t scores and confidence intervals
using the t distribution
What we will cover today:
t scores as estimates of z scores;
t curves as approximations of z curves
Estimated standard errors
CIs using the t distribution
Testing the no effect (null) hypothesis)
t scores indicate the distance and
direction of a score from the sample
mean.
• t scores are computed like Z scores.
• Simply substitute X for mu and s for sigma.
That is, substitute estimates for parameters.

X  mu 
Z
sigma

X X
t
s
More t scores
• As long as the raw scores are obtained from a
random sample, then t scores are least squares,
unbiased, consistent estimators of what the Z
scores would be if we had all the scores in the
population.
Any score can be translated to a t scores as long as
you can estimate mu and sigma with X-bar and s.
Calculating t scores
t=
score – estimated mean
Estimated standard deviation
What is the t score for someone 6’ tall, if the sample mean is
5’8” and the estimated standard deviation is 3 inches (s=3”)?
6’ - 5’8”
t=
3”
72 - 68
4
=
=
= 1.33
3
3
Body and tails of the curve
• The body of a curve is the area enclosed in a
symmetrical interval around the mean.
• The tails of a curve are the two regions of the curve
outside of the body.
• The critical values of the t curves are the number of
estimated standard deviations one must go from the
mean to reach the point where 95% or 99% of the
curve is enclosed in a symmetrical interval around the
mean (in the body) and 5% or 1% is in the two tails
combined.
• The critical values of the t curve change depending on
how many df there are for MSW and s.
We can define a curve by stating
its critical values
• The Z curve can be defined as one of the family of
t curves in which 95% of the curve falls within
1.960 standard deviations from the mean and 99%
falls within 2.576 standard deviations from the
mean.
• We can define t curves in terms of how many
estimated standard deviations you must go from
the mean before the body of the curve contains
95% and 99% of the curve and the combined
upper and lower tails contain 5% and 1%
respectively.
t curves
• t curves are used instead of the Z curve when you are
using samples to estimate sigma2 with MSW.
• Since we are estimating sigma instead of knowing it,
t curves are based on less information than the Z
curve.
• Therefore, t curves partake somewhat of the
rectangular (“I know nothing”) distribution and tend
to be flatter than the Z curve.
• The more degrees of freedom for MSW, the better our
estimate of sigma2.
• The better our estimate, the more t curves resemble Z
curves.
t curves and degrees of freedom
Z curve
F
r
e
q
u
e
n
c
y
Standard
deviations
5 df
1 df
score
3
2
1
0
1
2
To get 95% of the population in the body of the curve when
there are 5 df of freedom, you go out over 3 standard deviations.
To get 95% of the population in the body of the curve when there
is 1 df of freedom, you go out over 12 standard deviations.
3
Critical values of the t curves
• The following table defines t curves with
1 through 10,000 degrees of freedom
• Each curve is defined by how many estimated
standard deviations you must go from the mean to
define a symmetrical interval that contains a
proportions of .9500 and .9900 of the curve, leaving
proportions of .0500 and .0100 in the two tails of the
curve (combined).
• Values for .9500/.0500 are shown in plain print.
Values for .9900/.0900 and the degrees of freedom
for each curve are shown in bold print.
df
.05
.01
1
12.706
63.657
2
4.303
9.925
3
3.182
5.841
4
2.776
4.604
5
2.571
4.032
6
2.447
3.707
7
2.365
3.499
8
2.306
3.355
df
.05
.01
9
2.262
3.250
10
2.228
3.169
11
2.201
3.106
12
2.179
3.055
13
2.160
3.012
14
2.145
2.997
15
2.131
2.947
16
2.120
2.921
df
.05
.01
17
2.110
2.898
18
2.101
2.878
19
2.093
2.861
20
2.086
2.845
21
2.080
2.831
22
2.074
2.819
23
2.069
2.807
24
2.064
2.797
df
.05
.01
25
2.060
2.787
26
2.056
2.779
27
2.052
2.771
28
2.048
2.763
29
2.045
2.756
30
2.042
2.750
40
2.021
2.704
60
2.000
2.660
df
.05
.01
100
1.984
2.626
200
1.972
2.601
500
1.965
2.586
1000
1.962
2.581
2000
1.961
2.578
10000
1.960
2.576
Using the t table
You can answer things like:
• If we have 13 degrees of freedom then how far do we have to
go above the mean in order to have only 5% of the curve left
in the tails? (Answer: Out to a t score of 2.160)
• How many estimated standard deviations do we have to go out
in order to leave 1% of the scores in the tails with 3 degrees of
freedom?
(Answer: 5.841)
• With 10 degrees of freedom and a critical value of .05, is a t
score of -2.222 inside the body or the tail of the t curve? How
about with 11 df. (Answer: With 10 df, inside the body. With
11 df outside the body in the tail.)
