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Statistical Inference in Management - Lecturer's Desk

This lecture covers hypothesis testing, levels of significance, t-scores, and construction of brief summary statements. Homework on hypothesis testing with z-tests and t-tests is due. Exam grades will be posted soon.

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Statistical Inference in Management - Lecturer's Desk

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  1. Stage Screen Lecturer’s desk 11 10 9 8 7 6 5 2 14 13 12 4 3 1 Row A 14 13 12 11 10 9 6 8 7 5 4 3 2 1 Row B 28 27 26 23 25 24 22 Row C 7 6 5 Row C 2 4 3 1 21 20 19 18 17 16 13 Row C 15 14 12 11 10 9 8 22 27 28 26 25 24 23 Row D 1 Row D 6 21 20 19 18 17 16 13 7 5 4 3 2 15 14 12 11 10 9 8 Row D Row E 28 27 26 22 Row E 23 25 24 7 6 5 1 2 4 3 Row E 21 20 19 18 17 16 13 15 14 12 11 10 9 8 Row F 28 27 26 23 25 24 22 Row F 1 6 21 20 19 18 17 16 13 7 5 4 3 2 15 14 12 11 10 9 8 Row F Row G 22 27 28 26 25 24 23 7 6 5 1 Row G 2 4 3 Row G 21 20 19 18 17 16 13 15 14 12 11 10 9 8 Row H 28 27 26 22 23 25 24 Row H 6 21 20 19 18 17 16 13 7 5 4 3 2 1 15 14 12 11 10 9 8 Row H Row J 28 27 26 23 25 24 22 7 6 5 Row J 2 4 3 1 Row J 21 20 19 18 17 16 13 15 14 12 11 10 9 8 22 27 28 26 25 24 23 6 21 20 19 18 17 16 13 7 5 4 3 2 1 15 14 12 11 10 9 8 Row K Row K Row K 28 27 26 22 23 Row L 25 24 21 20 19 18 17 16 13 6 15 14 12 11 10 9 8 Row L 7 5 4 3 2 1 Row L 28 27 26 22 23 Row M 25 24 21 20 19 18 17 16 13 6 12 11 10 9 8 Row M 7 5 4 3 2 1 Row M table Projection Booth 14 13 2 1 table 3 2 1 3 2 1 Modern Languages R/L handed broken desk

  2. MGMT 276: Statistical Inference in ManagementSpring 2015 Welcome

  3. Schedule of readings Before our next exam (April 14th) Lind (10 – 12) Chapter 10: One sample Tests of Hypothesis Chapter 11: Two sample Tests of Hypothesis Chapter 12: Analysis of Variance Plous (2, 3, & 4) Chapter 2: Cognitive Dissonance Chapter 3: Memory and Hindsight Bias Chapter 4: Context Dependence

  4. By the end of lecture today 3/26/15 Logic of hypothesis testing Steps for hypothesis testing Levels of significance (Levels of alpha) what does p < 0.05 mean? what does p < 0.01 mean? Hypothesis testing with t-scores (one-sample) Hypothesis testing with t-scores (two independent samples) Constructing brief, complete summary statements

  5. Homework due – Tuesday (March 31st) On class website: Please print and complete homework worksheet #12 Hypothesis testing with z-tests and t-tests

  6. It went really well! Exam 2 – Thanks for your patience and cooperation The grades will be posted by Tuesday

  7. Remember… In a negatively skewed distribution: mean < median < mode Mode 80 = mode = tallest point 78 = median = middle score 77 = mean = balance point Frequency Mean Score on Exam Median Note: Always “frequency” Note: Label and Numbers

  8. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses Step 2: Decision rule • Alpha level? (α= .05 or .01)? • One or two tailed test? • Balance between Type I versus Type II error • Critical statistic (e.g. z or t or F or r) value? Step 3: Calculations Step 4: Make decision whether or not to reject null hypothesis If observed z (or t) is bigger then critical z (or t) then reject null Step 5: Conclusion - tie findings back in to research problem

  9. We lose one degree of freedom for every parameter we estimate Degrees of Freedom Degrees of Freedom (d.f.) is a parameter based on the sample size that is used to determine the value of the t statistic. Degrees of freedom tell how many observations are used to calculate s, less the number of intermediate estimates used in the calculation.

  10. Pop Quiz – Part 1 Standard deviation and Variance For Sample and Population These would be helpful to know by heart – please memorize these formula

  11. Pop Quiz – Part 1 Standard deviation and Variance For Sample and Population • Part 2: • When we move from a two-tailed test to a one-tailed test what happens to the critical z score (bigger or smaller?) - Draw a picture - • What affect does this have on the hypothesis test (easier or harder to reject the null?)

