Reasoning in Psychology Using Statistics

Reasoning in Psychology Using Statistics
2019

Quiz 5 is due Tue. Mar. 22nd at midnight
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Exam(s) 2 Combined Exam 2: 80.4% Lab Ex2, mean = 64.8/75 = 86.4%
Lecture Ex2, mean = 55.8/75 = 74.41% Combined Exam 2: 80.4% Exam(s) 2

Testing Hypotheses 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 H0 Null Hypothesis HA Alternative Hypothesis ⍺ Alpha level Distribution of Sample Means Note: In the labs I combine steps 1 & 2, so it is described as a 4 step program Statistics How To: Hypotheses (~4 mins) StatsLectures: Hypotheses and Error types (~4 mins) Hypothesis testing and Error types (~7 mins) Testing Hypotheses

Distribution of sample means
2 3 4 5 6 7 8 1 In long run, the random selection of tiles leads to a predictable pattern The distribution of sample means (DSM) mean mean mean 2 2 2 4 6 5 8 2 5 2 4 3 4 8 6 8 4 6 2 6 4 6 2 4 8 6 7 2 8 5 6 4 5 8 8 8 4 2 3 6 6 6 4 4 4 6 8 7 Population: 2 4 6 8 All 16 samples of n=2 from Distribution of sample means

Distribution of sample means
Distribution of sample means is a “virtual” distribution between the sample and population Note: There is a different one for each sample size Population Distribution of sample means Sample Shape Center Spread Distribution of sample means

Properties of the distribution of sample means
Shape If population is Normal, then the dist of sample means will be Normal If the sample size is large (n > 30), the DSM will be approximately Normal (regardless of shape of the population) Population Distribution of sample means n > 30 Properties of the distribution of sample means

Center The mean of the dist of sample means is equal to the mean of the population Population Distribution of sample means same numeric value different conceptual values Properties of the distribution of sample means

Center The mean of the dist of sample means is equal to the mean of the population Consider our earlier example 2 4 6 8 Population Distribution of sample means means 2 3 4 5 6 7 8 1 4 μ = = 5 16 = = 5 Properties of the distribution of sample means

Spread Standard deviation of the population Sample size The Stand. Dev. of the Distrib. of Sample Mean depends on 2 things Properties of the distribution of sample means

Spread Standard deviation of the population μ X 1 2 3 μ X 1 2 3 The smaller the population variability, the closer the sample means are to the population mean, so the smaller the spread of sample means Properties of the distribution of sample means

Spread Standard deviation of the population Sample size n = 1 μ X Properties of the distribution of sample means

Spread Standard deviation of the population Sample size n = 10 μ X Properties of the distribution of sample means

Spread Standard deviation of the population Sample size n = 100 μ - The larger the sample size the smaller the spread of sample means X Properties of the distribution of sample means

Spread Standard deviation of the population Sample size Putting them together we get the standard deviation of the distribution of sample means - The smaller the population variability, the smaller the spread - The larger the sample size the smaller the spread Commonly called the standard error Properties of the distribution of sample means

All three of these properties are combined to form the Central Limit Theorem MEMORIZE THIS For any population with mean μ and standard deviation σ, the distribution of sample means for sample size n will approach a normal distribution with a mean of μ and a standard deviation of as n approaches infinity (good approximation if n > 30). Properties of the distribution of sample means Central Limit Theorem (~13 mins)

The standard error is the average amount that you’d expect a sample (of size n) to deviate from the population mean In other words, it is an estimate of the error that you’d expect by chance (it is our estimate of the sampling error) Keep your distributions straight by taking care with your notation Population σ μ Distribution of sample means Sample s X Properties of the distribution of sample means

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

Statistical test decision tree
How do we know which test to use? The design of the research: how many groups, how many scores per person, etc. Start here Statistical test decision tree

Generic statistical test
How do we know which test to use? The design of the research: how many groups, how many scores per person, etc. Could be difference between a sample and a population, or between different samples Both of these parts change as a function of the design Based on standard error or an estimate of the standard error As a result, the test statistic changes Generic statistical test

Using the distribution of sample means
Test statistic Transform the distribution of sample means into the appropriate standardized distribution (as determined by the design features) Distribution of sample means Test statistic distribution We will use 2: z’s & t’s Using the distribution of sample means

One sample z-test statistic
Old z-formula New z-formula Same as before, with two differences: Uses the distribution of sample means Ask questions about samples rather than individual scores One sample z-test statistic

Old z-formula New z-formula mean of the distribution of sample means population mean raw score sample mean population standard deviation standard error One sample z-test statistic

Old z-formula New z-formula Same as before, with two differences: Uses the distribution of sample means Ask questions about samples rather than individual scores One sample z-test statistic

