setting Up the Hypothesis:
    H 0 :  1   2 H 1 :  1 >  2 H 0 :  1 -  2 = 0 H 1 :  1 -  2  0 H 0 :  1 =  2 H 1 :  1   2 H 0 :  1   2  H 0 :  1 -  2  0 H 1 :  1 -  2 > 0 H 0 :  1 -  2  H 1 :  1 -  2 < 0 OR OR OR Left Tail Right Tail Two Tail  H 1 :  1 <  2
  • 37. Developing the Pooled-Variance t Test (Part 2)
    • Calculate the Pooled Sample Variances as an Estimate of the Common Populations Variance:
    = Pooled-Variance = Variance of Sample 1 = Variance of sample 2 = Size of Sample 1 = Size of Sample 2
  • 38. t X X S n S n S n n df n n P                 1 2 1 2 2 1 1 2 2 2 2 1 2 1 2 1 1 1 1 2   Hypothesized Difference Developing the Pooled-Variance t Test (Part 3)
    • Compute the Test Statistic:
    ( ) ) ( ( ) ( ) ( ) ( ) n 1 n 2 _ _
  • 39.
    • You’re a financial analyst for Charles Schwab. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data:
    • NYSE NASDAQ Number 21 25
    • Mean 3.27 2.53
    • Std Dev 1.30 1.16
    • Assuming equal variances, is there a difference in average yield (  = 0.05 )?
    © 1984-1994 T/Maker Co. Pooled-Variance t Test: Example
  • 40. t X X S n n S n S n S n n P P                            1 2 1 2 2 1 2 2 1 1 2 2 2 2 1 2 2 2 3 27 2 53 0 1 510 21 25 2 03 1 1 1 1 21 1 1 30 25 1 1 16 21 1 25 1 1 510   . . . . . . . Calculating the Test Statistic: ( ( ( ( ( ( ( ( ( ( ( ) ) ) ) ) ) ) ) ) ) )
  • 41.
    • H 0 :  1 -  2 = 0 (  1 =  2 )
    • H 1 :  1 -  2  0 (  1   2 )
    •  = 0.05
    • df = 21 + 25 - 2 = 44
    • Critical Value(s):
    Test Statistic: Decision: Conclusion: Reject at  = 0.05 There is evidence of a difference in means. t 0 2.0154 -2.0154 .025 Reject H 0 Reject H 0 .025 t    3 27 2 53 1 510 21 25 2 03 . . . . Solution
  • 42. Z Test for Differences in Two Proportions
    • Assumption: Sample is large enough