• What are the critical values of a t curve with 20 df?
(Answer: 2.086 at .95/.05 and 2.845 at .99/.01.)
df
.05
.01
1
12.706
63.657
2
4.303
9.925
3
3.182
5.841
4
2.776
4.604
5
2.571
4.032
6
2.447
3.707
7
2.365
3.499
8
2.306
3.355
df
.05
.01
9
2.262
3.250
10
2.228
3.169
11
2.201
3.106
12
2.179
3.055
13
2.160
3.012
14
2.145
2.997
15
2.131
2.947
16
2.120
2.921
df
.05
.01
17
2.110
2.898
18
2.101
2.878
19
2.093
2.861
20
2.086
2.845
21
2.080
2.831
22
2.074
2.819
23
2.069
2.807
24
2.064
2.797
df
.05
.01
25
2.060
2.787
26
2.056
2.779
27
2.052
2.771
28
2.048
2.763
29
2.045
2.756
30
2.042
2.750
40
2.021
2.704
60
2.000
2.660
df
.05
.01
100
1.984
2.626
200
1.972
2.601
500
1.965
2.586
1000
1.962
2.581
2000
1.961
2.578
10000
1.960
2.576
A slightly harder problem type:
with a single sample (n=22), you
have 21 df for MSW and therefore
for the t curve. Say your t score =
+2.080, what percentile would
you be at? (Hint: the answer is
not the 95th percentile.)
df
.05
.01
1
12.706
63.657
2
4.303
9.925
3
3.182
5.841
4
2.776
4.604
5
2.571
4.032
6
2.447
3.707
7
2.365
3.499
8
2.306
3.355
df
.05
.01
9
2.262
3.250
10
2.228
3.169
11
2.201
3.106
12
2.179
3.055
13
2.160
3.012
14
2.145
2.997
15
2.131
2.947
16
2.120
2.921
df
.05
.01
17
2.110
2.898
18
2.101
2.878
19
2.093
2.861
20
2.086
2.845
21
2.080
2.831
22
2.074
2.819
23
2.069
2.807
24
2.064
2.797
df
.05
.01
25
2.060
2.787
26
2.056
2.779
27
2.052
2.771
28
2.048
2.763
29
2.045
2.756
30
2.042
2.750
40
2.021
2.704
60
2.000
2.660
df
.05
.01
100
1.984
2.626
200
1.972
2.601
500
1.965
2.586
1000
1.962
2.581
2000
1.961
2.578
10000
1.960
2.576
A slightly harder problem: with a
single sample (n=22), you have
21 df for MSW and therefore for
the t curve. Say your t score =
+2.080, what percentile would
you be at? 98th percentile
(50 + 47.5 = 97.5 = 98th
percentile)
Estimated distance of sample means from mu:
estimated standard errors of the mean
• We can compute the standard error of the mean
when we know sigma.
– We just have to divide sigma by the square root of n,
the size of the sample
• Similarly, we can estimate the standard error of
the mean, estimate the average unsquared distance
of sample means from mu.
– We just have to divide s by the square root of n, the size
of the sample in which we are interested
Note that the estimated standard error is
determined by only two factors: the estimated
average unsquared distance of scores from mu
(s) and the size of the sample (n).
sX  s / n
Look at the effects of increased
variation and increased sample
size in the next slide.
• In the first set of computations on the next
slide, n stays constant while s and the
estimated standard error increase.
• In the second set of computations on the
next slide, s stays constant while n increases
and the estimated standard error decreases.
•
•
•
•
•
•
•
•
•
s
n
sX  s / n
A 2.83
B 12.00
C 20.00
8
8
8
1.00 = 2.83/2.83
4.24 = 12.00/2.83
7.07 = 20.00/2.83
D
E
F
G
1
2
8
40
2.83
2.00
1.00
0.45
2.83
2.83
2.83
2.83
=
=
=
=
2.83/1.00
2.83/1.41
2.83/2.83
2.83/6.32
Confidence intervals using the
t distribution
There are two reasons to create confidence
intervals with the t distribution
• 1. To test a theory about what mu is. (We
call the theoretical population means muT).
• THAT’S THE IMPORTANT REASON
AND WE WILL LEARN ABOUT IT
FIRST
• 2. The other reason is to define an interval
in which we are confident mu would fall if
we knew it.
Confidence intervals around muT
and testing the null hypothesis
Confidence intervals and hypothetical means
• We frequently have a theory about what the
mean of a distribution should be.
• To be scientific, that theory about mu must
be able to be proved wrong (falsified).
• One way to test a theory about a mean is to
state a range where sample means should
fall if the theory is correct.
• We usually state that range as a 95%
confidence interval.
• To test our theory, we take a random sample from
the appropriate population and see if the sample
mean falls where the theory says it should, inside
the confidence interval.
