Lesson06_static11

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Lesson06_static11

  1. 1. Statistics for Management Fundamentals of Hypothesis Testing
  2. 2. Lesson Topics <ul><li>3. One Tail Test </li></ul><ul><li>t Test of Hypothesis for the Mean </li></ul><ul><li>4. Proportions </li></ul><ul><li>Z Test of Hypothesis for the Proportion </li></ul><ul><li>5. Comparing two independent samples </li></ul>
  3. 3. <ul><li>Assumptions </li></ul><ul><ul><li>Population is normally distributed </li></ul></ul><ul><ul><li>If not normal, only slightly skewed & a large sample taken </li></ul></ul><ul><li>Parametric test procedure </li></ul><ul><li>t test statistic </li></ul>t-Test:  Unknown
  4. 4. 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,
  5. 5. <ul><li> = 0.01 </li></ul><ul><li>n = 36, df = 35 </li></ul><ul><li>Critical Value: 2.4377 </li></ul>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
  6. 6. <ul><li>Involves categorical variables </li></ul><ul><li>Fraction or % of population in a category </li></ul><ul><li>If two categorical outcomes , binomial </li></ul><ul><li>distribution </li></ul><ul><ul><li>Either possesses or doesn’t possess the characteristic </li></ul></ul><ul><li>Sample proportion ( p s ) </li></ul>4. Proportions
  7. 7. Example:Z Test for Proportion <ul><li>Problem: A marketing company claims that it receives 4% responses from its Mailing. </li></ul><ul><li>Approach: To test this claim, a random sample of 500 were surveyed with 25 responses. </li></ul><ul><li>Solution: Test at the  = .05 significance level. </li></ul>
  8. 8. <ul><li> = .05 </li></ul><ul><li>n = 500 </li></ul>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
  9. 9. 5. Comparing two independent samples <ul><li>Comparing Two Means: </li></ul><ul><ul><li>Z Test for the Difference in Two Means </li></ul></ul><ul><ul><li>(Variances Known) </li></ul></ul><ul><ul><li>t Test for Difference in Two Means </li></ul></ul><ul><ul><li>(Variances Unknown) </li></ul></ul><ul><li>Comparing two proportions </li></ul><ul><li> Z Test for Differences in Two Proportions </li></ul>
  10. 10. <ul><li>Different Data Sources: </li></ul><ul><ul><li>Unrelated </li></ul></ul><ul><ul><li>Independent </li></ul></ul><ul><li>Sample selected from one population has no effect or bearing on the sample selected from the other population. </li></ul><ul><li>Use Difference Between the 2 Sample </li></ul><ul><li>Means </li></ul><ul><li>Use Pooled Variance t Test </li></ul>Independent Samples
  11. 11. <ul><li>Assumptions: </li></ul><ul><ul><li>Samples are Randomly and Independently </li></ul></ul><ul><ul><li>drawn </li></ul></ul><ul><ul><li>Data Collected are Numerical </li></ul></ul><ul><ul><li>Population Variances Are Known </li></ul></ul><ul><ul><li>Samples drawn are Large </li></ul></ul><ul><li>Test Statistic: </li></ul>Z Test for Differences in Two Means (Variances Known)
  12. 12. <ul><li>Assumptions: </li></ul><ul><ul><li>Both Populations Are Normally Distributed </li></ul></ul><ul><ul><li>Or, If Not Normal, Can Be Approximated by </li></ul></ul><ul><ul><li>Normal Distribution </li></ul></ul><ul><ul><li>Samples are Randomly and Independently </li></ul></ul><ul><ul><li>drawn </li></ul></ul><ul><ul><li>Population Variances Are Unknown But </li></ul></ul><ul><ul><li>Assumed Equal </li></ul></ul>t Test for Differences in Two Means (Variances Unknown)
  13. 13. Developing the Pooled-Variance t Test (Part 1) <ul><li>Setting Up the Hypothesis: </li></ul>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
  14. 14. Developing the Pooled-Variance t Test (Part 2) <ul><li>Calculate the Pooled Sample Variances as an Estimate of the Common Populations Variance: </li></ul>= Pooled-Variance = Variance of Sample 1 = Variance of sample 2 = Size of Sample 1 = Size of Sample 2
  15. 15. 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) <ul><li>Compute the Test Statistic: </li></ul>( ) ) ( ( ) ( ) ( ) ( ) n 1 n 2 _ _
  16. 16. <ul><li>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: </li></ul><ul><li> NYSE NASDAQ Number 21 25 </li></ul><ul><li>Mean 3.27 2.53 </li></ul><ul><li>Std Dev 1.30 1.16 </li></ul><ul><li>Assuming equal variances, is there a difference in average yield (  = 0.05 )? </li></ul>© 1984-1994 T/Maker Co. Pooled-Variance t Test: Example
  17. 17. 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: ( ( ( ( ( ( ( ( ( ( ( ) ) ) ) ) ) ) ) ) ) )
  18. 18. <ul><li>H 0 :  1 -  2 = 0 (  1 =  2 ) </li></ul><ul><li>H 1 :  1 -  2  0 (  1   2 ) </li></ul><ul><li> = 0.05 </li></ul><ul><li>df = 21 + 25 - 2 = 44 </li></ul><ul><li>Critical Value(s): </li></ul>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
  19. 19. Z Test for Differences in Two Proportions <ul><li>Assumption: Sample is large enough </li></ul>
  20. 20. Lesson Summary <ul><li>Made Connection to Confidence Interval Estimation </li></ul><ul><li>Performed One Tail and Two Tail Tests </li></ul><ul><li>Performed t Test of Hypothesis for the Mean </li></ul><ul><li>Performed Z Test of Hypothesis for the Proportion </li></ul><ul><li>Comparing two independent samples </li></ul>

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