Reasoning in Psychology Using Statistics

1. Reasoning in PsychologyUsing Statistics Psychology 138 2015

2. Quiz 5 • pushed back to due tonight at midnight • “bad” question (error when uploaded): • "A researcher wants to know if a new flu drug affects people, either by making them better or worse. Would the researcher use a one-tailed or two-tailed test and why?” • The correct answer as "Two-tailed because there is no predicted direction of the effect of the drug." • but ReggieNet has the right answer as "One-tailed because only one issue is discussed¬?the drug.” • I will correct by hand after quizzes are due Announcements

3. Hypothesis testing: a five step program • Step 1: State your hypotheses • Step 2: Set your decision criteria • Step 3: Collect your data from your sample • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Testing Hypotheses

4. Hypothesis testing: a five step program • Step 1: State your hypotheses • Step 2: Set your decision criteria • Step 3: Collect your data from your sample • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Testing Hypotheses

5. Population of memory patients • Conclusions: • The memory treatment sample are the same as those in the population of memory patients. • They aren’t the same as those in the population of memory patients We test this one Memory Test X H0: MemoryTest μ & σ known Memory patients Memory treatment HA: Compare these two means • What are we doing when we test the hypotheses? • Consider a variation of our memory experiment example Performing your statistical test

6. One population Two populations We test this one The memory treatment sample are the same as those in the population of memory patients. They are not the same as those in the population of memory patients XA XA • What are we doing when we test the hypotheses? Real world (‘truth’) H0: is true (no treatment effect) H0: is false (is a treatment effect) Performing your statistical test

7. Distribution of the test statistic • The generic test statistic distribution (a transformation of the distribution of sample means) • To reject the H0, you want a computed test statistics that is large • The probability of having a sample with that mean is very low • What is large enough? • The alpha level gives us the decision criterion α-level determines where these boundaries go “Generic” statistical test

8. Distribution of the test statistic If test statistic is here Reject H0 If test statistic is here Fail to reject H0 • The generic test statistic distribution (a transformation of the distribution of sample means) • To reject the H0, you want a computed test statistics that is large • The probability of having a sample with that mean is very low • What is large enough? • The alpha level gives us the decision criterion “Generic” statistical test

9. Reject H0 α = 0.05 0.025 split up into the two tails 0.025 Fail to reject H0 • The alpha level gives us the decision criterion Two -tailed • Go to the table (unit normal table for z-test) and find the z that has 0.050 in the tails. Zcritical = ±1.96 “Generic” statistical test

10. Reject H0 0.05 all of it in one tail α = 0.05 Fail to reject H0 • The alpha level gives us the decision criterion Two -tailed One -tailed Reject H0 Reject H0 Fail to reject H0 Fail to reject H0 • Go to the table (unit normal table for z-test) and find the z that has 0.050 in the tail. Zcritical = +1.645 “Generic” statistical test

11. Reject H0 α = 0.05 0.05 Fail to reject H0 • The alpha level gives us the decision criterion Two -tailed One -tailed all of it in one tail Reject H0 Reject H0 Fail to reject H0 Fail to reject H0 • Go to the table (unit normal table for z-test) and find the z that has 0.050 in the tail. Zcritical = -1.645 “Generic” statistical test

12. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. Population of memory patients Memory Test σ is known μ is known Memory Test X Memory patients Memory treatment Compare these two means 1-sample z-test

13. 1 sample • Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. Population of memory patients Memory Test σ is known μ is known Memory Test X Memory patients Memory treatment Compare these two means 1-sample z-test

14. 1 sample • Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. Population of memory patients Memory Test σ is known μ is known Memory Test X Memory patients Memory treatment Compare these two means • 1 score per subject 1-sample z-test

15. 1 sample 1-sample z-test • Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. Population of memory patients Memory Test σ is known μ is known Memory Test X Memory patients Memory treatment Compare these two means • 1 score per subject • Population mean (μ) and standard deviation (σ) are known (assume Normal dist) 1-sample z-test

16. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. One -tailed H0: the memory treatment sample are the same as those in the population of memory patients (or even worse). HA: the memory treatment sample perform better (fewer errors) than those in the population of memory patients • Step 1: State your hypotheses μTreatment > μpop = 60 • Step 2: Set your decision criteria • Step 3: Collect your data • Step 4: Compute your test statistics μTreatment < μpop = 60 • Step 5: Make a decision about your null hypothesis Performing your statistical test

17. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. μTreatment > μpop = 60 H0: α = 0.05 μTreatment < μpop = 60 HA: One -tailed • Step 1: State your hypotheses • Step 2: Set your decision criteria • Step 3: Collect your data • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Performing your statistical test

18. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. μTreatment > μpop = 60 H0: n = 16, X = 55 μTreatment < μpop = 60 HA: One -tailed α = 0.05 • Step 1: State your hypotheses • Step 2: Set your decision criteria • Step 3: Collect your data • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Performing your statistical test

19. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. μTreatment > μpop = 60 H0: n = 16, X = 55 μTreatment < μpop = 60 HA: One -tailed α = 0.05 • Step 1: State your hypotheses = -2.5 • Step 2: Set your decision criteria • Step 3: Collect your data • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Performing your statistical test

20. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors (while the typical memory patient averages μ = 60 errors, with σ = 8). Test using α = 0.05. μTreatment > μpop = 60 H0: n = 16, X = 55 μTreatment < μpop = 60 HA: 5% One -tailed α = 0.05 • Step 1: State your hypotheses = -2.5 • Step 2: Set your decision criteria • Step 3: Collect your data • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Reject H0 - Support for our HA, the evidence suggests that the treatment decreases the number of memory errors Performing your statistical test

21. The 1-sample t-test • Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). • Population standard deviation (s) is NOT known 1-sample t-test

22. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Population of memory patients MemoryTest σ is NOT known μ is known Memory Test X Memory patients Memory treatment Compare these two means • The 1-sample t-test • Hypotheses: • The memory treatment sample are the same as those in the population of memory patients (or even worse), that is, μA ≥ μ0. • The memory treatment sample perform better (fewer errors) than those in the population of memory patients, that is, μA < μ0. H0: HA: One sample t-test

23. Same Same • The 1-sample t-test • Hypothesis testing: a five step program • Step 1: State your hypotheses • Step 2: Set your decision criteria • Step 3: Collect your data • Step 4: Compute your test statistics • Step 5: Make a decision about your null hypothesis Different • Compute your estimatedstandard error • Compute your t-statistic • Compute your degrees of freedom Testing Hypotheses

24. Could be difference between a sample and a population, or between different samples Based on standard error or an estimate of the standard error • Computing a test statistic: Generic test • What are we doing when we test the hypotheses? Hypothesis testing

25. Could bedifference between a sample and a population, or between different samples Based on standard error oran estimate of the standard error • Computing a test statistic: 1-sample t-test • What are we doing when we test the hypotheses? 1-sample t-test

26. What are we doing when we test the hypotheses? • Computing a test statistic: 1-sample t-test Could bedifference between a sample and a population, or between different samples Based onandard error or estimate of standard error 1-sample t-test

27. identical 1-sample z 1-sample t Test statistic 1-sample t-test

28. 1-sample z 1-sample t different Standard error If don’t know this, so need to estimate it Test statistic Diff. Expected by chance s Use our best guess, sample standard deviation (s) XA 1-sample t-test

29. 1-sample z 1-sample t different Standard error If don’t know this, so need to estimate it Degrees of freedom Test statistic Diff. Expected by chance Estimated standard error Use our best guess, sample standard deviation (s) 1-sample t-test

