Student’s t-Distribution
The t-Distribution, t-Tests, & Measures of Effect Size
Sampling Distributions Redux
• Chapter 7 opens with a return to the concept of sampling distributions from chapter 4– Sampling distributions of the mean
Sampling Distribution of the Mean
• Because the SDotM is so important in statistics, you should understand it
• The SDotM is governed by the Central Limit Theorem
Given a population with a mean μ and a variance σ2, the sampling distribution of the mean (the distribution of sample means) will have a mean equal to μ, a variance equal to σ2/n, and a standard deviation equal to . The distribution will approach the normal distribution as n, the sample size, increases. (p. 178)
n/2σ
Sampling Distribution of the Mean
Translation:1. For any population with a given mean and
variance the sampling distribution of the mean will have:
• μx = μ• σx
2 = σ2/n• σx = σ/√n
2. As n increases, the sampling distribution of the mean (μx) approaches a normal curve
Sampling Distribution of the Mean
• Analysis:– Although μx and μwill tend to be similar to one
another…– The relationships between…
• σx2 and σ2
• σx and σ– …will differ as a function of the sample size
• We saw this in our sampling distribution of the mean example from chapter 4…
So, you wanna test a hypothesis, do ya?
• Our understanding of sampling and sampling distributions now allows us to test hypotheses
• How we test a hypothesis depends on the information we have available
Choosing a Test• μ?
– σ?– s?
• Number of data sets:– 1– 2
• Number of Groups– 1– 2
1. Which variables are available?
2. How many data sets are you presented with?
3. Do your data sets come from 1 or 2 groups?
Testing Hypotheses about Means: The Rare Case of Knowing σ
• So far, to test the probability of finding a particular score, we’ve used the Standard Normal Distribution– IQ = 83– μ = 100– σ = 15
σ)( xxz −
=
15)10083( −
=z
15)17(−
=z
33.1−=z
-1.96 < z < 1.96 Fail to reject H0
…The Rare Case of Knowing σ
• Remember: we rarely know the population mean and standard deviation
• This test can ONLY be used in situations where the population mean and standard deviation are known!
…The Rare Case of Knowing σ:
• Stick with IQ:– μ = 100– σ = 15
• However, we want to test a group of children against the population values for IQ– n = 5 (a group of 5 children)
…The Rare Case of Knowing σ:
• Research Hypothesis:– The children’s IQ scores are different from
the population IQ scores • H1: μc ≠ μp
• Null Hypothesis – The children’s IQ scores do not differ from
the population IQ scores • H0: μc = μp
• Test the students (x-bar = 122)
…The Rare Case of Knowing σ:
• Select:• Rejection region
• α = .05• “Tail” or directionality
• We don’t know exactly how the students will score: we just expect them to show scores differing from the population values
…The Rare Case of Knowing σ: The z-Test
• Generate sampling distribution of the mean assuming H0 is true
• z-Test• Given our sampling distribution:
• Conduct the statistical test
Conducting the z-Test
515
)100122( −=z
24.215
)22(=z
70.6)22(
=z
n
xz σμ)( −
=Note: this equation is a modification of the original z-score formula
This formula adjusts z for sample size according to the rules of the central limit theorem
28.3=z
z = 3.28 > 1.96 : Reject H0
How the z-Test Works
10015
)100122( −=z
1015
)22(=z
5.1)22(
=z 67.14=z
215
)100122( −=z
41.115
)22(=z
64.10)22(
=z 07.2=z
115
)100122( −=z
115
)22(=z
15)22(
=z 47.1=z
n = 100
n = 2
n = 1
How the z-Test Works
• Large samples reduce the amount of random variance (sampling error)
– More confidence that the sample mean = population mean
• Larger samples improve our ability to detect differences between samples and populations
• For n = 1=
n
xz σμ)( −
=σμ)( −
=xz
Statistical Tests We Have Learned
1. z-Test
Testing Hypotheses: When σ Is Unknown
• Generally, the population standard deviation, σ, is unknown to us
• Occasionally, we will know the population mean, μ, when we don’t know σ
• In these situations, the standard normal distribution no longer meets our needs
Testing Hypotheses: When σ Is Unknown
• Knowing μ…– We can produce an estimate of σ from s– Changes the nature of the test we are
conducting, as s is not distributed in the same fashion as σ
• Sampling distribution of the sample standard deviation is NOT normally distributed
– Strong positive skew
Testing Hypotheses: When σ Is Unknown
Sampling distribution of s
Sampling distribution of σ
So How Does s Estimate σ?
