Reasoning in Psychology Using Statistics

Reasoning in Psychology Using Statistics
2018

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

Related-samples t-test: Matched Pairs
The related samples t-test can be used when: Related-samples t-test: Matched Pairs

Related-samples t-test: Within Persons/Repeated Measures
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 25 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 5 fewer errors after the treatment. Average the pair-wise differences should be 0 if no effect Memory Test scores Pre-test Memory Test scores Post-test Memory patients Memory treatment Compare pair-wise differences Hypotheses: Memory performance at the post-test is equal to memory performance at the pre-test, that is, H0: HA: Memory performance at the post-test is NOT equal to memory performance at the pre-test, that is, Related-samples t-test: Within Persons/Repeated Measures

Related-samples t-test: Matched Pairs
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 recruits 50 patients and matches them into 2 related samples. He then gives one sample the new treatment;, while the other sample comprises the no-treatment control group.) Following the treatment period, all participants take a memory test. Treatment participants averaged 5 fewer errors than their matched partners. Memory Test scores No Memory treatment Average the pair-wise differences; should be 0 if no effect Memory patients related Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the matched no-treatment group Memory performance by the treatment group is NOT equal to memory performance by the matched no-treatment group HA: Related-samples t-test: Matched Pairs

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 One-sample t Related-samples t These are computed differently than last time Testing Hypotheses

Related-samples t-test
What are all of these “D’s” referring to? Mean of differences Estimated standard deviation of differences Estimated standard error of differences Degrees of freedom of difference scores Related-samples t-test

Related-samples t-test: Within Persons
(Pre-test) - (Post-test) What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 1 2 3 4 45 55 40 60 43 49 35 51 2 6 5 9 If a 1-tailed test, be careful about the order in forming Ds. Pre-Post = positive if Pre score higher, so positive if lower score is improvement (fewer memory errors, as above) Post-Pre = positive if Post score higher, so positive if higher score is improvement (more items correct) Note: SPSS always calculates D = Gr1 – Gr2 Related-samples t-test: Within Persons

You can think of this as doing a 1-sample t-test, where your sample is the difference scores, and your null hypothesis is that the average of the differences is equal to 0 Once you do the subtraction, you can ignore these numbers Difference scores 1-sample t Related-samples t Person Pre-test Post-test 45 55 40 60 43 49 35 51 1 2 2 6 3 5 4 9 Is this a likely sample if μD = 0? Related-samples t-test: Within Persons

(Pre-test) - (Post-test) What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 1 2 3 4 45 55 40 60 43 49 35 51 2 6 5 9 22 = 5.5 Related-samples t-test: Within Persons

Related-samples t-test Within Persons
What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 1 45 43 2 2 55 49 6 3 40 35 5 4 60 51 9 22 D = 5.5 Related-samples t-test Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 - 5.5 = -3.5 12.25 2 6 0.5 - 5.5 = 0.25 3 5 -0.5 - 5.5 = 0.25 4 9 3.5 - 5.5 = 12.25 22 25 = SSD D = 5.5 Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 -3.5 12.25 2 6 0.5 0.25 3 5 -0.5 0.25 4 9 3.5 12.25 22 25 = SSD D = 5.5 Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 -3.5 12.25 2 6 0.5 0.25 3 5 -0.5 0.25 4 9 3.5 12.25 22 25 = SSD D = 5.5 2.9 = sD Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 ? 1 2 -3.5 12.25 2 6 0.5 0.25 3 5 -0.5 0.25 4 9 3.5 12.25 Think back to null hypothesis 22 25 = SSD D = 5.5 2.9 = sD 1.45 = sD Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 -3.5 12.25 2 6 0.5 0.25 H0: No difference in memory performance. 3 5 -0.5 0.25 4 9 3.5 12.25 22 25 = SSD D = 5.5 2.9 = sD 1.45 = sD Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 -3.5 12.25 2 6 0.5 0.25 tobs 3 5 -0.5 0.25 4 9 3.5 12.25 22 25 = SSD D = 5.5 2.9 = sD 1.45 = sD Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 -3.5 12.25 2 6 0.5 0.25 tobs 3 5 -0.5 0.25 tcrit α = 0.05 2-tailed 4 9 3.5 12.25 22 25 = SSD D = 5.5 2.9 = sD 1.45 = sD Note how large tcrit is when df = 3. Related-samples t-test: Within Persons

What are all of these “D’s” referring to? Difference scores Person Pre-test Post-test 45 55 40 60 43 49 35 51 D - D (D - D)2 1 2 -3.5 12.25 2 6 0.5 0.25 tobs 3 5 -0.5 0.25 tcrit α = 0.05 2-tailed 4 9 3.5 12.25 +3.18 = tcrit 22 25 = SSD tobs=3.8 D = 5.5 2.9 = sD - Reject H0 1.45 = sD Why is tobs in upper tail (critical region)? Because positive difference (improvement) Related-samples t-test: Within Persons

Hypothesis testing: Summary So Far – and next
Design Test statistic (Estimated) Standard error One sample, σ known One sample, σ unknown df Two related samples, σ unknown One more to go Hypothesis testing: Summary So Far – and next

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

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

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

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

Related-samples t Sample means Test statistic 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

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

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

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

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

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

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

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

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 ( )2 + ( )2 + ( )2 + ( )2 = 155 SS = B 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

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

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

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

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

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

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? Goldstein: Choosing a Statistical Test (~12 mins) MathMeeting: z-test vs. t-test (~8 mins) KahnAcademy: z-statistic vs t-statistic (~6 mins) How2Stats: paired samples t-test in SPSS (~8 mins) Repeated Measures t Tests Part I Introduction (~14 mins) Independent vs. Paired samples t-tests (~4 mins) 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

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

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

Graphing Independent-Samples T Tests

Reasoning in Psychology Using Statistics

Similar presentations

Presentation on theme: "Reasoning in Psychology Using Statistics"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Reasoning in Psychology Using Statistics

Similar presentations

Presentation on theme: "Reasoning in Psychology Using Statistics"— Presentation transcript:

Similar presentations

About project

Feedback