# Statistics for the Social Sciences

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Statistics for the Social Sciences
Psychology 340 Spring 2010 Using t-tests (independent samples)

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

Independent samples What are we doing when we test the hypotheses?
Consider a new variation of our memory experiment example Memory Test Memory placebo Compare these two means Memory patients Memory Test Memory treatment 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:

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

Independent-samples t One-sample t Observed (sample) means Test statistic

Independent-samples t One-sample t Hypothesized population means from the Null hypothesis Test statistic

Independent-samples t One-sample t Test statistic Hypothesized population means from the Null hypothesis H0: Memory performance by the treatment group is equal to memory performance by the no treatment group. So:

One-sample t Test statistic Estimated standard error (difference expected by chance) We have two samples, so the estimate is based on two samples estimate is based on one sample The Estimate of the Standard Error is based on the variability of both samples

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

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 Compute your estimated standard error Compute your t-statistic Compute your degrees of freedom This is the one you use to look up your tcrit

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 = 0.05. Need to compute the mean and variability for each sample Exp. group Control group 45 55 40 60 43 49 35 51

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 = 0.05. Need to compute the mean and variability 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

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 = 0.05. Need to compute the mean and variability 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

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 = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95

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 = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95 Tobs= 0.95 Tcrit= ±2.447 α = 0.05 Two-tailed

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 = 0.05. Exp. group Control group 45 55 40 60 43 49 35 51 = 0.95 Tobs= 0.95 Tcrit= ±2.447 α = 0.05 Two-tailed +2.45 = tcrit - Fail to Reject H0 tobs=0.95

Assumptions: Independent samples t
Each of the population distributions follows a normal curve (this is an assumption of all t-tests) T-tests are fairly ‘robust’ against this assumption This means that the results generally still hold even if this assumption is violated Homogeneity of variance: The two populations have the same variance SPSS tests this using Levene’s Test Two rows in the SPSS output Us the top row if the p-value for the Levene’s test is greater than 0.05 Use the bottom row if the p-value for the Levene’s test is less than 0.05 Tests the Null hypothesis that the two groups have equal variances

Effect Size for the t Test for Independent Means
Estimated effect size after a completed study “pooled standard deviation” not “pooled variance,” so take the square root of sP2

Power for the t Test for Independent Means (.05 significance level)
8-5 8-4

Approximate Sample Size Needed for 80% Power (.05 significance level)
8-5

Statistical Tests Summary
Design Statistical test (Estimated) Standard error One sample, σ known One sample, σ unknown Next time Two related samples, σ unknown Two independent samples, σ unknown

Using SPSS: Independent samples t
Person Cntrl-grp Exp-grp Entering the data Different groups of observations go into SAME column e.g., Exp grp and control grp in a single column 1 45 43 2 55 49 3 40 35 4 60 51 Separate column defines the group membership for each observation e.g., exp grp = 0, control grp = 1 Performing the analysis Analyze -> Compare means -> independent samples t-test Identify which columns have the observations (test variable) and which column has the group membership defined (grouping variable) Define groups: what numbers correspond to the two groups? Reading the output Means of the different groups, the mean difference, the computed-t, degrees of freedom, p-value (Sig.), Levene’s test