30. If test statistic is beyond here Reject H0 If test statistic is here, fail to reject H0 • The t-statistic distribution (a transformation of the distribution of sample means) • To reject the H0, you want a computed test statistics that is large • The alpha level gives us the decision criterion • New table: the t-table; set up to find critical t-value for given p-value Distribution of the t-statistic Note: Bottom row: same as z. The t-distribution becomes the Normal distribution when df = ∞ The t-distribution

31. α levels 1-tailed - or - 2-tailed Degrees of freedom df Critical values of t tcrit • The t distributions are like the z distribution (μ = 0; σ =1). • New table: the t-table Each row corresponds to a different curve The t-distribution

32. New table: the t-table • As df gets smaller, need larger tcrit, shown here for α = .05, 2-tailed. Curve flattens (becomes platykurdic). df= df = df = The t-distribution

33. tcrit = +2.571 • What is the tcrit for a 2-tailed hypothesis test with a sample size of n = 6 and an α level of 0.05? α = 0.05 2-tailed df = n - 1 = 5 The t-distribution

34. tcrit = +2.015 • What is the tcrit for a 1-tailed hypothesis test with a sample size of n = 6 and an α level of 0.05? α = 0.05 n = 6 1-tailed df = n - 1 = 5 1-tailed, larger critical region in 1-tail, so smaller critical value needed. tcrit = 2.015 vs. ±2.571 The t-distribution

35. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Test this with α level = 0.05. H0: Memory treatment sample same as or make more errors than population of memory patients. 1-tailed HA: Sample make fewer errors than population of memory patients An example: 1-sample t-test H0: μA > μ0 = 60 HA: μA < μ0 = 60 • Step 1: Hypotheses 1-sample t-test

36. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Test this with α level = 0.05. α = 0.05 An example: 1-sample t-test H0: μA> μ0 = 60 HA: μA < μ0 = 60 1-tailed • Step 1: Hypotheses • Step 2: Criterion for decision 1-sample t-test

37. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. . His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Test this with α level = 0.05. = 55, s = 8 An example: 1-sample t-test H0: μA> μ0 = 60 HA: μA < μ0 = 60 α = 0.05 1-tailed • Step 1: Hypotheses • Step 2: Criterion for decision • Step 3: Sample statistics 1-sample t-test

38. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. . His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Test this with α level = 0.05. = 55, s = 8 An example: 1-sample t-test H0: μA> μ0 = 60 HA: μA < μ0 = 60 α = 0.05 1-tailed • Step 1: Hypotheses = -2.5 • Step 2: Criterion for decision • Step 3: Sample statistics • Step 4: Test statistic 1-sample t-test

39. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. . His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Test this with α level = 0.05. = 55, s = 8 An example: 1-sample t-test H0: μA> μ0 = 60 HA: μA < μ0 = 60 α = 0.05 1-tailed • Step 1: Hypotheses • Step 2: Criterion for decision • Step 3: Sample statistics • Step 4: Test statistic • Step 5: Compare observed to critical value & make a decision about your null hypothesis tcrit = -1.753 1-sample t-test

40. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He hypothesizes that the treatment will improve memory performance. To test it he collects a sample of 16 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with s = 8 (while for the typical memory patient μ = 60 errors). Test this with α level = 0.05. = 55, s = 8 tobs=-2.5 An example: 1-sample t-test H0: μA> μ0 = 60 HA: μA < μ0 = 60 α = 0.05 1-tailed • Step 1: Hypotheses • Step 2: Criterion for decision • Step 3: Sample statistics • Step 4: Test statistic • Step 5: Compare observed to critical value & make a decision about your null hypothesis -1.753 = tcrit “Evidence supports the hypothesis that the memory treatment improved performance” 1-sample t-test

41. = 8 – 10 = -5 .4 Cover later: Estimate difference between -3 and -1 In SPSS, compare observed & critical p-values. 1-tail p = .0005 Less than α = .05? SPSS output for 1-sample t-test

42. In lab: Practice using 1-sample t-tests and the hypothesis testing framework • Questions? Wrap up