Upload
gisela
View
37
Download
0
Embed Size (px)
DESCRIPTION
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
Citation preview
One Sample t TestsKarl L. Wuensch
Department of PsychologyEast 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
325
15 NM
p = .0198, two-tailed
33.23
100107
M
MZ
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
Confidence Interval
MM CVMCVMCI
88.11212.101)515(96.1107)515(96.1107
95.
CI
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)
NsMt
NMZ
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.
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
William Gosset
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.
•
We Reject That Null
df = N – 1 = 113 p = .034
746.8114385.93
NssM
147.2746.8
51678.5340
Ms
Mt
CI.95
• From the t table for df = 100, CV = 1.984.
MM sCVMsCVMCI
13.55243.517)746.8(984.178.534)746.8(984.178.534
95.
CI
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.
20.385.9378.18
s
Md
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 - ).
The t Family