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
M 
Z

15

3
N
25
M  
M
107  100

 2.33
3
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
Confidence Interval
CI  M  CV   M  M  CV   M
CI.95 
107  1.96(15 5)  107  1.96(15 5)
 101.12  112.88
The Fly in the Ointment
• How could we
know the value
of  but not
know the value
of  ?
Student’s t
Z
M  

N
t
M  
s
N
• 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 df increase, t approaches the normal
distribution.
CV for t,  = .05, 2-tailed
Degrees of Freedom
1
2
3
10
30
100

Critical Value for t
12.706
4.303
3.182
2.228
2.042
1.984
1.960
William Gosset
SAT-Math
• For the entire nation, between 2000 and
2004,  = 516.
• For my students in undergrad stats:
M = 534.78
s = 93.385
N = 114
• H0: For the population from which my
students came,  = 516.
•
We Reject That Null
s
93.385
sM 

 8.746
N
114
M  0 534.78  516
t

 2.147
sM
8.746
df = N – 1 = 113
p = .034
CI.95
CI  M  CV  sM  M  CV  sM
• From the t table for df = 100, CV = 1.984.
CI.95 
534.78  1.984(8.746)  534.78  1.984(8.746)
 517.43  552.13
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 
d
M  
s
18.78

 .20
93.385
• 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 - ).
The t Family