  12. Pop Quiz – Part 3 1.  When do we use a t-test and when do we use a z-test?         (Be sure to write out the formulae) 2.  How many steps in hypothesis testing   (What are they?) 3.  What is our formula for degrees of freedom in one sample t-test? 4.  We lose one degree of freedom for every ________________ 5.  What are the three parts to the summary (below) The mean response time for following the sheriff’s new plan was 24 minutes, while the mean response time prior to the new plan was 30 minutes. A t-test was completed and there appears to be no significant difference in the response time following the implementation of the new plan t(9) = -1.71; n.s.

  13. Pop Quiz Standard deviation and Variance For Sample and Population Critical value gets smaller • Part 2: • When we move from a two-tailed test to a one-tailed test what happens to the critical z score (bigger or smaller?) - Draw a picture - • What affect does this have on the hypothesis test (easier or harder to reject the null?) Gets easier to reject the null

  14. Pop Quiz Writing Assignment 1.  When do we use a t-test and when do we use a z-test?         (Be sure to write out the formulae) Use the t-test when you don’t know the standard deviation of the population, and therefore have to estimate it using the standard deviation of the sample Population versus sample standard deviation Population versus sample standard deviation

  15. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses How is a t score similar to a z score? How is a t score different than a z score? Same logic and same steps Step 2: Decision rule: find “critical z” score • Alpha level? (α= .05 or .01)? • One versus two-tailed test Step 3: Calculate observed z score Step 4: Compare “observed z” with “critical z” If observed z > critical z then reject null p < 0.05 and we have significant finding Step 5: Conclusion - tie findings back in to research problem

  16. Writing Assignment 3.  What is our formula for degrees of freedom in one sample t-test? One sample t-test Degrees of freedom =(df) = (n – 1) First Sample Second Sample Two sample t-test Degrees of freedom (df ) = (n1 - 1) + (n2 – 1) 4.  We lose one degree of freedom for every parameter we estimate Use the word "parameter” when describing a whole population (not just a sample). Usually we don’t know about the whole population so we have guess by using what we know about our sample. A short-hand way to let the reader know it we are describing a population (a parameter) is to use a Greek letter – for example, σ for populations standard deviation, and an s for the sample. In a t-test we never know the population standard deviation (parameter σ) we have to estimate this one parameter (using “s”), so we lose one df our degree of freedom is n-1 Sample standard deviation Parameter: Population standard deviation

  17. Writing Assignment 5.  What are the three parts to the summary (below) Finish with statistical summaryt(4) = 1.96; ns Start summary with two means (based on DV) for two levels of the IV Or if it *were* significant: t(9) = 3.93; p < 0.05 The mean response time for following the sheriff’s new plan was 24 minutes, while the mean response time prior to the new plan was 30 minutes. A t-test was completed and there appears to be no significant difference in the response time following the implementation of the new plan t(9) = -1.71; n.s. Describe type of test (t-test versus anova) with brief overview of results n.s. = “not significant” p<0.05 = “significant” n.s. = “not significant” p<0.05 = “significant” Type of test with degrees of freedom Value of observed statistic

  18. Hypothesis testing:one sample t-test Let’s jump right in and do a t-test Is the mean of my observed sample consistent with the known population mean or did it come from some other distribution? We are given the following problem: 800 students took a chemistry exam. Accidentally, 25 students got an additional ten minutes. Did this extra time make a significant difference in the scores? The average number correct by the large class was 74. The scores for the sample of 25 was Please note: In this example we are comparing our sample mean with the population mean (One-sample t-test) 76, 72, 78, 80, 73 70, 81, 75, 79, 76 77, 79, 81, 74, 62 95, 81, 69, 84, 76 75, 77, 74, 72, 75

  19. Hypothesis testing H1: = 74 µ = 74 Ho: Step 1: Identify the research problem / hypothesis Did the extra time given to this sample of students affect their chemistry test scores Describe the null and alternative hypotheses One tail or two tail test? µ

  20. Hypothesis testing Step 2: Decision rule = .05 n = 25 Degrees of freedom (df) = (n - 1) = (25 - 1) = 24 two tail test