• Given the differences in distribution shape, it is easy to conclude that s ≠ σ– s is an unbiased estimator of σ over repeated
samplings – However, a SINGLE value of s is likely to
underestimate σ• Because of this fact, small samples will systematically
underestimate σ as a function of s
– This leads to any given statistic calculated from this distribution to be < a comparable value of z
– We cannot use z any longer t
t and the t-Distribution
• Developed by Student while he was working for the Guinness Brewing Co.
1. The shape of the t-distribution is a direct function of the size of the sample we are examining
2. For small samples, the t-distribution is somewhat flatter than the standard normal distribution, with a lower peak and fatter tails
t and the t-Distribution
3. As sample size increases:• The t-distribution approaches a normal
distribution• Theoretically, we mean that the closer that our
sample comes to infinity, the more it looks like a normal distribution
• Practically, when n ~ 100 – 120
t and the t-Distribution
t and the t-Distribution
4. Identifying values of t associated with a given rejection region depends on:– α– the number of tails associated with the test– the degrees of freedom available in the analysis
– For this one-sample test, (df = n-1) because we used one degree of freedom calculating s using the sample mean and not the population mean.
One-Sample t-Test
xsxt )( μ−
=
ns
xtx
)( μ−=
ns
xtx
2
)( μ−=or or
z-Test vs. One-Sample t-Test
n
xz σμ)( −
=
ns
xtx
)( μ−=
Note the similarities between these tests: ONLY the source of “variance” and the distribution you test against have changed!
Using the One-Sample t-Test
• You are one the admissions board for a graduate school of Psychology.
• You are attempting to determine if the GRE scores for the students applying to your program is competitive with the national average. – μVerbal = 569– x-bar = 643– s = 82– n = 24
Using the One-Sample t-Test
• Research Hypothesis:– The GRE scores from your applicants differ
from the population norms• H1: μa ≠ μp
• Null Hypothesis – The GRE scores from your applicants do not
differ from the population norms • H0: μa = μp
• Evaluate the students’ GRE-V scores
Using the One-Sample t-Test
• Select:• Rejection region
• α = .05• “Tail” or directionality
• We don’t know exactly how the students will score: we just expect them to show scores differing from the population values
• Might predict higher scores…
Using the One-Sample t-Test
• Generate sampling distribution of the mean assuming H0 is true
• One-Sample t-test• Given our sampling distribution:
• Conduct the statistical test
Using the One-Sample t-Test
ns
xtx
)( μ−=
2482
)569643( −=t
73.16)74(
=t
90.482
)74(=t 42.4=t
μVerbal = 569x-bar = 643s = 82n = 24
This numerical value is called tobt
tobt(23) = 4.42
Evaluating Statistical Significance of the t-Test
• First note:– α = .05– Tail or directionality: two-tailed– t-Value = 4.42– Degrees of freedom (df)
• For the One-Sample t-Test, df = n-1 (24-1 = 23)• Estimating s from x-bar (not σ from μ)
p. 747 in Howell Text
1) Find row for TAIL
2) In the ROW for the correct tail, find α
3) Find df
4) Track ROW of dfacross to COLUMN of α
The numerical value you obtain is called tcrit
tcrit(23) = 2.069
Evaluating Statistical Significance of the t-Test
• Compare tcrit to our tobt value• If tobt falls into the rejection region
identified by tcrit, then we reject H0
• If tobt does not fall into the rejection region identified by tcrit, then we fail to reject H0
Evaluating Statistical Significance of the t-Test
tcrit = 2.069tcrit = - 2.069
0
tobt = 4.42
Because tobt falls within the rejection region identified by tcrit we reject H0
Statistical Tests We Have Learned
1. z-Test• 1 group• μ & σ known
2. One-Sample t-Test• 1 group• μ known• Estimate σ with s using x-bar
Testing Hypotheses: Two Matched (Repeated) Samples
• Sometimes, we’re interested in how a single set of scores change over time– Psychotherapy tx influences depression– Patients respond to medication– Consumer attitudes before and after an
advertisement• When we look at two sets of scores collected
from a single sample at different time points, we need to use a matched samples test
Matched Samples
• Matched samples– Use the same participants at two or more
different time points to collect similar data• MUST BE THE SAME SAMPLE!
