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Research Methods

William G. Zikmund

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- 1. BusinessResearch Methods William G. Zikmund Chapter 21:Univariate Statistics
- 2. Univariate Statistics• Test of statistical significance• Hypothesis testing one variable at a time
- 3. Hypothesis• Unproven proposition• Supposition that tentatively explains certain facts or phenomena• Assumption about nature of the world
- 4. Hypothesis• An unproven proposition or supposition that tentatively explains certain facts or phenomena – Null hypothesis – Alternative hypothesis
- 5. Null Hypothesis• Statement about the status quo• No difference
- 6. Alternative Hypothesis• Statement that indicates the opposite of the null hypothesis
- 7. Significance Level• Critical probability in choosing between the null hypothesis and the alternative hypothesis
- 8. Significance Level• Critical Probability• Confidence Level• Alpha• Probability Level selected is typically .05 or .01• Too low to warrant support for the null hypothesis
- 9. The null hypothesis that the mean isequal to 3.0: H o : µ = 3.0
- 10. The alternative hypothesis that themean does not equal to 3.0: H 1 : µ ≠ 3 .0
- 11. A Sampling Distribution µ=3.0 x
- 12. A Sampling Distributionα=.025 α=.02 5 µ=3.0 x
- 13. A Sampling DistributionLOWER UPPERLIMIT LIMIT µ=3.0
- 14. Critical values of µCritical value - upper limit S = µ + ZS X or µ + Z n 1.5 = 3.0 + 1.96 225
- 15. Critical values of µ = 3.0 + 1.96( 0.1) = 3.0 + .196 = 3.196
- 16. Critical values of µCritical value - lower limit S = µ - ZS X or µ - Z n 1.5 = 3.0 - 1.96 225
- 17. Critical values of µ = 3.0 − 1.96( 0.1) = 3.0 − .196 = 2.804
- 18. Region of RejectionLOWER UPPER µ=3.0LIMIT LIMIT
- 19. Hypothesis Test µ =3.02.804 3.196 3.78 µ=3.0
- 20. Type I and Type II Errors Accept null Reject nullNull is true Correct- Type I no error errorNull is false Type II Correct- error no error
- 21. Type I and Type II Errors in Hypothesis TestingState of Null Hypothesis Decisionin the Population Accept Ho Reject HoHo is true Correct--no error Type I errorHo is false Type II error Correct--no error
- 22. Calculating Zobs x−µz = sxobs
- 23. Alternate Way of Testing the Hypothesis X −µ Z obs = SX
- 24. Alternate Way of Testing the Hypothesis 3.78 − µ 3.78 − 3.0Z obs = = SX .1 0.78 = = 7.8 .1
- 25. Choosing the Appropriate Statistical Technique• Type of question to be answered• Number of variables – Univariate – Bivariate – Multivariate• Scale of measurement
- 26. PARAMETRIC NONPARAMETRIC STATISTICS STATISTICS
- 27. t-Distribution• Symmetrical, bell-shaped distribution• Mean of zero and a unit standard deviation• Shape influenced by degrees of freedom
- 28. Degrees of Freedom• Abbreviated d.f.• Number of observations• Number of constraints
- 29. Confidence Interval Estimate Using the t-distribution µ = X ± tc.l . S X S Upper limit = X + tc.l .or n S Lower limit = X − tc.l . n
- 30. Confidence Interval Estimate Using the t-distribution µ = population mean X = sample meantc.l . = critical value of t at a specified confidence levelSX = standard error of the meanS = sample standard deviationn = sample size
- 31. Confidence Interval Estimate Using the t-distribution µ = X ± t cl s x X = 3.7 S = 2.66 n = 17
- 32. upper limit = 3.7 + 2.12(2.66 17 )= 5.07
- 33. Lower limit = 3.7 − 2.12( 2.66 17 )= 2.33
- 34. Hypothesis Test Using the t-Distribution
- 35. Univariate Hypothesis Test Utilizing the t-DistributionSuppose that a production manager believesthe average number of defective assemblieseach day to be 20. The factory records thenumber of defective assemblies for each of the25 days it was opened in a given month. Themean X was calculated to be 22, and thestandard deviation, S ,to be 5.
- 36. H 0 : µ = 20H1 : µ ≠ 20
- 37. SX = S / n = 5 / 25 =1
- 38. Univariate Hypothesis Test Utilizing the t-DistributionThe researcher desired a 95 percentconfidence, and the significance levelbecomes .05.The researcher must then findthe upper and lower limits of the confidenceinterval to determine the region of rejection.Thus, the value of t is needed. For 24 degreesof freedom (n-1, 25-1), the t-value is 2.064.
- 39. Lower limit : (µ − tc.l . S X = 20 − 2.064 5 / 25 ) = 20 − 2.064(1) = 17.936
- 40. Upper limit : (µ + t c.l . S X = 20 + 2.064 5 / 25 ) = 20 + 2.064( 1) = 20.064
- 41. Univariate Hypothesis Test t-Test X −µ 22 − 20tobs = = SX 1 2 = 1 =2
- 42. Testing a Hypothesis about a Distribution• Chi-Square test• Test for significance in the analysis of frequency distributions• Compare observed frequencies with expected frequencies• “Goodness of Fit”
- 43. Chi-Square Test (Oi − Ei )²x² = ∑ Ei
- 44. Chi-Square Testx² = chi-square statisticsOi = observed frequency in the ith cellEi = expected frequency on the ith cell
- 45. Chi-Square TestEstimation for Expected Number for Each Cell Ri C j Eij = n
- 46. Chi-Square TestEstimation for Expected Number for Each CellRi = total observed frequency in the ith rowCj = total observed frequency in the jth columnn = sample size
- 47. Univariate Hypothesis Test Chi-square ExampleX 2 = ( O1 − E1 ) 2 + ( O2 − E2 ) 2 E1 E2
- 48. Univariate Hypothesis Test Chi-square ExampleX 2 = ( 60 − 50) 2 + ( 40 − 50) 2 50 50 =4
- 49. Hypothesis Test of a Proportionπ is the population proportionp is the sample proportionπ is estimated with p
- 50. Hypothesis Test of a Proportion H0 : π = . 5 H1 : π ≠ . 5
- 51. Sp = ( 0.6)( 0.4) = .24 100 100 = .0024 = .04899
- 52. p −π .6 − . 5Zobs = = Sp .04899 .1 = = 2.04 .04899
- 53. Hypothesis Test of a Proportion: Another Example n = 1,200 p = .20 pq Sp = n (.2)(.8) Sp = 1200 .16 Sp = 1200 Sp = .000133 Sp = .0115
- 54. Hypothesis Test of a Proportion: Another Example n = 1,200 p = .20 pq Sp = n (.2)(.8) Sp = 1200 .16 Sp = 1200 Sp = .000133 Sp = .0115
- 55. Hypothesis Test of a Proportion: Another Example p−πZ= Sp .20 − .15Z= .0115 .05 Z= .0115 Z = 4.348 The Z value exceeds 1.96, so the null hypothesis should be rejected at the .05 level.Indeed it is significant beyond the .001 t

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