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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  - PowerPoint PPT Presentation
One Sample t Tests Karl L. Wuensch Department of Psychology East Carolina University
Page 1: One Sample  t  Tests

One Sample t TestsKarl L. Wuensch

Department of PsychologyEast Carolina University

Page 2: One Sample  t  Tests

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

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H0: IQ = 100 N = 25, M = 107, = 15


15 NM

p = .0198, two-tailed





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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

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Confidence Interval





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The Fly in the Ointment

• How could we know the value of but not know the value of ?

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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




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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 df increase, t approaches the normal distribution.

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CV for t, = .05, 2-tailedDegrees of Freedom Critical Value for t

1 12.7062 4.3033 3.182

10 2.22830 2.042

100 1.984 1.960

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William Gosset

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SAT-Math• For the entire nation, between 2000 and

2004, = 516.• For my students in undergrad stats:

M = 534.78s = 93.385N = 114

• H0: For the population from which my students came, = 516.

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We Reject That Null

df = N – 1 = 113 p = .034







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• From the t table for df = 100, CV = 1.984.





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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]

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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

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• 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.




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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).

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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


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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

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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

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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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The t Family