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Lecturer’s desk

Screen. Cabinet. Cabinet. Lecturer’s desk. Table. Computer Storage Cabinet. Row A. 3. 4. 5. 19. 6. 18. 7. 17. 16. 8. 15. 9. 10. 11. 14. 13. 12. Row B. 1. 2. 3. 4. 23. 5. 6. 22. 21. 7. 20. 8. 9. 10. 19. 11. 18. 16. 15. 13. 12. 17. 14. Row C. 1. 2.

jonas-goodwin
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Lecturer’s desk

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  1. Screen Cabinet Cabinet Lecturer’s desk Table Computer Storage Cabinet Row A 3 4 5 19 6 18 7 17 16 8 15 9 10 11 14 13 12 Row B 1 2 3 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 Row C 1 2 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 Row D 1 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 Row E 1 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 Row F 27 1 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 28 Row G 27 1 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 29 10 19 11 18 16 15 13 12 17 14 28 Row H 27 1 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 Row I 1 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 1 Row J 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 28 27 1 Row K 26 2 25 3 24 4 23 5 6 22 21 7 20 8 9 10 19 11 18 16 15 13 12 17 14 Row L 20 1 19 2 18 3 17 4 16 5 15 6 7 14 13 INTEGRATED LEARNING CENTER ILC 120 9 8 10 12 11 broken desk

  2. Introduction to Statistics for the Social SciencesSBS200, COMM200, GEOG200, PA200, POL200, or SOC200Lecture Section 001, Fall, 2014Room 120 Integrated Learning Center (ILC)10:00 - 10:50 Mondays, Wednesdays & Fridays. Welcome http://www.youtube.com/watch?v=oSQJP40PcGI

  3. A noteon doodling Reminder

  4. Schedule of readings Before next exam (November 21st) Please read chapters 7 – 11 in Ha & Ha Please read Chapters 2, 3, and 4 in Plous Chapter 2: Cognitive Dissonance Chapter 3: Memory and Hindsight Bias Chapter 4: Context Dependence

  5. Use this as your study guide By the end of lecture today10/29/2014 Logic of hypothesis testing Steps for hypothesis testing Levels of significance (Levels of alpha) Hypothesis testing with t-scores (one-sample) Constructing brief, complete summary statements

  6. Homework due Assignment 16 One-sample z and t hypothesis tests Due date extended: Friday, October 31st

  7. Lab sessions Labs continue this week with Project 2

  8. Standard deviation and Variance For Sample and Population These would be helpful to know by heart – please memorize these formula Review

  9. 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 • Part 3: • What are the five steps for hypothesis testing?

  10. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses 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 If we change from two-tailed to one tailed test does critical z score get bigger or smaller? 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 Review

  11. Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses How is a t score different than a z score? Step 2: Decision rule: find “critical z” score • Alpha level? (α= .05 or .01)? • One versus two-tailed test 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 Population versus sample standard deviation Population versus sample standard deviation Step 5: Conclusion - tie findings back in to research problem

  12. 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

  13. Hypothesis testing Independent Variable? Step 1: Identify the research problem / hypothesis Extra time vs. no extra time Dependent Variable? Test Scores Did the extra time given to this sample of students affect their chemistry test scores? IV: Nominal Ordinal Interval or Ratio? IV: Nominal DV: Nominal Ordinal Interval or Ratio? DV: Ratio One tail or two tail test? Two-tailed test Null: The extra time did not affect their chemistry test scores Alternative: The extra time did affect their chemistry test scores

  14. Hypothesis testing Step 2: Decision rule What is formula for degres of freedom? n – 1 = .05 n = 25 Degrees of freedom (df) = (n - 1) = (25 - 1) = 24 two tail test

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

  16. . 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 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 Step 3: Calculations 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 = -0.44 = -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 µ = 74 N = 25 Σx 1911 = = 25 N = 6.01 868.16 24 Σx = 1911 Σ(x- x) = 0 Σ(x- x)2 = 868.16

  17. . Hypothesis testing = 76.44 76.44 - 74 = = 2.03 1.20 Step 3: Calculations µ = 74 N = 25 Σx 1911 = = 25 N s = 6.01 76.44 - 74 6.01 critical t critical t 25 Observed t(24) = 2.03