• If the sample mean falls outside the 95%
confidence interval established by the theory, the
evidence suggests that our theoretical population
mean and the theory that led to its prediction is
wrong.
• When that happens our theory has been falsified.
We must discard it and look for an alternative
explanation of our data.
For example:
• For example, let’s say that we had a new
antidepressant drug we wanted to peddle.
Before we can do that we must show that
the drug is safe.
• Drugs like ours can cause problems with
body temperature. People can get chills or
fever.
• We want to show that body temperature is
not effected by our new drug.
Testing a theory
• “Everyone knows” that normal body temperature
for healthy adults is 98.6oF.
• Therefore, it would be nice if we could show that
after taking our drug, healthy adults still had an
average body temperature of 98.6oF.
• So we might test a sample of 16 healthy adults,
first giving them a standard dose of our drug and,
when enough time had passed, taking their
temperature to see whether it was 98.6oF on the
average.
Testing a theory - 2
• Of course, even if we are right and our drug
has no effect on body temperature, we
wouldn’t expect a sample mean to be
precisely 98.600000…
• We would expect some sampling fluctuation
around a population mean of 98.6oF.
Testing a theory - 3
• So, if our drug does not cause change in
body temperature, the sample mean should
be close to 98.6. It should, in fact, be within
the 95% confidence interval around a
theoretical mean of 98.6.
• SO WE MUST CONSTRUCT A 95%
CONFIDENCE INTERVAL AROUND 98.6o
AND SEE WHETHER OUR SAMPLE MEAN
FALLS INSIDE OR OUTSIDE THE CI.
To create a confidence interval around muT,
we must estimate sigma from a sample.
• We randomly select a group of 16 healthy
individuals from the population.
• We administer a standard clinical dose of our new
drug for 3 days.
• We carefully measure body temperature.
• RESULTS: We find that the average body
temperature in our sample is 99.5oF with an
estimated standard deviation of 1.40o (s=1.40).
• IS 99.5oF. IN THE 95% CI AROUND MUT???
Knowing s and n we can easily compute
the estimated standard error of the mean.
• Let’s say that s=1.40o and n = 16:
• sX  s / n
= 1.40/4.00 = 0.35
• Using this estimated standard error we can
construct a 95% confidence interval for the
body temperature of a sample of 16 healthy
adults.
We learned how to create confidence intervals with
the Z distribution in Chapter 4.
95% of sample means will fall in a symmetrical
interval around mu that goes from 1.960 standard
errors below mu to 1.960 standard errors above mu
• A way to write that fact in statistical language is:
CI.95: mu + ZCRIT* sigmaX-bar
or
CI.95: mu - ZCRIT* sigmaX-bar < X-bar < mu + ZCRIT* sigmaX-bar
For a 95% CI, ZCRIT = 1.960
But when we must estimate sigma with s, we
must use the t distribution to define critical
intervals around mu or muT.
Here is how we would write the formulae
substituting t for Z and s for sigma
CI95: muT + tCRIT* sX-bar
or
CI.95: muT - tCRIT* sX-bar < X-bar <
muT + tCRIT* sX-bar
Notice that the critical value of t
that includes 95% of the sample
means changes with the number
of degrees of freedom for s, our
estimate of sigma, and must be
taken from the t table.
If n= 16 in a single sample,
dfW=n-k=15.
df
.05
.01
1
12.706
63.657
2
4.303
9.925
3
3.182
5.841
4
2.776
4.604
5
2.571
4.032
6
2.447
3.707
7
2.365
3.499
8
2.306
3.355
df
.05
.01
9
2.262
3.250
10
2.228
3.169
11
2.201
3.106
12
2.179
3.055
13
2.160
3.012
14
2.145
2.997
15
df
.05
.01
17
2.110
2.898
18
2.101
2.878
19
2.093
2.861
20
2.086
2.845
21
2.080
2.831
22
2.074
2.819
23
2.069
2.807
24
2.064
2.797
df
.05
.01
25
2.060
2.787
26
2.056
2.779
27
2.052
2.771
28
2.048
2.763
29
2.045
2.756
30
2.042
2.750
40
2.021
2.704
60
2.000
2.660
df
.05
.01
100
1.984
2.626
200
1.972
2.601
500
1.965
2.586
1000
1.962
2.581
2000
1.961
2.578
10000
1.960
2.576
16
2.131 2.120
2.947 2.921
So, muT=98.6, s=1.40, n=16, df = 15,
tCRIT=2.131 Here is the confidence interval:
CI.95: muT + tCRIT* sX-bar =
n
= 98.6 + (2.131)*(1.40/
)=
= 98.6 + (2.131)*(1.40/4)
= 98.6 + (2.131)(0.35) = 98.60+ 0.75
CI.95: 97.85 < X-bar < 99.35
CI.95: 97.85 < X-bar < 99.35
:
The confidence interval consistent with the
theory that our drugndoes not casues a
change in body temperature goes from
97.85o to 99.35o. Our sample mean was
99.5o F. So, our sample mean falls outside
the CI.95 and falsifies the theory that our
drug has no effect on body temperature. Our
drug may cause a slight fever.
