Transcript General Psychology 1

Introduction to the t-statistic
Introduction to Statistics
Chapter 9
Oct 13-15, 2009
Classes #16-17
The Problem with Z-scores
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Z-scores have a shortcoming as an inferential
statistic:
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The computation of the standard error requires
knowing the population standard deviation ().
In reality, we rarely know the value of .
When we don’t know , can’t compute standard
error.
Therefore, we use the t statistic, rather than the
Z-score, for hypothesis testing.
How to test?
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Still need to know about standard deviation of the
population
To figure this out…
 Use formula learned back in chapter 4:
S
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2
(
X

X
)

n 1
Use this to calculate estimated standard error (sM)
We calculate t statistic similarly to how we calculated the zstatistic, but now we will use the estimated standard error of
the mean (rather than the standard error of the mean)
The t Statistic
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Use the sample variance (s2) to estimate the population
variance
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s2 = SS/df = SS/(n-1)
Use variance s2 in the formula to get the estimated
standard error:
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Provides an estimate of the standard distance between M
and  when  is unknown
estimated standard error =
sM = s
n
=
s2
n
The t Statistic
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Finally, we replace the standard error in the
z-score formula with the estimated standard
error to get the t statistic formula:
t = M-
sM
Illustration
In chapter 8 for Z-scores we used:
plugging it into z-score
formula:
M = 
n
Z = M-
M
Now we are using an estimate of
standard error by using the sample
SD
sM = s
n
Population
SD
plugging it into t statistic
formula:
t = M-
sM
Another change…
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Up until this chapter we have been using
formulas that used the standard deviation as
part of the standard error formula
Now, we shift our focus to the formula based
on variance
On page, 234 the book gives reasoning for
this.
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Main reason: sample variance (s2) provides an
accurate and unbiased estimate of the population
variance (²)
One more change…
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Although the definitional formula for sum of squares
is the most direct for computing SS it has its
problems
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When the mean is not a whole number the deviations will
contain decimals and thus calculations become more
difficult leading to rounding error
Therefore, from now on we will use the SS
computational formula which can be found on page 93 in
the textbook
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“From now on” means all future tests and final exam
Sample Variance
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Therefore, we will use sample variance rather
than sample standard deviation to compute sM
Sample standard deviation is a descriptive statistic
rather than a inferential statistic
 Sample variance will provide the most accurate
way to estimate the standard error
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We are now using variance-based
formula in these equations
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Why?
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Inferential purpose, rather than descriptive
Drawing inferences about the population
estimated standard error =
sM = s
n
=
s2
n
t statistic
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Definition:
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Used to test hypotheses about an unknown
population mean  when the value of  is
unknown.
The formula for the t statistic has the same
structure as the z-score formula, except the t
statistic uses the estimated standard error in the
denominator.
t = M-
sM
Z-score vs. T-Score
Z-distribution stays the same, regardless of
sample size
 T-distribution changes, depending on how
many pieces of information you have: degrees
of freedom
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here, df = n-1
Everything else stays the same
Have an alpha level
 Have one-tailed and two-tailed tests
 Determine boundaries of critical region
 Determine whether t-statistic falls in critical
region
 If it does, reject null and know that p<alpha
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Degrees of Freedom
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How well does s approximate ?
Depends on the size of the sample.
 The larger the sample, the better the
approximation.
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Degrees of Freedom (df) = n-1
Measures the number of scores that are free to vary
when computing SS for sample data.
 The value of df also describes how well a
t
statistic estimates a normal curve.
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Degrees of Freedom
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Degrees of Freedom = df = n-1
As df (sample size) gets larger, 3 things
result:
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1) s2 (sample variance) better represents 2
(population variance).
2) t better approximates z.
3) in general, the sample better represents the
population.
The t-distribution
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T-distribution
 The set of all possible t statistics obtained by selecting
all possible samples of size n from a given population
How well the t distribution approximates a normal
distribution is determined by the df.
In general, the greater n (and df), the more normal the t
distribution becomes.
t distribution more variable and flatter than normal z-score
distribution – why is this the case? Both mean and standard
error can vary in t-distribution – only the mean varies in the zdistribution
Distributions of the t statistic
The Versatility of the t test
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You do not need to know  when testing
with t
The t test permits hypothesis testing in
situations in which  is unknown
All you really need to compute t is a
sensible null hypothesis and a sample
drawn from the unknown population
Hypothesis Testing with t (two tails)
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Same four steps, with a few differences:
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Now estimating the standard error, so
compute t rather than z
Consult t-distribution table rather than
Unit Normal Table to find critical value
for t (this will involve the calculation of
the df)
Hypothesis Testing w/ t-statistic
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Instead of the Unit Normal Table, we now have
the t-table p. 531-532
Similar in form to the Unit Normal Table
 Pay attention to the df column!!
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Let’s think about this table for a minute
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Looking at the two-tail, p=0.05 column:
 What
is value at 10 df?
 What is value at 20 df?
 What is value at 30 df?
 What is value at 120 df?
A portion of the t-distribution table
Hypothesis Testing with t (two tails)
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Step 1: State the hypotheses.
Step 2: Set  and locate the critical region.
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You will need to calculate the df to do this, and use the t
distribution table.
Step 3: Graph (shade) the critical region.
Step 4: Collect sample data and compute t.
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This will involve 3 calculations, given SS, n, , and M:
 a) the sample variance (s2)
 b) the estimated standard error (sM)
 c) the t statistic
Hypothesis Testing with t (two tails)
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Step 5: Go back to graph and see if tcalc falls in the
critical region
Step 6: Make a decision. Compare t computed in
Step 3 tCALC with tCRIT found in the t table:
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If tCALC > tCRIT (ignoring signs)  Reject HO
If tCALC < tCRIT (ignoring signs)  Fail to Reject HO
One-Tailed Hypothesis Testing with t
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Same as with z, only steps 1 and 2 change.
Step 1:
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Now use directional hypotheses.
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H0: = ? and H1: ? (predicts decrease) OR
H0: = ? and H1: ? (predicts increase).
Step 2:
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Now the critical region located in only one tail of the
distribution (sign of tCRIT represents the direction of
the predicted effect).
You will have to use a different column on the t
distribution table.
Example1
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Do eye-spot patterns affect behavior?
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If eye-spots do affect behavior, birds should spend more or
less time in chamber w/ eye-spots painted on the walls.
Sample of n=16 birds.
Allowed to wander between the 2 chambers for 60 minutes.
If eye-spots do not affect behavior, we’d expect they’d
spend about 30 minutes in each chamber.
We’re told the sample mean =39, SS = 540.
Example1
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Step 1: State the hypotheses
Ho: µplain side = 30 min.
 H1: µplain side ≠ 30 min.
 Two-tailed Alpha = 0.05
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Step 2: Locate the critical region
Based on df.
 What are df here?
 What is the critical value?
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Step 3: Shade in critical region
Example 1
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Step 4: Calculate the t-statistic.
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First calculate the sample variance
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s2 = SS/n-1 , 540/15 = 36.
Next use the sample variance (s2) to calculate the estimated
standard error
sM
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s2
36
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 2.25  1.50
n
16
Finally, compute the t-statistic:
M   39  30
9
t


