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Please start portfolios

Please start portfolios. MGMT 276: Statistical Inference in Management Room McClelland Hall, Room 129 Summer II, 2011. Welcome. Please read: Chapters 5, 7, 8, 9 in Lind book and Chapters 10, 11, 12 and 14 in Plous book : (Before the next exam) Lind

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Please start portfolios

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  1. Please start portfolios

  2. MGMT 276: Statistical Inference in ManagementRoom McClelland Hall, Room 129Summer II, 2011 Welcome

  3. Please read: Chapters 5, 7, 8, 9 in Lind book and Chapters 10, 11, 12 and 14 in Plous book: (Before the next exam) Lind Chapter 5: Probability Concepts Chapter 7: Probability Distributions Chapter 8: Sampling and Central Limit Theorem Chapter 9: Estimation and Confidence Intervals Plous Chapter 10: The Representativeness Heuristic Chapter 11: The Availability Heuristic Chapter 12: Probability and Risk Chapter 14: The Perception of Randomness

  4. Use this as your study guide By the end of lecture today7/26/11 Logic of hypothesis testing Steps for hypothesis testing Hypothesis testing with t-scores (two independent samples) Interpreting excel output of hypothesis tests Constructing brief, complete summary statements

  5. Questions from the homework assignment?

  6. Questions from the homework assignment? 95% 26.08 < µ< 33.92 mean + z σ = 30 ± (1.96)(2) 24 26 28 30 32 34 36 99% 24.84 < µ< 35.16 mean + z σ = 30 ± (2.58)(2) 24 26 28 30 32 34 36

  7. Melvin Melvin Mark Difference not due sample size because both samples same size Difference not due population variability because same population Yes! Difference is due to sloppiness and random error in Melvin’s sample Melvin

  8. √ z- score : because we know the population standard deviation Ho: µ = 5 Bags of potatoes from that plant are not different from other plants Ha: µ ≠ 5 Bags of potatoes from that plant are different from other plants Two tailed test (α = .05) 1.96 1 1 = .25 = 6 – 5 4 16 = 4.0 .25 4.0 -1.96 1.96

  9. Because theobserved z (4.0 ) is bigger than critical z (1.96) These three will always match Yes Probability of Type I error is always equal to alpha Yes Yes .05 1.64 No Because observed z (4.0) is still bigger than critical z (1.64) 2.58 No Because observed z (4.0) is still bigger than critical z(2.58) there is not there is a difference there is there is no difference 1.96 2.58

  10. -2.13 2.13 √ t- score : because we don’t know the population standard deviation Two tailed test (α = .05) n – 1 =16 – 1 = 15 Critical t(15) = 2.131 89 - 85 2.667 6 16

  11. Because theobserved z (2.67) is bigger than critical z (2.13) These three will always match Yes Probability of Type I error is always equal to alpha Yes Yes .05 1.753 No Because observed t (2.67) is still bigger than critical t (1.753) 2.947 Yes Because observed t (2.67) is not bigger than critical t(2.947) No These three will always match No No she did not consultant did improve morale she did consultant did not improve morale 2.131 2.947

  12. Finish with statistical summaryz = 4.0; p < 0.05 Or if it *were not* significant: z = 1.2 ; n.s. Start summary with two means (based on DV) for two levels of the IV Describe type of test (z-test versus t-test) with brief overview of results n.s. = “not significant” p<0.05 = “significant” The average weight of bags of potatoes from this particular plant is 6 pounds, while the average weight for population is 5 pounds. A z-test was completed and this difference was found to be statistically significant. We should fix the plant. (z = 4.0; p<0.05) Value of observed statistic

  13. Finish with statistical summaryt(15) = 2.67; p < 0.05 Or if it *were not* significant: t(15) = 1.07; n.s. Start summary with two means (based on DV) for two levels of the IV Describe type of test (z-test versus t-test) with brief overview of results n.s. = “not significant” p<0.05 = “significant” The average job-satisfaction score was 89 for the employees who went On the retreat, while the average score for population is 85. A t-test was completed and this difference was found to be statistically significant. We should hire the consultant. (t(15) = 2.67; p<0.05) Value of observed statistic df

  14. “Between Groups”Differences . Difference between means Difference between means Difference between means Variabilityof curve(s) “Within Groups”Variability Variabilityof curve(s) Variabilityof curve(s)