  18. . 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

  19. . 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

  20. . What if we had chosen a one-tail test? = .05 Step 1: Identify the research problem / hypothesis Did the extra time given to this sample of students improve their chemistry test scores? Null: The extra time did not improve their chemistry test scores Alternative: The extra time did improve their chemistry test scores Step 2: Decision rule Degrees of freedom (df) = (n - 1) = (25 - 1) = 24 Critical t (24) = 1.711

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

  22. . Hypothesis testing (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 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 Step 3: Calculations 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 = -0.44 = -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 µ = 74 N = 25 Σx 1911 = = 25 N = 6.01 868.16 24 Σx = 1911 Σ(x- x) = 0 Σ(x- x)2 = 868.16

  23. . Hypothesis testing (exactly same as two-tail test) = 76.44 76.44 - 74 = = 2.03 1.20 One-tailed test has no effect on calculations stage Step 3: Calculations µ = 74 N = 25 Σx 1911 = = 25 N s = 6.01 76.44 - 74 6.01 critical t critical t 25

  24. . Hypothesis testing Step 4: Make decision whether or not to reject null hypothesis Observed t(24) = 2.03 Critical t(24)= 1.711 2.03 is not farther out on the curve than 1.711, so, we do not reject the null hypothesis Step 6: Conclusion: The extra time did 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 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” Type of test with degrees of freedom Value of observed statistic

  26. Comparing z score distributions with t-score distributions z-scores Similarities include: Using bell-shaped distributions to make confidence interval estimations and decisions in hypothesis testing Use table to find areas under the curve (different table, though – areas often differ from z scores) t-scores • Summary of 2 main differences: • We are now estimating standard deviation • from the sample • (We don’t know population standard deviation) • We have to deal with degrees of freedom

  27. Comparing z score distributions with t-score distributions Differences include: • We use t-distribution when we don’t know standard deviation • of population, and have to estimate it from our sample Critical t (just like critical z) separates common from rare scores Critical t used to define both common scores “confidence interval” and rare scores “region of rejection

  28. Comparing z score distributions with t-score distributions Differences include: • We use t-distribution when we don’t know standard deviation • of population, and have to estimate it from our sample 2) The shape of the sampling distribution is very sensitive to small sample sizes (it actually changes shape depending on n) Please notice: as sample sizes get smaller, the tails get thicker. As sample sizes get bigger tails get thinner and look more like the z-distribution

  29. Comparing z score distributions with t-score distributions Please note: Once sample sizes get big enough the t distribution (curve) starts to look exactly like the z distribution (curve) scores Differences include: • We use t-distribution when we don’t know standard deviation • of population, and have to estimate it from our sample 2) The shape of the sampling distribution is very sensitive to small sample sizes (it actually changes shape depending on n) 3) Because the shape changes, the relationship betweenthe scores and proportions under the curve change (So, we would have a different table for all the different possible n’s but just the important ones are summarized in our t-table)

  30. Interpreting t-table We use degrees of freedom (df) to approximate sample size Technically, we have a different t-distribution for each sample size This t-table summarizes the most useful values for several distributions This t-table presents useful values for distributions (organized by degrees of freedom) Each curve is based on its own degrees of freedom (df) - based on sample size, and its own table tying together t-scores with area under the curve n = 17 n = 5 . Remember these useful values for z-scores? 1.64 1.96 2.58

  31. Area betweentwo scores Area between two scores Area beyond two scores (out in tails) Area beyond two scores (out in tails) Area in each tail (out in tails) Area in each tail (out in tails) df

  32. Area betweentwo scores Area between two scores Area beyond two scores (out in tails) Area beyond two scores (out in tails) Area in each tail (out in tails) Area in each tail (out in tails) df Notice with large sample size it is same values as z-score . Remember these useful values for z-scores? 1.64 1.96 2.58

  33. Comparison of z and t • For very small samples, t-values differ substantially from the normal. • As degrees of freedom increase, the t-values approach the normal z-values. • For example, for n = 31, the degrees of freedom are: • What would the t-value be for a 90% confidence interval? n - 1= 31 – 1 = 30 df

  34. 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.

  35. 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

  36. Questions from the homework assignment?

  37. 95% 26.08 < µ< 33.92 mean + z σ = 30 ± (1.96)(2) 99% 24.84 < µ< 35.16 mean + z σ = 30 ± (2.58)(2)

  38. 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

  39. √ 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

  40. 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

  41. -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

  42. 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

  43. 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

  44. 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

  45. Thank you! See you next time!!

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