Testing the no-effect (null) hypothesis
• Specify what you believe the value of a statistic will be if
your idea of the situation is correct. In this case, you
specified muT=98.6o F.
• Define a 95% Confidence Interval around that value of
your test statistic.
• See if the test statistic falls inside the CI.95
• If it falls within the CI.95, you retain the no-effect
hypothesis
• If it falls outside the CI.95, you reject the no effect
hypothesis. Further, if you then have to guess what the
true value of the test statistic is in the population as a
whole, you chose the value that was found in the
random sample.
• THE FOLLOWING MATERIAL IS LESS
IMPORTANT AND WILL ONLY BE
GONE OVER IF THERE IS EXTRA
TIME. MAKE SURE YOU
UNDERSTAND TESTING THE NO
EFFECT HYPOTHESIS BEFORE GOING
ON.
Interval estimates for mu based
on a random sample
Since Chapter 1 you have been making
least squared, unbiased estimates
• You have learned to predict that everyone
would score at a specific point. This is
called “making point estimates.”
• The point estimates have been wrong in a
least squared, consistent, unbiased way.
• If you want to be right, not wrong, you can’t
make point estimates, you must make
interval estimates.
Interval estimates of mu
• As you know, X-bar is our least squared,
unbiased, consistent estimate of mu.
• To create an interval estimate of mu, we
create a symmetrical interval around X-bar.
• We usually create 95% and/or 99% CIs to
define that interval.
The generic formula for
confidence intervals for mu
CI: X-bar + (tCRIT* sX-bar )
or
CI: X-bar – tCRIT* sX-bar < mu< X-bar+ tCRIT* sX-bar
If we use the critical values of t at .95 we will have
an interval that includes mu if our sample is one
of the 95% that falls within a 95% CI around mu.
If we use the critical values of t at .99 we will have
an interval that includes mu if our sample is one
of the 95% that falls within a 99% CI around mu.
Here is an example
• Let’s say (because professors know everything)
that I knew that the average height of all male
Rutgers Juniors is 70.0 inches (5’10”)
• I assign you, the student to take a sample of 16
male Rutgers Juniors and measure their height.
• Then you must define an interval in which mu
should be found.
• You find that X-bar = 69.3” and s = 2.00 inches.
df
.05
.01
1
12.706
63.657
2
4.303
9.925
3
3.182
5.841
4
2.776
4.604
5
2.571
4.032
6
2.447
3.707
7
2.365
3.499
8
2.306
3.355
df
.05
.01
9
2.262
3.250
10
2.228
3.169
11
2.201
3.106
12
2.179
3.055
13
2.160
3.012
14
2.145
2.997
15
df
.05
.01
17
2.110
2.898
18
2.101
2.878
19
2.093
2.861
20
2.086
2.845
21
2.080
2.831
22
2.074
2.819
23
2.069
2.807
24
2.064
2.797
df
.05
.01
25
2.060
2.787
26
2.056
2.779
27
2.052
2.771
28
2.048
2.763
29
2.045
2.756
30
2.042
2.750
40
2.021
2.704
60
2.000
2.660
df
.05
.01
100
1.984
2.626
200
1.972
2.601
500
1.965
2.586
1000
1.962
2.581
2000
1.961
2.578
10000
1.960
2.576
16
2.131 2.120
2.947 2.921
So, X-bar=69.3, s=2.00, n=16,df = 15,
tCRIT=2.131. Here is the 95% confidence interval.
CI.95: X-bar + tCRIT* sX-bar =
= 69.3 + (2.131)*(2.00/ 61 ) =
= 69.3 + (2.131)*(2.00/4)
= 69.3 + (2.131)(0.50) = 69.30+ 1.16
CI.95: 68.14 < mu< 70.46
In this example, we knew mu ahead of time (70.0)
and the computations were just an exercise. But
exercise or not, our interval does include mu.
Here is a 99% CI with the same data
X-bar=69.3, s=2.00, n=16, df = 15,
tCRIT=2.947
CI.99: X-bar + tCRIT* sX-bar =
= 69.3 + (2.947)*(2.00/4)
= 69.3 + (2.947)(0.50) = 69.30+ 1.47
CI.99: 67.83 < mu< 70.77
Of course, since we have to go further out into the
tails of the curve to include 99% of the sample
means, the interval for the CI99 is wider (less
precise) that the interval for the CI.95 .