  15. . Rejecting the null hypothesis Difference between means Variability of curve(s) • If the observed stat is more extreme than the critical stat in the distribution (curve): • then it is so rare, (taking into account the variability) we conclude it must be from some other distribution • decision considers effect size and variability • then we reject the null hypothesis – we have a significant result • then we have support for our alternative hypothesis • p < 0.05 (p < α) • If the observed stat is NOT more extreme than the critical stat in the distribution (curve): • then we know it is a common score (either because the effect size is too small or because the variability is to big) and is likely to be part of this null distribution, • we conclude it must be from this distribution • decision considers effect size and variability – could be overly variable • then we do not reject the null hypothesis • then we do not have support for our alternative hypothesis • p not less than 0.05 (p not less than α) p is n.s. Difference between means Difference between means critical statistic critical statistic Variability of curve(s) (within group variability) Variability of curve(s)

  16. Independent samples t-test Are the two means significantly different from each other, or is the difference just due to chance? Donald is a consultant and leads training sessions. As part of his training sessions, he provides the students with breakfast. He has noticed that when he provides a full breakfast people seem to learn better than when he provides just a small meal (donuts and muffins). So, he put his hunch to the test. He had two classes, both with three people enrolled. The one group was given a big meal and the other group was given only a small meal. He then compared their test performance at the end of the day. Please test with an alpha = .05 Small meal 19 23 21 Big Meal 22 25 25 Mean= 21 Mean= 24

  17. Hypothesis testing two tailed test Step 1: Identify the research problem Did the size of the meal affect the learning / test scores? Step 2: Describe the null and alternative hypotheses α= .05 Step 3: Decision rule Degrees of freedom total (dftotal) = (n1 - 1) + (n2 – 1) = (3 - 1) + (3 – 1) = 4 Critical t(4) = 2.776

  18. Mean= 21 Mean= 24 Big Meal Deviation From mean -2 1 1 Small Meal Deviation From mean -2 2 0 SquaredDeviation 4 4 0 Squared deviation 4 1 1 Big Meal 22 25 25 Small meal 19 23 21 Σ = 6 Σ = 8 = 1.732 6 Notice: s2 = 3.0 Notice: Simple Average = 3.5 1 2 1 = 2.0 8 Notice: s2 = 4.0 2 2 2 (n1 – 1) s12 + (n2 – 1) s22 S2pooled = n1 + n2 - 2 (3 – 1) (1.732) 2 + (3 – 1) (2)2 = 3.5 S2pooled = 31+ 32- 2

  19. 24 – 21 = 1.5275 Mean= 21 Mean= 24 Big Meal Deviation From mean -2 1 1 Small Meal Deviation From mean -2 2 0 SquaredDeviation 4 4 0 Squared deviation 4 1 1 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 Σ = 6 Σ = 8 S2p= 3.5 24 - 21 = 1.964 3.5 3.5 3 3 Observed t

  20. Hypothesis testing Step 5: Make decision whether or not to reject null hypothesis Observed t = 1.964 Critical t = 2.776 1.964 is not farther out on the curve than 2.776 so, we do not reject the null hypothesis t(4) = 1.964; n.s. Step 6: Conclusion: There appears to be no difference in test scores between the two groups

  21. How to report the findingsfor a t-test Mean of small meal was 21 Mean of big meal was 24 One paragraph summary of this study. Describe the IV & DV. Present the two means, which type of test was conducted, and the statistical results. Finish with statistical summaryt(4) = 1.96; ns Observed t = 1.964 Start summary with two means (based on DV) for two levels of the IV df = 4 Or if it *were* significant: t(9) = 3.93; p < 0.05 Describe type of test (t-test versus anova) with brief overview of results We compared test scores for large and small meals. The mean test scores for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there appears to be no significant difference in test scores between the two types of meals t(4) = 1.964; n.s. 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

  22. Mean= 21 Figuring out the two means using Excel Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3

  23. Mean= 21 Figuring out the two means Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3

  24. Mean= 21 Figuring out the two means Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 Mean= 21 Mean= 24

  25. Mean= 21 Complete a t-test Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 May want to remove means so don’t include in the t-test

  26. Mean= 21 Complete a t-test Mean= 24 Big Meal 22 25 25 Small meal 19 23 21 Participant 1 2 3 If checked you’ll want to include the labels in your variable range If checked, you’ll want to include the labels in your variable range If checked you’ll want to include the labels in your variable range

  27. Finding Means Finding Means

  28. Finding degrees of freedom

  29. Finding Observed t

  30. Finding Critical t

  31. Finding p value (Is it less than .05?)

  32. Hypothesis testing α= .05 • Step 4: Make decision whether or not to reject null hypothesis • Reject when: • observed stat > critical stat • 1.96396 is not bigger than 2.776 • “p” is less than 0.05 (or whatever alpha is) • p = 0.121 is not less than 0.05 Step 5: Conclusion - tie findings back in to research problem There was no significant difference, there is no evidence that size of meal affected test scores

