Upcoming SlideShare
×

# Research Methods William G. Zikmund, Ch21

1,373 views

Published on

Research Methods
William G. Zikmund

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
1,373
On SlideShare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
175
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Research Methods William G. Zikmund, Ch21

1. 1. BusinessResearch Methods William G. Zikmund Chapter 21:Univariate Statistics
2. 2. Univariate Statistics• Test of statistical significance• Hypothesis testing one variable at a time
3. 3. Hypothesis• Unproven proposition• Supposition that tentatively explains certain facts or phenomena• Assumption about nature of the world
4. 4. Hypothesis• An unproven proposition or supposition that tentatively explains certain facts or phenomena – Null hypothesis – Alternative hypothesis
5. 5. Null Hypothesis• Statement about the status quo• No difference
6. 6. Alternative Hypothesis• Statement that indicates the opposite of the null hypothesis
7. 7. Significance Level• Critical probability in choosing between the null hypothesis and the alternative hypothesis
8. 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. 9. The null hypothesis that the mean isequal to 3.0: H o : µ = 3.0
10. 10. The alternative hypothesis that themean does not equal to 3.0: H 1 : µ ≠ 3 .0
11. 11. A Sampling Distribution µ=3.0 x
12. 12. A Sampling Distributionα=.025 α=.02 5 µ=3.0 x
13. 13. A Sampling DistributionLOWER UPPERLIMIT LIMIT µ=3.0
14. 14. Critical values of µCritical value - upper limit S = µ + ZS X or µ + Z n  1.5  = 3.0 + 1.96   225 
15. 15. Critical values of µ = 3.0 + 1.96( 0.1) = 3.0 + .196 = 3.196
16. 16. Critical values of µCritical value - lower limit S = µ - ZS X or µ - Z n  1.5  = 3.0 - 1.96     225 
17. 17. Critical values of µ = 3.0 − 1.96( 0.1) = 3.0 − .196 = 2.804
18. 18. Region of RejectionLOWER UPPER µ=3.0LIMIT LIMIT
19. 19. Hypothesis Test µ =3.02.804 3.196 3.78 µ=3.0
20. 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. 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. 22. Calculating Zobs x−µz = sxobs
23. 23. Alternate Way of Testing the Hypothesis X −µ Z obs = SX
24. 24. Alternate Way of Testing the Hypothesis 3.78 − µ 3.78 − 3.0Z obs = = SX .1 0.78 = = 7.8 .1
25. 25. Choosing the Appropriate Statistical Technique• Type of question to be answered• Number of variables – Univariate – Bivariate – Multivariate• Scale of measurement
26. 26. PARAMETRIC NONPARAMETRIC STATISTICS STATISTICS
27. 27. t-Distribution• Symmetrical, bell-shaped distribution• Mean of zero and a unit standard deviation• Shape influenced by degrees of freedom
28. 28. Degrees of Freedom• Abbreviated d.f.• Number of observations• Number of constraints
29. 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. 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. 31. Confidence Interval Estimate Using the t-distribution µ = X ± t cl s x X = 3.7 S = 2.66 n = 17
32. 32. upper limit = 3.7 + 2.12(2.66 17 )= 5.07
33. 33. Lower limit = 3.7 − 2.12( 2.66 17 )= 2.33
34. 34. Hypothesis Test Using the t-Distribution
35. 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. 36. H 0 : µ = 20H1 : µ ≠ 20
37. 37. SX = S / n = 5 / 25 =1
38. 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. 39. Lower limit : (µ − tc.l . S X = 20 − 2.064 5 / 25 ) = 20 − 2.064(1) = 17.936
40. 40. Upper limit : (µ + t c.l . S X = 20 + 2.064 5 / 25 ) = 20 + 2.064( 1) = 20.064
41. 41. Univariate Hypothesis Test t-Test X −µ 22 − 20tobs = = SX 1 2 = 1 =2
42. 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. 43. Chi-Square Test (Oi − Ei )²x² = ∑ Ei
44. 44. Chi-Square Testx² = chi-square statisticsOi = observed frequency in the ith cellEi = expected frequency on the ith cell
45. 45. Chi-Square TestEstimation for Expected Number for Each Cell Ri C j Eij = n
46. 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. 47. Univariate Hypothesis Test Chi-square ExampleX 2 = ( O1 − E1 ) 2 + ( O2 − E2 ) 2 E1 E2
48. 48. Univariate Hypothesis Test Chi-square ExampleX 2 = ( 60 − 50) 2 + ( 40 − 50) 2 50 50 =4
49. 49. Hypothesis Test of a Proportionπ is the population proportionp is the sample proportionπ is estimated with p
50. 50. Hypothesis Test of a Proportion H0 : π = . 5 H1 : π ≠ . 5
51. 51. Sp = ( 0.6)( 0.4) = .24 100 100 = .0024 = .04899
52. 52. p −π .6 − . 5Zobs = = Sp .04899 .1 = = 2.04 .04899
53. 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. 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. 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