  21. two tail test α= .05 (df) = 24 Critical t(24) = 2.064

  22. Hypothesis testing (x - x) (x - x)2 = 76.44 x 76 72 78 80 73 70 81 75 79 76 77 79 81 74 62 95 81 69 84 76 75 77 74 72 75 76 – 76.44 72 – 76.44 78 – 76.44 80 – 76.44 73 – 76.44 70 – 76.44 81 – 76.44 75 – 76.44 79 – 76.44 76 – 76.44 77 – 76.44 79 – 76.44 81 – 76.44 74 – 76.44 62 – 76.44 95 – 76.44 81 – 76.44 69 – 76.44 84 – 76.44 76 – 76.44 75 – 76.44 77 – 76.44 74 – 76.44 72 – 76.44 75– 76.44 0.1936 19.7136 2.4336 12.6736 11.8336 41.4736 20.7936 2.0736 6.5536 0.1936 0.3136 6.5536 20.7936 5.9536 208.5136 344.4736 20.7936 55.3536 57.1536 0.1936 2.0736 0.3136 5.9536 19.7136 2.0736 = -0.44 =-4.44 =+1.56 =+ 3.56 =-3.44 =-6.44 =+4.56 =-1.44 =+2.56 =-0.44 =+0.56 =+2.56 =+4.56 =-2.44 =-14.44 =+18.56 =+4.56 =-7.44 =+7.56 =-0.44 =-1.44 =+0.56 =-2.44 =-4.44 =-1.44 Step 3: Calculations µ = 74 Σx 1911 = = 25 N N = 25 = 6.01 868.16 24 Σx = 1911 Σ(x- x) = 0 Σ(x- x)2 = 868.16

  23. . Hypothesis testing = 76.44 76.44 - 74 = = 2.03 1.20 Step 3: Calculations µ = 74 N = 25 s = 6.01 76.44 - 74 6.01 critical t critical t 25 Observed t(24) = 2.03

  24. Hypothesis testing Step 4: Make decision whether or not to reject null hypothesis Observed t(24) = 2.03 Critical t(24)= 2.064 2.03 is not farther out on the curve than 2.064, so, we do not reject the null hypothesis Step 6: Conclusion: The extra time did not have a significant effect on the scores

  25. Hypothesis testing: Did the extra time given to these 25 students affect their average test score? Start summary with two means (based on DV) for two levels of the IV notice we are comparing a sample mean with a population mean: single sample t-test Finish with statistical summaryt(24) = 2.03; ns Describe type of test (t-test versus z-test) with brief overview of results Or if it had been different results that *were* significant:t(24) = -5.71; p < 0.05 The mean score for those students who where given extra time was 76.44 percent correct, while the mean score for the rest of the class was only 74 percent correct. A t-test was completed and there appears to be no significant difference in the test scores for these two groups t(24) = 2.03; n.s. n.s. = “not significant” p<0.05 = “significant” Type of test with degrees of freedom n.s. = “not significant” p<0.05 = “significant” Value of observed statistic

  26. What if we had chosen a one-tail test? µ ≤ 74 Ho: Step 1: Identify the research problem Did the extra time given to this sample of students increase their chemistry test scores Prediction is uni-directional Describe the null and alternative hypotheses One tail or two tail test? Prediction is uni-directional µ > 74 H1: Step 2: Decision rule α= .05 Degrees of freedom (df) = (n - 1) = (25 - 1) = 24 α is all at one end so “critical t” changes Critical t (24) = ?????

  27. one tail test α= .05 (df) = 24 Critical t(24) = 1.711

  28. What if we had chosen a one-tail test? µ ≤ 74 Ho: = .05 Step 1: Identify the research problem Did the extra time given to this sample of students increase their chemistry test scores Describe the null and alternative hypotheses One tail or two tail test? µ > 74 H1: Step 2: Decision rule Degrees of freedom (df) = (n - 1) = (25 - 1) = 24 Critical t (24) = 1.711

  29. Calculations (exactly same as two-tail test) (x - x)2 (x - x) = 76.44 x 76 72 78 80 73 70 81 75 79 76 77 79 81 74 62 95 81 69 84 76 75 77 74 72 75 76 – 76.44 72 – 76.44 78 – 76.44 80 – 76.44 73 – 76.44 70 – 76.44 81 – 76.44 75 – 76.44 79 – 76.44 76 – 76.44 77 – 76.44 79 – 76.44 81 – 76.44 74 – 76.44 62 – 76.44 95 – 76.44 81 – 76.44 69 – 76.44 84 – 76.44 76 – 76.44 75 – 76.44 77 – 76.44 74 – 76.44 72 – 76.44 75– 76.44 0.1936 19.7136 2.4336 12.6736 11.8336 41.4736 20.7936 2.0736 6.5536 0.1936 0.3136 6.5536 20.7936 5.9536 208.5136 344.4736 20.7936 55.3536 57.1536 0.1936 2.0736 0.3136 5.9536 19.7136 2.0736 = -0.44 =-4.44 =+1.56 =+ 3.56 =-3.44 =-6.44 =+4.56 =-1.44 =+2.56 =-0.44 =+0.56 =+2.56 =+4.56 =-2.44 =-14.44 =+18.56 =+4.56 =-7.44 =+7.56 =-0.44 =-1.44 =+0.56 =-2.44 =-4.44 =-1.44 Step 3: Calculations µ = 74 Σx 1911 = = 25 N N = 25 = 6.01 868.16 24 Σx = 1911 Σ(x- x) = 0 One-tailed test has no effect on calculations stage Σ(x- x)2 = 868.16

