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One Sample t Tests. Karl L. Wuensch Department of Psychology East Carolina University. Nondirectional Test. Null: = some value Alternative: that value We have a sample of N scores Somehow we magically know the value of the population

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## One Sample t Tests

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**One Sample t Tests**Karl L. Wuensch Department of Psychology East Carolina University**Nondirectional Test**• Null: = some value • Alternative: that value • We have a sample of N scores • Somehow we magically know the value of the population • We trust that the population is normally distributed • Or invoke the Central Limit Theorem**H0: IQ = 100**N = 25, M = 107, = 15 p = .0198, two-tailed**Directional Test**• For z = 2.33 • If predicted direction in H1 is correct, then p = .0099 • If predicted direction in H1 is not correct, then p = 1 - .0099 = .9901**The Fly in the Ointment**• How could we know the value of but not know the value of ?**Student’s t**• The sampling distribution of 2 is unbiased but positively skewed. • Thus, more often than not, s2 < 2 • And | t | > | z |, giving t fat tails (high kurtosis)**Fat-Tailed t**• Because of those fat tails, one will need go out further from the mean to get to the rejection region. • How much further depends on the df, which are N-1. • The fewer the df, the further out the critical values. • As dfincrease, t approaches the normal distribution.**SAT-Math**• For the entire nation, between 2000 and 2004, = 516. • For my students in undergrad stats: • M = 534.78 • s = 93.385 • N = 114 • H0: For the population from which my students came, = 516.**We Reject That Null**df = N – 1 = 113 p = .034**CI.95**• From the t table for df = 100, CV = 1.984.**Effect Size**• Estimate by how much the null is wrong. • Point estimate = M – null value • Can construct a CI. • For our data, take the CI for M and subtract from each side the null value • [517.43 – 516, 552.13 – 516] = • [1.43, 36.13]**Standardized Effect Size**• When the unit of measure is not intrinsically meaningful, • As is often case with variables studied by psychologists, • Best to estimate the effect size in standard deviation units. • The parameter is**Estimated **• We should report a CI for • Constructing it by hand in unreasonably difficult. • Professor Karl will show how to use SAS or SPSS to get the CI.**Assumptions**• Only one here, that the population is normally distributed. • If that is questionable, one might use nonlinear transformations, especially if the problem is skewness. • Or, use analyses that make no normality assumption (nonparametrics and resampling statistics).**Summary Statements**• who or what the research units were (sometimes called “subjects” or “participants”) • what the null hypothesis was (implied) • descriptive statistics such as means and standard deviations • whether or not you rejected the null hypothesis**Summary Statements 2**• if you did reject the null hypothesis, what was the observed direction of the difference between the obtained results and those expected under the null hypothesis • what test statistic (such as t) was employed • the degrees of freedom**Summary Statements 3**• if not obtainable from the degrees of freedom, the sample size • the computed value of the test statistic • the p value (use SPSS or SAS to get an exact p value) • an effect size estimate • and a confidence interval for the effect size parameter**Example Summary Statements**• Carefully study my examples in my document One Mean Inference. • Pay special attention to when and when not to indicate a direction of effect. • and also when the CI would more appropriately be with confidence coefficient (1 - 2) rather than (1 - ).

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