Tuesday, September 24, 2013 Independent samples t-test

Exam I Results, 2013 M = S = ,9,8,8,8,7,6,5,4,4,3,1,1 79,8,6,5,4 69,8,2, Good work; room to improve Caution: More effort needed Trouble: Seek extra help asap

Next Homework (due Tuesday, October 1) Chapter 10: 1-4, 6, 21, 22 Chapter 11: 1, 5, 9, 10, 18, 20

Monitor Your Understanding Please click on the above link, and type (or paste) where it says “classroom”: _9/24/13 Then click buttons “got it” “unsure” or “lost me” to provide anonymous feedback about your understanding throughout the lecture

Last Time Introduction to t-tests (starting with the one- sample t-test) Use of s M (the estimated standard error in the denominator) Knowing when to use the one-sample t-test and when to use the z-score test Questions about these items before we move on?

Statistical analysis follows design The one-sample z-test can be used when: –1 sample –One score per subject –Population mean (μ) and standard deviation (  ) are known

Statistical analysis follows design The one-sample t-test can be used when: –1 sample –One score per subject –Population mean (μ) is known –but standard deviation (  ) is NOT known

Monitor your understanding Please provide feedback on teacher tap to let me know if you understand. Classroom: _9/24/13

Independent samples What are we doing when we test the hypotheses? – Consider a new variation of our memory experiment example Memory treatment Memory patients Memory Test the memory treatment population is the same as those in the population of untreated memory patients (μ A = μ B ). they aren’t the same as those in the population of memory patients (μ A ≠ μ B ) H0:H0: HA:HA: Memory placebo Memory Test Compare these two means

Statistical analysis follows design The independent samples t-test can be used when: –2 samples –Samples are independent

Performing your statistical test Test statistic Diff. Expected by chance Standard error Estimated standard error One sample z One sample t different Don’t know this, so need to estimate it Degrees of freedom

Statistical Tests Summary DesignStatistical test(Estimated) Standard error One sample, σ known One sample, σ unknown Two independent samples, σ unknown (n A – 1) + (n B -1) Degrees of freedom n – 1

Performing your statistical test Estimate of the standard error based on the variability of both samples

Performing your statistical test Test statistic One-sample t Independent-samples t Sample means

Performing your statistical test Test statistic One-sample t Independent-samples t Population means from the null hypothesis

Performing your statistical test Test statistic One-sample t Independent-samples t Population means from the hypotheses H0:H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. So:

Monitor your understanding Please provide feedback on teacher tap to let me know if you understand. Classroom: _9/24/13

Performing your statistical test Test statistic One-sample t 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

Monitor your understanding Please provide feedback on teacher tap to let me know if you understand. Classroom: _9/24/13

Performing your statistical test “pooled variance” We combine the variance from the two samples Number of subjects in group A Number of subjects in group B

variance Performing your statistical test “pooled variance” We combine the variance from the two samples Recall “weighted means,” need to use “weighted variances” here Variance (s 2 ) * degrees of freedom (df)

Performing your statistical test Independent-samples t Compute your estimated standard error Compute your t-statistic Compute your degrees of freedom This is the one you use to look up your t crit

Monitor your understanding Please provide feedback on teacher tap to let me know if you understand. Classroom: _9/24/13

Performing your statistical test Person Exp. group Control group H 0 : µ A = µ B (µ A - µ B = 0) H A : µ A ≠ µ B 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 α = α = tailed test

Performing your statistical test Person Exp. group Control group Need to compute the mean and variability for each sample 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 α = H 0 : µ A = µ B (µ A - µ B = 0); H A : µ A ≠ µ B; α = 0.05; 2-tailed

Performing your statistical test Person Exp. group Control group Need to compute the mean and variability for each sample Control group = 50 (45-50) 2 + (55-50) 2 + (40-50) 2 + (60-50) 2 = 250 SS = A 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 α = H 0 : µ treatment = µ control H A : µ treatment ≠ µ control, α = tailed

Performing your statistical test Exp. group ( ) 2 + ( ) 2 + ( ) 2 + ( ) 2 = 155 SS = B Person Exp. group Control group Need to compute the mean and variability for each sample 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 α = = 44.5 H 0 : µ A = µ B (µ A - µ B = 0); H A : µ A ≠ µ B; α = 0.05; 2-tailed

Performing your statistical test Person Exp. group Control group 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 α = = 0.95 H 0 : µ A = µ B (µ A - µ B = 0); H A : µ A ≠ µ B; α = 0.05; 2-tailed

Performing your statistical test T obs = 0.95 T crit = ±2.447 α = 0.05 Two-tailed Person Exp. group Control group 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 α = = 0.95

Performing your statistical test T obs = 0.95 α = 0.05 Two-tailed T crit = ±2.447 Person Exp. group Control group 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 α = = t crit - Fail to Reject H 0 t obs =0.95 = 0.95

Performing your statistical test T obs = 0.95 α = 0.05 Two-tailed T crit = ±2.447 T obs T crit Compare < Fail to reject the H 0 and conclude that  A =  B Person Exp. group Control group 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 α= = 0.95

Monitor your understanding Please provide feedback on teacher tap to let me know if you understand. Classroom: _9/24/13

Assumptions of the independent samples t-test Each of the population distributions follows a normal curve (but test is robust to violations of this assumption if the sample is large) The two populations have the same variance If the variance is not equal and the samples are very different in size, use the corrected degrees of freedom provided after Levene’s test (see spss output)

Practice problem You are conducting an experiment about methods for teaching reading. You have access to a sample of 10 3 rd graders. You randomly assign half of the group to an experimental reading intervention, and the other half receives instruction as usual in their classrooms. After the intervention, you measure the number of words each child can read correctly in one minute, and obtain the following results. Group 1 (experimental) scores: 30, 35, 40, 20, 32 Group 2 scores: 25, 30, 20, 18, 18 Conduct a t-test to find out whether the groups are different in reading ability at the end of the study. t table

Using spss to conduct t-tests One-sample t-test: Analyze =>Compare Means =>One sample t-test. Select the variable you want to analyze, and type in the expected mean based on your null hypothesis. Independent samples t-test: Analyze =>Compare Means =>Independent samples t-test. Specify test variable and grouping variable, and click on define groups to specify how grouping variable will identify groups.

Practice problem Use SPSS to conduct a one-sample t-test to see if the students in this class are taller than the average college student ( μ = 66.5 inches). Use SPSS to conduct an independent samples t-test comparing the heights of men and women, based on the data from this class.

Using excel to compute t-tests =t-test(array1,array2,tails,type) Select the arrays that you want to compare, specify number of tails (1 or 2) and type of t-test (1=dependent, 2=independent w/equal variance assumed, 3=independent w/unequal variance assumed). Returns the p-value associated with the t- test.

Effect Size for the t Test for Independent Means Estimated effect size after a completed study

Power for the t Test for Independent Means (.05 significance level)

Approximate Sample Size Needed for 80% Power (.05 significance level)

Next Homework (due Tuesday, October 1) Chapter 10: 1-4, 6, 21, 22 Chapter 11: 1, 5, 9, 10, 18, 20