  30. . Calculations(exactly same as two-tail test) One-tailed test has no effect on calculations stage = 76.44 76.44 - 74 = = 2.03 1.20 Step 3: Calculations µ = 74 N = 25 s = 6.01 76.44 - 74 6.01 critical t critical t 25 Observed t(24) = 2.03

  31. Hypothesis testing = 76.44 Step 3: Calculations µ = 74 t(24) = 2.03 N = 25 s = 6.01 Step 4: Make decision whether or not to reject null hypothesis Observed t(24)= 2.03 Critical t(24) = 1.711 2.0 is farther out on the curve than 1.711, so, we do reject the null hypothesis Step 5: Conclusion: The extra time did have a significant effect on the scores

  32. Hypothesis testing: Did the extra time given to these 25 students affect their average test score? Start summary with two means (based on DV) for two levels of the IV notice we are comparing a sample mean with a population mean: single sample t-test Finish with statistical summaryt(24) = 2.03; ns Describe type of test (t-test versus z-test) with brief overview of results Or if it had been different results that *were* significant:t(24) = -5.71; p < 0.05 The mean score for those students who where given extra time was 76.44 percent correct, while the mean score for the rest of the class was only 74 percent correct. A one-tailed t-test was completed and there appears to be significant difference in the test scores for these two groups t(24) = 2.03; p < 0.05. n.s. = “not significant” p<0.05 = “significant” n.s. = “not significant” p<0.05 = “significant” Type of test with degrees of freedom Value of observed statistic

  33. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses Step 2: Decision rule • Alpha level? (α= .05 or .01)? • One or two tailed test? • Balance between Type I versus Type II error • Critical statistic (e.g. z or t or F or r) value? Step 3: Calculations Step 4: Make decision whether or not to reject null hypothesis If observed z (or t) is bigger then critical z (or t) then reject null Step 5: Conclusion - tie findings back in to research problem

  34. Hypothesis testing Is this a single sample or two sample test? Is it a z or a t test or an ANOVA? Is it a one-tail test or a two tail test? Step 1: Identify the research problem Did the sheriff keep her promise to change response times from the previous average of 30 minutes? Step 2: Describe the null and alternative hypotheses Ho: The response times are not changed H1: The response times did change As the new chief of police, I am going to change response times for traffic accidents. Before I started the average response time was 30 minutes

  35. Hypothesis testing Gather the data: We measured the time for police to respond to 10 accidents Two-tailed test n = 10 • One or two tailed test? • What is our sample size • What is size of our degrees of freedom? • What is our alpha • What is our critical t value? • n = 10 (df = 9) • Alpha = .05 • Decision rule: critical t = 1.83 df = 9 alpha = 0.05

  36. Hypothesis testing Gather the data: We measured the time for police to respond to 10 accidents • One or two tailed test? • What is our sample size • What is size of our degrees of freedom? • What is our alpha • What is our critical t value? • Decision rule: critical t = 2.262 Two-tailed test Alpha of 0.05 Critical t (9) = 2.262

  37. Step 3: Calculations: • Average time for response before 30 minutes • Average time for response after 24 minutes • Observed t = - 2.71 Step 4: Make decision whether or not to reject null hypothesis Observed t = - 2.71 Critical t = 2.262 -2.71 is farther out on the curve than 2.262 so, we do not reject the null hypothesis Step 5: Conclusion: There appears to be a significant difference between the sheriff’s times and 30 minutes

  38. Hypothesis testing: Did the sheriff keep her promise to reduce response times to less than 30 minutes? Start summary with two means (based on DV) for two levels of the IV notice we are comparing a sample mean with a population mean: single sample t-test Finish with statistical summaryt(9) = -5.71; p < 0.05 Describe type of test (t-test versus anova) with brief overview of results Or if it had been different results that *were not* significant: t(9) = -1.71; ns The mean response time for following the sheriff’s new plan was 24 minutes, while the mean response time prior to the new plan was 30 minutes. A t-test was completed and there appears to be a significant difference in the response time following the implementation of the new plan t(9) = -2.262; p < 0.05 n.s. = “not significant” p<0.05 = “significant” n.s. = “not significant” p<0.05 = “significant” Type of test with degrees of freedom Value of observed statistic

  39. . . A note on z scores, and t score: • Numerator is always distance between means • (how far away the distributions are or “effect size”) • Denominator is always measure of variability • (how wide or much overlap there is between distributions) Difference between means Difference between means Variability of curve(s)(within group variability) Variabilityof curve(s)

  40. Thank you! See you next time!!

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