Welcome to
the Real
World…
T
The World
of T
Confidence Intervals and Significance Testing in the
World of T
T is used for testing means (averages)…
Looking for Statistically Significant
differences
Aquaculture
Calculating and comparing Tank
Flow Rate to Ideal Flow Rate
HMMM… T – Testing in
Research????
Wow! A tailor made T for each
Check
Me
Out!!
sample!!
is a density
curve
me break
ForLet
T- Testing,
we’re
Let’s
See
Symmetric about Zero,
single
peaked,
still
testing
for“bell” shaped
this down
What
Makes
Population
Means,
but
reallyonsimply
T’s variation depends
sample
size
T,
T…SAMPLE
we only
need
for
you…
Remember, samples become less variable as they get
data!!
larger
T
Degrees of Freedom
T makes an adjustment for each sample size by
changing the degrees of freedom
Basically gives us a new T to work with for each sample size!!
One Sample T Statistic
__
This is that
personal touch
for each
different
sample…
Sample
Standard
Dev.
x 
t
s
n
Standard Error
With n-1 Degrees of Freedom
This is for the TTable… Let’s Practice
P-Value
Degrees of Freedom (df)
T-Statistic
Area
to
the
right
of
t
•Left
hand
column
of
chart
•Located in MIDDLE
•Different
T-Distribution
for
Area
to
the
left
of
–t
of chart
each
sample size
2(P)
for the
two-sided
•Larger
the closer
•Leads
to sample,
the p-value
to
Normal
the
T
distribution
or vice versa
What happens if you get a T
 Find
thenot
t-statistic
the following:
that’s
on yourfor
table?
 1) 5 dof;
p = .05
(right)t = 2.015
Then
What?
 2) n = 22; p = .99 (left)t = 2.518
 3) 80% CI; n = 18
t = 1.333
 Find the p-value for the following:
p = .01
.025 < p <.05
 2) n = 12; t = 1.856
.02 < p <.025
 1) 5 dof; t = 3.365
 3) n = 67; t = 2.056
Notice the t-statistic is
You will simply say you’re p-value
limited to certain values on
is BETWEEN 2 values!!
your table!!!
With these tests you are given an alpha
level against which you test your p-value
*(Standard level = .05):
p ≤ a – Reject the null; accept the Ha
p > a – Fail to reject the null
Ha: µ ≠ µ0
Ha: µ > µ0
Ha: µ < µ0
Sample Size
Distribution
Proof
If
your
HISTOGRAM
is
n <15
Needs
to
be
Graph
You need
yourtodistribution
show thisHistogram or
skewed,
either
normalscrap Stem Plot
forcheck
samples
of conditions
less than 40
the
t-test
or
talk
after
you
write
level
your
ofHistogram or
15 < n <to
40determine
No STRONG
about
the
hypotheses…
normality!!!
outliers or
Stem Plot
questionability
of the
skewness
results!!
n < 40
No restrictions
Not Needed
For T – Tests (testing for population
mean with sample mean and standard
deviation…
Compare
State
Find the
theyour
p-value
Ho and
p-value
from
Ha to
Find
the T-Statistic
the
int-statistic
symbols
w/ n-1
the
specified
a,and
and
make
your
degrees
decision
context
of freedom
in context
We use the Same Basic
steps as in all Hypothesis
Testing
Since phas
is less
than the
.0005,
which
Mrs. Luniewski
claimed
ideal
flow rate for fish
less than
.05, we
have
growth is 22isL/min
of water
flow.
You’ve decided to check
significantsignificant
evidence difference
to see ifstatistically
there’s a statistically
thatflow
our flow
is flow rate of your tank.
between the ideal
rate rate
and the
than
You takeSIGNIFICANTLY
a SRS of 50 ratesless
from
thethe
past week and find
rate.
Thisflow
would
help
your tank ideal
has an
average
rate
ofus14 with a standard
withlevel, is your flow
deviationidentify
of 1.36.potential
At a 5%problems
significance
thethan
tank…
rate significantly less
the ideal flow rate?
14  22
Ho: µ = 22 L/min
t
 41.6 (n – 1) df = 49
Ha: µ < 22L/m
1.36
(round down to
40 for table)
50
t = -2.4072
P is less than .0005
You use this difference
column to get your
MatchedSample
PairsMean
Testand
Sample Standard
Used when taking same measurements on
Deviation…
same media over different time period
“Difference”
between
data
Flow
Rate of
Flow
Rate ofvalues
Tank is THE
Difference
Tank PrePost-treatment
(Post – Pre)
H
=
µ
=
0
[µ
=
(µ
µ
)]
o
diff
diff
1
2
treatment
H12
≠
a = µdiff < or > or9.5
11
8
9
5
2.5
3
4
Because the measurements were
the same a
tank,
at thetreatment to the
Now you taken
haveonapplied
water
sameto
time,
this would
be
tanks, hoping
make
a difference
in the
considered a matched pairs test.
average flow
rate of your tank. To check to see
You would need to adjust your
if the difference
statistically
hypothesesisaccordingly
and usesignificant, you
collect 3themeasurements
really
enough!!)
difference between (not
the data
as
from the tanks
atdata
thesource.
exact same times they
your
were collected pre-treatment. Test to see if
there is a significant difference post-treatment.
Ho: µdiff = 0
Ha: µdiff < 0 (Increase in flow rate)
Flow Rate of
Tank Pretreatment
Flow Rate of Tank
Post-treatment
Difference
(Post – Pre)
REJECT Ho and conclude the
12
9.5
2.5
treatment
made
a significant
11
8
3
increase
in
flow
rate…
9
5
4
Sample Average Difference = 3.17
Sample Std Dev = .7638
3.17  0
 7.19
.7638
3
DF = 2 .005 < p < .01