Reasoning in Psychology Using Statistics 2017

Announcements Quiz 7 Due Friday at midnight. It has a section on independent-t, and another the has the rest to help with reviewing for exam Exam 3 is 1 week from today (both in-class and in-lab exams) Distribution of Sample Means (including Central Limit Theorem) Hypothesis testing 1 sample z 1 sample t related sample t independent sample t Announcements

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. 2 samples The groups are independent People in one group are not related to those in the other group Samples are independent No Memory treatment Random Assignment Memory patients Memory treatment Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. 2 samples Samples are independent Memory Test X One score per subject No Memory treatment Compare these two means Memory patients Random Assignment Memory treatment Memory treatment patients No Memory Test scores related Compare pair-wise differences D Compare: Related-samples t-test (from last lecture) Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. 2 samples Samples are independent One score per subject Independent sample t-test Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Memory Test X No Memory treatment Compare these two means Memory patients Random Assignment Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group, that is, μA = μB. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group, that is, μA ≠ μB. HA: Independent-samples t-test

Testing Hypotheses 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 Compute your estimated standard error Compute your t-statistic Compute your degrees of freedom Step 5: Make a decision about your null hypothesis Independent-samples t Related-samples t These are computed differently than last time Testing Hypotheses

Independent-samples t-test Related-samples t Sample means Test statistic Independent-samples t-test

Independent-samples t-test Related-samples t Population means from the hypotheses Test statistic Note that population mean is 0 (difference between group means), just as in related-samples design (average difference) H0: Memory performance by the treatment group is equal to memory performance by the no treatment group, that is, μA = μB. Independent-samples t-test

Independent-samples t-test Related-samples t Test statistic Population means from the hypotheses H0: Memory performance by treatment group is equal to that by no treatment group or Difference in memory performance is 0. HA: Memory performance by treatment group is not equal to to that by no-treatment group or Difference in memory is not 0. Independent-samples t-test

Independent-samples t-test Related-samples t Test statistic Estimated standard error (difference expected by chance) estimate is based on one sample We have two samples, so the estimate is based on two samples Independent-samples t-test

Independent-samples t-test We combine the variance from the two samples “pooled variance” Number of subjects in group A Number of subjects in group B Independent-samples t-test

Independent-samples t-test Recall “weighted means,” need to use “weighted variances” here We combine the variance from the two samples “pooled variance” Variance (s2) * degrees of freedom (df) Independent-samples t-test

Independent-samples t-test We combine the variance from the two samples Recall “weighted means,” need to use “weighted variances” here “pooled variance” Variance (s2) * degrees of freedom (df) variance Independent-samples t-test

Performing your statistical test Independent-samples t Compute your estimated standard error Pick whichever one is easier given what you know Compute your t-statistic Compute your degrees of freedom This is the one you use to look up your tcrit Performing your statistical test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Need to compute the mean and SS for each sample Exp. group Control group 45 55 40 60 43 49 35 51 Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Need to compute the mean and SS for each sample Exp. group Control group 45 55 40 60 43 49 35 51 Control group = 50 (45-50)2 + (55-50)2 + (40-50)2 + (60-50)2 = 250 SS = A Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Need to compute the mean and SS for each sample Exp. group Control group 45 55 40 60 43 49 35 51 Exp. group = 44.5 (43-44.5)2 + (49- 44.5)2 + (35- 44.5)2 + (51- 44.5)2 = 155 SS = B Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95 Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95 tobs= 0.95 tcrit= ±2.447 α = 0.05 2-tailed Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95 tobs= 0.95 tcrit= ±2.447 α = 0.05 2-tailed Fail to Reject H0 +2.45 = tcrit tobs=0.95 Independent-samples t-test

Independent-samples t-test Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95 tobs= 0.95 tcrit= ±2.447 α = 0.05 2-tailed Fail to Reject H0 “The evidence suggests that there is no difference between those who got the treatment and those who did not.” Independent-samples t-test

Comparing statistical tests Independent-samples t Related-samples t Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he collects a sample of 4 patients and gives them his new treatment. Before the treatment he gives them a pre-treatment memory test and after the treatment a post-treatment memory test. The data are presented below. Use α = 0.05. Similar numerators: both involve differences Between sample means or mean of pair-wise difference scores Null population mean of no difference = = Differences between sample means, mean of pair-wise differences Comparing statistical tests

Comparing statistical tests Independent-samples t Related-samples t Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use α = 0.05. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he collects a sample of 4 patients and gives them his new treatment. Before the treatment he gives them a pre-treatment memory test and after the treatment a post-treatment memory test. The data are presented below. Use α = 0.05. Dissimilar denominators (standard or sampling error): Based on pooling 2 samples (n’s can differ) vs. 1 sample of differences Comparing statistical tests

Comparing statistical tests Independent-samples t Related-samples t Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use  = 0.05. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he collects a sample of 4 patients and gives them his new treatment. Before the treatment he gives them a pre-treatment memory test and after the treatment a post-treatment memory test. The data are presented below. Use  = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 Difference scores 2 6 5 9 Pair Group A Group B 1 3 4 What happens if data are the same? Comparing statistical tests

Comparing statistical tests Independent-samples t Related-samples t Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use  = 0.05. Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he collects a sample of 4 patients and gives them his new treatment. Before the treatment he gives them a pre-treatment memory test and after the treatment a post-treatment memory test. His sample averaged 60 errors before the treatment and 55 errors after the treatment. Exp. group Control group 45 55 40 60 43 49 35 51 Difference scores 2 6 5 9 Pair Group A Group B 1 3 4 Although data are exactly the same, designs are different, so use different statistical tests, which may produce different conclusions = 0.95 Fail to Reject H0 Reject H0 Note: that pair-wise differences in related samples t-test reduce difference expected by chance (taking out variability from individual differences) Comparing statistical tests

Hypothesis testing formulas summary (Estimated) Standard error df Design Test statistic One sample, σ known One sample, σ unknown Two related samples, σ unknown Two independent samples, σ unknown Hypothesis testing formulas summary

In lab: Practice using independent-sample t-tests, by hand and using SPSS Questions? Wrap up

Assumptions of t-tests All t-tests: Independence: No influence of some scores on other scores (e.g., participants are not from the same family). Normality: Population is normally distributed. Only extreme violations of this assumption influence t-tests. Independent-samples t-test: Homogeneity of Variance: Samples have equal variances. If Levene’s Test of Equality of Variance finds significant difference, SPSS provides corrected values of observed t and degrees of freedom. Independence: It is impossible to detect violations of this assumption statistically. It is avoided by good research design. Assumptions of t-tests

Independent-Samples T Test in SPSS Dependent Variable Independent Variable Note: First design requiring data column for independent variable Type in whatever numbers used to code 2 groups Independent-Samples T Test in SPSS

Independent-Samples T Test in SPSS If Levene’s Test is not significant (small F, p > .05), use upper row to find t-test p-value. p-value testing Homogeneity of Variance. If Levene’s Test is significant (large F, p <.05), use lower row to find t-test p-value. Independent-Samples T Test in SPSS

Independent-Samples T Test in SPSS Fail to reject H0 t-value too small, p > .05 Independent-Samples T Test in SPSS

Graphing Independent-Samples T Tests

Graphing Independent-Samples T Tests