Transcript Reasoning of significance tests

T-tests
• Computing a t-test
 the t statistic
 the t distribution
• Measures of Effect Size
 Confidence Intervals
 Cohen’s d
Part I
• Computing a t-test
 the t statistic
 the t distribution
Z as test statistic
• Use a Z-statistic only if you know the population standard
deviation (σ).
•Z-statistic converts a sample mean into a z-score from the null
distribution.
Z test 
X  H0
SE

X  H0

n
• p-value is the probability of getting a Ztest as extreme as yours
under the null distribution
One tail
Reject H0
Fail to reject H0
.05
Zcrit
-1.65
Two tail
Reject H0
Reject H0
Fail to reject H0
.025
Zcrit
-1.96
.025
Zcrit
1.96
t as a test statistic
Sample
_
X
Sample
_
X, s
Population

z-test
Population
t-test
t as a test statistic
• t-test: uses sample data to evaluate a hypothesis about a
population mean when population stdev () is unknown
• We use the sample stdev (s) to estimate the standard error
standard
error x
z
=

n
X  H0
X
estimated
standard
error
t
sx =
s
n
X  H0
sX
t as a test statistic
• Use a t-statistic when you don’t know the population standard
deviation (σ).
•t-statistic converts a sample mean into a t-score (using the null
hypothesis)
ttest 
X  H0
SE

X  H0
s
n
• p-value is the probability of getting a ttest as extreme as yours
under the null distribution
t distribution
• You can use s to approximate σ, but then the sampling
distribution is a t distribution instead of a normal distribution
•Why are Z-scores normally distributed, but t-scores are not?
normal
normal
Random
variable
Random
variable
constant
ztest 
X  H0
X
constant
ttest 
X  H0
sX
non normal
constant
Random
variable
t distribution
•With a very large sample, the estimated standard error will be very
near the true standard error, and thus t will be almost exactly the
same as Z.
•Unlike the standard normal (z) distribution, t is a family of curves.
As n gets bigger, t becomes more normal.
•For smaller n, the t distribution is platykurtic (narrower peak, fatter
tails)
•We use “degrees of freedom” to identify which t curve to use. For
a basic t-test, df = n-1
Comparing a t (df=5)
with the Standard Normal Distribution
Zcrit Tcrit
1.96 2.57
Degrees of freedom
the number of scores in a sample that are free to vary
e.g. for 1 sample, sample mean restricts one value so df = n-1
As df approaches infinity, t-distribution approximates a normal curve
At low df, t-distribution has “fat tails”, which means tcrit is going to
be a bit larger than Zcrit. Thus the sample evidence must be more
extreme before we get to reject the null. (a tougher test).
t distribution
Too many different curves to put each one in a table. Table E.6
shows just the critical values for one tail at various degrees of
freedom and various levels of alpha.
Table E.6
Level of significance for a one-tailed test
df
.05
1
2
3
4
6.314
2.920
2.353
2.132
.025
.01
.005
tcrit
12.706 32.821 63.657
4.303 6.965 9.925
3.182 4.541 5.841
2.776 3.747 4.604
This makes it harder to get exact p-values. You have to estimate.
Practice with Table E.6
With a sample of size 6, what is the degrees of freedom? For a onetailed test, what is the critical value of t for an alpha of .05? For an
alpha of .01?
df=5, t =2.015; t =3.365
crit
crit
For a sample of size 25, doing a two-tailed test, what is the degrees
of freedom and the critical value of t for an alpha of .05 and for an
alpha of .01?
df=24, tcrit=2.064; tcrit=2.797
You have a sample of size 13 and you are doing a one-tailed test.
Your tcalc = 2. What do you approximate the p-value to be?
p-value between .025 and .05
What if you had the same data, but were doing a two-tailed test?
p-value between .05 and .10
Illustration
In a study of families of cancer patients, Compas et al (1994)
observed that very young children report few symptoms of anxiety
on the CMAS. Contained within the CMAS are nine items that
make up a “social desirability scale”. Compas wanted to know if
young children have unusually high social desirability scores.
Illustration
He got a sample of 36 children of families with a cancer parent.
The mean SDS score was 4.39 with a standard deviation of 2.61.
Previous studies indicated that a population of elementary school
children (all ages) typically has a mean of 3.87 on the SDS.
Is there evidence that Compas’s sample of very young children was
significantly different than the general child population?
tcalc=1.195, df = 35
two tailed p-value = btwn .20 and .30
What should he conclude?
What can he do now?
Factors that affect the magnitude of t and
the decision
• the actual obtained difference X  
• the magnitude of the sample variance (s2)
• the sample size (n)
• the significance level (alpha)
• whether the test is one-tail or two-tail
How could you increase your chances of
rejecting the null?
Part II
• Measures of Effect Size
 Confidence Intervals
 Cohen’s d
Hypothesis Tests vs Effect Size
• Hypothesis Tests
•Set up a null hypothesis about µ. Reject (or fail to reject) it.
•Only indicates direction of effect (e.g. >μH0)
•Says nothing about effect size except “large enough to be
significant”.
• Effect Size
•Tells you about the magnitude of the phenomenon
•Helpful in deciding “importance”
•Not just “which direction” but “how far”
P-value: Bad Measure of Effect
Size
• “Significant” does not mean important or large
• Significance is dependent on sample size
• “The null hypothesis is never true in fact. Give me a
large enough sample and I can guarantee a significant
result. ” -Abelson
Confidence Interval
•
We could estimate effect size with our observed sample
deviation X   H
0
•
But we want a window of uncertainty around that
estimate.
•
So we provide a “confidence interval” for our observed
deviation
•
We say we are xx% confident that the true effect size
lies somewhere in that window
Finding the Window