What is the probability of getting a 630 or better on the SAT? Old z-formula μ = 500, σ = 100, Normal From the table: z(1.3) =.0968 So the probability is What is the probability of getting a sample of n = 4 students with an average of 630 or better on the SAT? New z-formula μ = 500, σ = 100, Normal From the table: z(2.6) =.0047 So the probability is One sample z-test statistic

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

Performing your statistical test
What are we doing when we test the hypotheses? Consider a variation of our memory experiment example Population of memory patients MemoryTest μ & σ known Memory treatment patients Test X Compare these two means We test this one 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 H0: HA: Performing your statistical test

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

“Generic” statistical test
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’s large enough? The alpha level gives us the decision criterion Distribution of the test statistic α-level determines where these boundaries go “Generic” statistical test

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’s large enough? The alpha level gives us the decision criterion Distribution of the test statistic If test statistic is here Reject H0 If test statistic is here Fail to reject H0 “Generic” statistical test

The alpha level gives us the decision criterion Two -tailed z .00 .01 … .06 -3.4 -3.3 : -1.9 1.0 1.9 3.4 0.0003 0.0005 0.0287 0.5000 0.8413 0.9713 0.9997 0.0281 0.5040 0.8438 .9719 .0250 .9750 α = 0.05 0.025 split up into the two tails Reject H0 Fail to reject H0 Go to the table (unit normal table for z-test) and find the z that has in the tails. Zcritical = ±1.96 “Generic” statistical test

The alpha level gives us the decision criterion Two -tailed One -tailed Reject H0 Fail to reject H0 α = 0.05 0.05 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 in the tail. Zcritical = “Generic” statistical test

The alpha level gives us the decision criterion Two -tailed One -tailed Reject H0 Fail to reject H0 α = 0.05 all of it in one tail 0.05 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 in the tail. Zcritical = “Generic” statistical test

1-sample z-test Population of memory patients Memory Test σ is known
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 treatment patients Test X Compare these two means 1-sample z-test

1-sample z-test 1 sample Population of memory patients Memory 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. 1 sample Population of memory patients Memory Test σ is known μ is known Memory treatment patients Test X Compare these two means 1-sample z-test

1-sample z-test 1 sample 1 score per subject Population of
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. 1 sample 1 score per subject Population of memory patients Memory Test σ is known μ is known Memory treatment patients Test X Compare these two means 1-sample z-test

1-sample z-test 1-sample z-test 1 sample 1 score per subject
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. 1 sample 1 score per subject Population mean (μ) and standard deviation (σ) are known (assume Normal dist) Population of memory patients Memory Test σ is known μ is known Memory treatment patients Test X Compare these two means 1-sample z-test 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. One -tailed Step 1: State your hypotheses μTreatment > μpop = 60 Step 2: Set your decision criteria H0: the memory treatment sample are the same as those in the population of memory patients (or even worse). Step 3: Collect your data Step 4: Compute your test statistics HA: the memory treatment sample perform better (fewer errors) than those in the population of memory patients μTreatment < μpop = 60 Step 5: Make a decision about your null hypothesis Performing your statistical 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. H0: μTreatment > μpop = 60 HA: μTreatment < μpop = 60 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

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 60 errors, with a σ = 8). Test using α = 0.05. H0: μTreatment > μpop = 60 HA: μTreatment < μpop = 60 n = 16, X = 55 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

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 60 errors, with a σ = 8). Test using α = 0.05. H0: μTreatment > μpop = 60 HA: μTreatment < μpop = 60 One -tailed α = 0.05 n = 16, X = 55 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

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 60 errors, with a σ = 8). Test using α = 0.05. H0: μTreatment > μpop = 60 HA: μTreatment < μpop = 60 One -tailed α = 0.05 n = 16, X = 55 Step 1: State your hypotheses = -2.5 Step 2: Set your decision criteria Step 3: Collect your data Step 4: Compute your test statistics 5% 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

Wrap up In lab Make hypotheses (both null & alternative)
Test hypotheses using 1-sample z-test Questions? Wrap up 45

If time allows: The following pages give examples of situations that require different statistical tests. 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 collects a sample of 25 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with a s = 8 (while the typical memory patient averages 60 errors). Population of memory patients MemoryTest  is NOT known is known Memory treatment patients Test X Compare these two means Hypotheses: 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 H0: HA: 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 collects a sample of 25 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with a s = 8 (while the typical memory patient averages 60 errors). 1 sample Population of memory patients MemoryTest  is NOT known is known Memory treatment patients Test X Compare these two means Hypotheses: 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 H0: HA: 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 collects a sample of 25 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with a s = 8 (while the typical memory patient averages 60 errors). 1 sample One score per subject Population of memory patients MemoryTest  is NOT known is known Memory treatment patients Test X Compare these two means Hypotheses: 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 H0: HA: 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 collects a sample of 25 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with a s = 8 (while the typical memory patient averages 60 errors). 1 sample One score per subject Population mean (μ) is known Population of memory patients MemoryTest  is NOT known is known Memory treatment patients Test X Compare these two means Hypotheses: 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 H0: HA: 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 collects a sample of 25 patients and gives them his new treatment. Following the treatment he gives them a standard memory test. His sample averaged 55 errors, with a s = 8 (while the typical memory patient averages 60 errors). 1 sample One score per subject Population mean (μ) is known Population standard deviation (s) is NOT known Population of memory patients MemoryTest  is NOT known is known Memory treatment patients Test X Compare these two means Hypotheses: 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 H0: HA: Performing your statistical test