6
sM
1.50
1.50
Example 1
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Step 5: Make a decision.
T-calculated = 6.00
 t-critical = + 2.131
 We observe that our t-value is in the region of
rejection.
 We conclude that eye-spots have an effect on
predatory behavior.
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Example 2
A teacher was trying to see whether a new teaching
method would increase the Test of English as Foreign
Language (TOFEL) scores of students. She received a
report which included a partial list of previous scores
on the exam. Unfortunately, most of the records were
burned in a fire that occurred in the school’s Records’
Department. From the available data, students taught
by old methods had  = 580. She tested her method in
a class of 20 students and got a mean of 595 and
variance of 225. Is this increase statistically
significant at the level of 0.05 in a 2-tailed test?
Example 2
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Step 1:
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Step 2:
Example 2: Step 3
Example 2
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Step 4:
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Step 5:
Example 3
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A researcher believes that children in poverty-stricken regions
are undernourished and underweight. Past studies show the
mean weight of 6-year olds is normally distributed with a  
20.9 kg. However, the exact mean and standard deviation of
the population is not available. The researcher collects a
sample of 9 children from these poverty-stricken regions, with
a sample mean of 17.3 kg & s = 2.51 kg.
Using a one-tailed test and a 0.01 level of significance,
determine if this sample is significantly different from what
would be expected for the population of 6-year olds.
Example 3
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Step 1
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Step 2
Example 3: Step 3
Example 3
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Step 4
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Step 5
Example 4
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A researcher has developed a new formula (Sunblock
Extra) that she claims will help protect against the harmful
rays of the sun. In a recent promotion for the new formula
she is quoted as saying she is sure her new formula is
better than the old one (Sunscreen).
Her prediction: The “improved” Sunblock Extra will score
higher than the previous Sunscreen score of 12?
She decides to use the .05 significance level to test for
differences. To the right are the Sunblock Extra scores for
participants in her study.
In notation form:
H0:
HA:
Determine if there is a significant difference between the
new product and the old one (make your decision and
interpret).
X
X2
12 144
13 169
6
36
11 121
12 144
8
64
11 121
7
49
10 100
16 256
10 100
7
49
14 196
15 225
16 256
168 2030
Example 4
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Step 2:
Example 4: Step 3
Example 4
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Step 4
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Step 5
Example 5
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Scientists believe that the “Monstro Motors”
new model will get the highest gas mileage
of any car on their lot. Although, not much
data is available on the older cars, from a
review of previous models they estimate that
the best of the rest of their cars achieved 67
m.p.g. They using an alpha level α = .01.
H0:
H A:
Determine if there is a significant difference
between the MPG of the new car and the
best old model on their lot (make your
decision and interpret).
X
X2
65
4225
76
5776
69
4761
71
5041
74
5476
78
6084
77
5929
68
4624
72
5184
75
5625
74
5476
64
4096
69
4761
63
3969
82
6724
1077 77751
Example 5
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Step 2:
Example 5: Step 3
Example 5
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Step 4
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Step 5
Steps
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Step 2?
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Step 3?
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Step 4?
Effect Size
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estimated d
estimated d 
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r squared
r2 
t2
t 2  df
M   
mean difference

sample standard deviation
s
Effect Size
estimated d 
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Example 4
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d
M     11.2  12 
s
3.25
 .8
 -0.24
3.25
Small effect (small to medium)
Example 5
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M   
mean difference

sample standard deviation
s
Large effect

M    71.8  67
d
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s
5.49
4.8
 0.87
5.49
Credits
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http://myweb.liu.edu/~nfrye/psy53/ch9.ppt#9
http://homepages.wmich.edu/~malavosi/Chapt9PPT_S_05.ppt#2
http://faculty.plattsburgh.edu/alan.marks/Stat%20206/Introduction%20to%2
0the%20t%20Statistic.ppt#4
http://home.autotutor.org/hiteh/Stats%20S04/Statistics04onesamplettest1.ppt#7