Reasoning in Psychology Using Statistics

Reasoning in Psychology Using Statistics
2018

Quiz 6 is due this Monday at Midnight (moved back a few days)
Exam 3 is a week and a half away (Mon. Apr. 9) Announcements

1-sample z test vs. t-test
1-sample t Test statistic Diff. Expected by chance Standard error Estimated standard error Degrees of freedom 1-sample z test vs. t-test

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 If test statistic is beyond here Reject H0 If test statistic is here, fail to reject H0 Note: Bottom row: same as z. The t-distribution becomes the Normal distribution when df = ∞ The t-distribution

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

The t-distribution 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

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 tcrit = The t-distribution

α = 0.05 n = 6 1-tailed df = n - 1 = 5 tcrit = 1-tailed, larger critical region in 1-tail, so smaller critical value needed. tcrit = vs. ±2.571 The t-distribution

1-sample t-test An example: 1-sample t-test H0: HA:
Memory treatment sample same as or make more errors than population of memory patients. 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. 1-tailed H0: μA > μ0 = 60 HA: Sample make fewer errors than population of memory patients Step 1: Hypotheses HA: μA < μ0 = 60 1-sample t-test

1-sample t-test An example: 1-sample t-test H0: μA > μ0 = 60
HA: μA < μ0 = 60 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. 1-tailed α = 0.05 Step 1: Hypotheses Step 2: Criterion for decision 1-sample t-test

1-sample t-test An example: 1-sample t-test = 55, s = 8
H0: μA > μ0 = 60 HA: μA < μ0 = 60 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. 1-tailed α = 0.05 = 55, s = 8 Step 1: Hypotheses Step 2: Criterion for decision Step 3: Sample statistics 1-sample t-test

1-sample t-test An example: 1-sample t-test = 55, s = 8
H0: μA > μ0 = 60 HA: μA < μ0 = 60 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. 1-tailed α = 0.05 = 55, s = 8 Step 1: Hypotheses = -2.5 Step 2: Criterion for decision Step 3: Sample statistics Step 4: Test statistic 1-sample t-test

1-sample t-test An example: 1-sample t-test = 55, s = 8 tcrit = -1.753
H0: μA > μ0 = 60 HA: μA < μ0 = 60 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. 1-tailed α = 0.05 = 55, s = 8 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-sample t-test

1-sample t-test An example: 1-sample t-test = 55, s = 8 tobs=-2.5
H0: μA > μ0 = 60 HA: μA < μ0 = 60 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. 1-tailed α = 0.05 = 55, s = 8 tobs=-2.5 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 “Evidence supports the hypothesis that the memory treatment improved performance” 1-sample t-test

SPSS output for 1-sample t-test
H0: μA = μ0 = 10 HA: μA ≠ μ0 = 10 10 = 8 – 10 = -4.9 .4082 df=n-1=9-1=8 In SPSS, compare observed & critical p-values. 1-tail p = .0005 Less than α = .05? SPSS output for 1-sample t-test

Studying Suggestion: Make flash cards
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. Side 1: Write out the problem One sample t-test Reject H0 One-tailed “improve” α-level =0.05 Evidence suggests that the treatment improved memory H0: μA > μ0 = 60 HA: μA < μ0 = 60 = -2.5 tcrit = Side 2: Write out the solution Studying Suggestion: Make flash cards

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 Hypothesis testing: Summary So Far – and next

Related-samples t-test
The related samples t-test can be used with 2 different designs What are all of these “D’s” referring to? Difference scores Note: Different names Repeated-measures t-test Within-persons t-test Matched-pairs t-test Related-samples t-test Related-samples t-test

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. Memory patients treatment Test scores Pre-test Post-test Compare pair-wise differences 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. 1 sample Memory Test scores Pre-test Memory Test scores Post-test Memory patients Memory treatment Compare pair-wise differences 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. 1 sample 2 scores per subject Memory Test scores Pre-test Memory Test scores Post-test Memory patients Memory treatment Compare pair-wise differences Hypotheses: Related-samples t-test: Within Persons/Repeated Measures

The related-samples t-test can be used when: 1 sample 2 scores per subject 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. No Memory treatment Memory patients Memory treatment 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. 2 samples, matched related On a pair-by-pair basis every person in the No Treatment group is matched (related) to a person in the Memory Treatment group & the average difference is calculated No Memory treatment Memory patients Memory treatment Not Independent groups participants matched Pair 1, condition A = male, 26, Age = 21 Pair 1, condition B= male, 24, Age = 21 pre existing relationship twins - one in each group couples - one in each group matched Red Short 21yrs Blue tall 23yrs Green average 22yrs Brown Related-samples t-test: Matched Pairs

The related samples t-test can be used when: 2 samples matched 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 All positive Ds since lower on post-test 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

α = 0.05 2-tailed df = n - 1 = 5 tcrit = The t-distribution

What is the tcrit for a 1-tailed hypothesis test (looking for higher scores) 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 tcrit = 1-tailed, larger critical region in 1-tail, so smaller critical value needed. tcrit = vs. ±2.571 The t-distribution

In lab: Practice using related 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

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, sD = 2. Test this with α = 0.05. Memory Test scores Pre-test Memory Test scores Post-test Memory patients Memory treatment Compare pair-wise differences Hypotheses: Memory performance at 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

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, sD = 4. Test this with α = 0.05. H0: HA: 2-tailed α = 0.05 Step 1: Hypotheses Step 2: Criterion for decision Step 3: Sample statistics Step 4: Test statistic = - 5, sD = 4 df = n-1 = 24 Related-samples t-test: Within Persons

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, sD = 4. Test this with α = 0.05. H0: HA: 2-tailed, α = 0.05 t = df = 24 Step 1: Hypotheses Step 2: Criterion for decision Step 3: Sample statistics Step 4: Test statistic Step 5: Compare observed to critical value = 5, sD = 4 -6.25 > |2.064| Related-samples t-test: Within Persons

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, sD = 4. Test this with α = 0.05. H0: HA: 2-tailed, α = 0.05 t = df = 24 Step 1: Hypotheses Step 2: Criterion for decision Step 3: Sample statistics Step 4: Test statistic Step 5: Compare observed to critical value Decide about hypotheses Conclude about relationship tobs = -6.25 = 5, sD = 4 tcrit = ± 2.064 “Reject H0: Evidence supports the hypothesis that the memory treatment improved performance.” Related-samples t-test: Within Persons

Reasoning in Psychology Using Statistics

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