Which test do we use? The single sample t-test can be used when:
One score per subject Population mean (m) is known Population standard deviation (s) is NOT known Which test do we use?

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 60 errors before the treatment and 55 errors after the treatment. Memory Test X Pre-test Memory Test X Post-test Memory patients Memory treatment Compare these two means Hypotheses: Memory performance at the post-test is equal to memory performance at the pre-test. H0: HA: Memory performance at the post-test is NOT equal to memory performance at the pre-test 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 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 60 errors before the treatment and 55 errors after the treatment. 1 sample Memory Test X Pre-test Memory Test X Post-test Memory patients Memory treatment Compare these two means Hypotheses: Memory performance at the post-test is equal to memory performance at the pre-test. H0: HA: Memory performance at the post-test is NOT equal to memory performance at the pre-test 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 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 60 errors before the treatment and 55 errors after the treatment. 1 sample Two scores per subject Memory Test X Pre-test Memory Test X Post-test Memory patients Memory treatment Compare these two means Hypotheses: Memory performance at the post-test is equal to memory performance at the pre-test. H0: HA: Memory performance at the post-test is NOT equal to memory performance at the pre-test Performing your statistical test

Which test do we use? The related sample t-test can be used when:
Two scores per subject Which test do we use?

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 matches them to a sample of similar individuals. He then gives them one sample the new treatment (but not the other). Following the treatment period he gives both groups a memory test. His treatment sample averaged 55 errors after the treatment and his matched sample averaged 60 errors over the same time period. related On a pair-by-pair basis every person in the No Treatment group is related to or matched to a person in the Memory Treatment group No Memory treatment Memory patients Memory treatment 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 collects a sample of 25 patients and matches them to a sample of similar individuals. He then gives them one sample the new treatment (but not the other). Following the treatment period he gives both groups a memory test. His treatment sample averaged 55 errors after the treatment and his matched sample averaged 60 errors over the same time period. Memory Test X No Memory treatment Compare these two means Memory patients related Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: 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 collects a sample of 25 patients and matches them to a sample of similar individuals. He then gives them one sample the new treatment (but not the other). Following the treatment period he gives both groups a memory test. His treatment sample averaged 55 errors after the treatment and his matched sample averaged 60 errors over the same time period. 2 samples Memory Test X No Memory treatment Compare these two means Memory patients related Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: 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 collects a sample of 25 patients and matches them to a sample of similar individuals. He then gives them one sample the new treatment (but not the other). Following the treatment period he gives both groups a memory test. His treatment sample averaged 55 errors after the treatment and his matched sample averaged 60 errors over the same time period. 2 samples Samples are matched with one score per subject Memory Test X No Memory treatment Compare these two means Memory patients related Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: Performing your statistical test

Which test do we use? The related sample t-test can be used when:
2 samples Samples are matched with one score per subject Which test do we use?

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 50 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. His treatment sample averaged 55 errors after the treatment and his control sample averaged 60 errors over the same time period. Memory Test X No Memory treatment Compare these two means Memory patients Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: 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 50 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. His treatment sample averaged 55 errors after the treatment and his control sample averaged 60 errors over the same time period. 2 samples Memory Test X No Memory treatment Compare these two means Memory patients Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: 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 50 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. His treatment sample averaged 55 errors after the treatment and his control sample averaged 60 errors over the same time period. 2 samples Samples are independent Memory Test X No Memory treatment Compare these two means Memory patients Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: 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 50 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. His treatment sample averaged 55 errors after the treatment and his control sample averaged 60 errors over the same time period. 2 samples Samples are independent Memory Test X One score per subject No Memory treatment Compare these two means Memory patients Memory treatment Hypotheses: H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. Memory performance by the treatment group is NOT equal to memory performance by the no treatment group. HA: Performing your statistical test

Which test do we use? The independent sample t-test can be used when:
2 samples Samples are independent One score per subject Which test do we use?

Wrap up In lab Make hypotheses (both null & alternative)
Test hypotheses using 1-sample z-test Questions? Wrap up 67

Reasoning in Psychology Using Statistics

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