  33. The mean test score for participants who ate the big meal was 24, while the mean test score for participants who ate the small meal was 21. A t-test was completed and there appears to be no significant difference in the test scores as a function of the size of the meal, t(4) = 1.96; n.s. Type of test with degrees of freedom n.s. = “not significant” p<0.05 = “significant” Value of observed statistic Start summary with two means (based on DV) for two levels of the IV Describe type of test (t-test versus anova) with brief overview of results Finish with statistical summaryt(4) = 1.96; ns

  34. Graphing your t-test results

  35. Graphing your t-test results

  36. Graphing your t-test results Chart Layout

  37. Graphing your t-test results Fill out titles

  38. Independent samples t-test What if we ran more subjects? We are given the following problem: The size of the training sessions were increased from 3 people per class up to 9 people per class. One group ate a big meal and the other group ate the small meal. They then took a test. Does eating the size of the meal affect test scores? Please test with an alpha = .05 Small meal 19 23 21 19 23 21 19 23 21 Big Meal 22 25 25 22 25 25 22 25 25 Mean= 21 Mean= 24

  39. Hypothesis testing Step 1: Identify the research problem Did the size of the meal affect the test scores? Step 2: Describe the null and alternative hypotheses Ho: The size of the meal has no effect on test scores H1: The size of the meal does have an effect on test scores One tail or two tail test?

  40. Hypothesis testing Step 1: Identify the research problem Did the size of the meal affect the learning / test scores? Step 2: Describe the null and alternative hypotheses Ho: The size of the meal has no effect on test scores H1: The size of the meal does have an effect on test scores Step 3: Decision rule α= .05 None of this will change with more subjects

  41. Hypothesis testing Step 3: Decision rule α= .05 n1 = 9; n2 = 9 Degrees of freedom total (df total) = (n1 - 1) + (n2 – 1) = (9 - 1) + (9 – 1) = 16 Degrees of freedom total (dftotal) = (n total - 2) = 18 – 2 = 16 two tailed test Notice: Two different ways to think about it Critical t(16) = 2.12

  42. two tail test α= .05 (df) = 16 Critical t(16) = 2.12

  43. Mean= 21 Mean= 24 Big Meal Deviation From mean 2 -1 -1 2 -1 -1 2 -1 -1 Small Meal Deviation From mean 2 -2 0 2 -2 0 2 -2 0 SquaredDeviation 4 4 0 4 4 0 4 4 0 Squared deviation 4 1 1 4 1 1 4 1 1 Big Meal 22 25 25 22 25 25 22 25 25 Small meal 19 23 21 19 23 21 19 23 21 Σ = 18 Σ = 24 = 1.50 18 Notice: Simple Average = 2.625 Notice: s2 = 2.25 1 8 1 = 1.732 24 Notice: s2 = 3.0 2 8 2

  44. Mean= 21 Mean= 24 Big Meal 22 25 25 22 25 25 22 25 25 Small meal 19 23 21 19 23 21 19 23 21 Sp = 2.625 S22 = 3.00 S21 = 2.25 S2 = 1.732 S1 = 1.5 (n1 – 1) s12 + (n2 – 1) s22 S2pooled = n1 + n2 - 2 (9 – 1) (1.50)2 + (9 – 1) (1.732)2 = 2.625 S2pooled = 9 + 9 - 2

  45. 24 – 21 = 0.7638 Mean= 21 Mean= 24 Big Meal 22 25 25 22 25 25 22 25 25 Small meal 19 23 21 19 23 21 19 23 21 Sp2= 2.625 S2 = 1.732 S1 = 1.5 24 - 21 = 3.928 2.625 2.625 9 9

  46. Hypothesis testing Step 5: Make decision whether or not to reject null hypothesis Observed t = 3.928 Critical t = 2.120 3.928 is farther out on the curve than 2.120 so, we do reject the null hypothesis t(16) = 3.928; p < 0.05 Step 6: Conclusion: There appears to be a difference in the test scores between the two groups

  47. How to report the findingsfor a t-test Mean of small meal was 21 Mean of big meal was 24 One paragraph summary of this study. Describe the IV & DV. Present the two means, which type of test was conducted, and the statistical results. Start summary with two means (based on DV) for two levels of the IV Observed t = 3.928 df = 16 Describe type of test (t-test versus anova) with brief overview of results Finish with statistical summaryt(9) = 3.93; p < 0.05 p < 0.05 We compared test scores for large and small meals. The mean test score for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there appears to be a significant difference in test scores between the two types of meals t(4) = 3.928; p < 0.05

  48. What if we had run more subjects?

  49. What if we had run more subjects? Finding Means Finding Means

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