__
X
Finding the Window

__
X
Finding the Window

could be…
__
X
If we want alpha = .05, what is the lowest µH0 we would accept?
.05

X  1.96( SE )
__
X
If we want alpha = .05, what is the highest µH0 we would accept?
.05
__
X

X  1.96(SE )
Let’s generalize.
Our Window!
For any particular level of alpha, the confidence interval is
X  tcrit (SE )
X  1.96( SE )
to
__
X
X  tcrit (SE )
X  1.96(SE )
Confidence intervals
Confidence level = 1 - 
If alpha is .05, then the confidence level is 95%
95% confidence means that 95% of the time, this procedure will
capture the true mean (or the true effect) somewhere within the
range.
Constructing confidence intervals
Choose level of confidence (90%, 95%, 99%…)
Find critical t-value (compare with two-tailed alphas)
Find standard error
Get Confidence Interval
C.I. for mean
C.I. for effect
X  tcrit ( s X )
( X  H0 )  tcrit (s X )
Exercise in constructing CI
We have a sample of 10 girls who, on average, went on their 1st
dates at 15.5 yrs, with a standard deviation of 4.2 years.
What range of values can we assert with 95% confidence
contains the true population mean?
Margin = 3 years
CI = (12.50, 18.50)
• Using an alpha=.05, would
we reject the null hypothesis
that µ=10?
•What about that µ=17?
yes
no
Exercise in constructing CI
We have a sample of 10 girls who, on average, went on their 1st
dates at 15.5 yrs, with a standard deviation of 4.2 years.
Let’s say we were comparing this sample (of girls from New
York) to the general American population μ = 13 years
What is our C.I. estimate of the effect size for being from New
York?
Margin = 3 years
CI = (-0.50, 5.50)
Factors affecting a CI
1. Level of confidence
1. (higher confidence ==> wider interval)
2. Sample size
1. (larger n ==> narrower interval)
Confidence Intervals
Pros
• Gives a range of likely
values for effect in original
units
• Has all the information of a
significance test and more
• builds in the level of
certainty
Cons
• Units are specific to sample
•Hard to compare across
studies
• No reference point (is this a
big effect?)
Cohen’s D
•
A standardized way to estimate effect size
•
Compares the size of the effect to the size of the
standard deviation
d
X  H0
s
Exercise in constructing d
We have a sample of 10 girls who, on average, went on their 1st
dates at 15.5 yrs, with a standard deviation of 4.2 years.
Let’s say we were comparing this sample (of girls from New
York) to the general American population μ = 13 years
What is our d estimate of the effect size for being from New
York?
15.5  13
d
 .595
4.2
Exercise in constructing d
What is our d estimate of the effect size for being from New
York?
15.5  13
d
 .595
4.2
Is this big?
.2
small
.5
moderate
.8
large
>1
a very big deal
Cohen’s D
Pros
• Uses an important reference
point (s)
• Is standardized
• Can be compared across
studies
Cons
• Loses raw units
• Provides no estimate of
certainty
Review
Hypothesis Tests
• t-test
ttest 
X 
sX
Confidence Interval
• t interval
X  tcrit s X
0
Effect Size
• Cohen’s d
or ( X  H )  tcrit s X
X  uH0
s