BDI - II BDI - II
Time 1 Wait 30 Days Time 2
Matched Samples Test
• With a matched samples test, you are testing the change in scores between the two administrations of the test– H0: μ1 = μ2
– H0: μ1 - μ2 = 0• This is truly the null hypothesis for the matched
samples test• There is a difference…
Matched Samples Test
• Essentially, the group means at each time point mean little to us– Change in scores is the key– Conduct this test by obtaining the average
difference score between the two time points
Matched Samples Test
ns
DtD
0−=
D-bar represents average difference scores between time points
sD is the standard deviation of the difference scores
-0 may seem redundant, but isn’t!
Calculating the Matched Samples t-Test
• You are a researcher examining the impact of a new therapy intervention on the incidence of self-injurious behavior (SIB)
• You collect a measure of the frequency of self-injurious acts when clients enter your treatment (time 1)
• You collect a measure of the frequency of self-injurious acts two weeks later (time 2)
Calculating the Matched Samples t-Test
• Research Hypothesis:– The new treatment will change SIB scores
• H1: μ1 ≠ μ2
• Null Hypothesis – The SIB scores at time 2 will be the same as
the scores at time 1 (no change)• H0: μ1 = μ2
• H0: μ1 - μ2 = 0
• Evaluate SIB at time 1 & time 2
Using the One-Sample t-Test
• Select:• Rejection region
• α = .05• “Tail” or directionality
• We don’t know exactly how the treatment will work, so we’d better use a two-tailed test
Using the One-Sample t-Test
• Generate sampling distribution of the mean assuming H0 is true
• Matched Samples t-test• Given our sampling distribution:
• Conduct the statistical test
Calculating the Matched Samples t-Test
54274413445D 2516449161619161625D2
261791191074108Time 2
71019161513111081413Time 1
∑D = 43D-bar = 3.91∑D2 = 193
(∑D)2 = 1849
Calculating the Matched Samples t-Test
)1(
)( 22
2
−
−=∑ ∑
nnx
xs )111(
1143193
2
2
−
−=s
)10(11
18491932
−=s )10(
09.1681932 −=s
)10(91.242 =s 49.22 =s 49.2=s 58.1=s
Calculating the Matched Samples t-Test
ns
DtD
0−=
1158.1
091.3 −=t
32.358.191.3
=t
48.91.3
=t 15.8=t tobt = 8.15
Evaluating Statistical Significance of the t-Test
• First note:– α = .05– Tail or directionality: two-tailed– t-Value = 8.15– Degrees of freedom (df)
• For the Matched Samples t-Test:– df = number of PAIRS of scores -1– df = 11 - 1 = 10
p. 747 in Howell Text
1) Find row for TAIL
2) In the ROW for the correct tail, find α
3) Find df
4) Track ROW of dfacross to COLUMN of α
The numerical value you obtain is called tcrit
tcrit(10) = 2.228
Evaluating Statistical Significance of the t-Test
tcrit = 2.228tcrit = - 2.228
0
tobt = 8.15
Because tobt falls within the rejection region identified by tcrit we reject H0
Statistical Tests We Have Learned
1. z-Test• 1 group• 1 set of data• μ & σ known
2. One-Sample t-Test• 1 group• 1 set of data• μ known• Estimate σ with s using x-bar
3. Matched Samples t-Test• 1 group• 2 sets of data• μ & σ unknown• Estimate σD with sD using D-bar
Testing Hypotheses: Two Independent Samples
• Probably the most common use of the t-Test and the t-distribution
• Compare the mean scores of two groups on a single variable– IV: Groups– DV: Variable of interest
• Groups must be independent of one another– Scores in 1 group cannot influence scores in
the other group
Testing Hypotheses: Two Independent Samples
• Actually uses a different sampling distribution– Sampling distribution of differences between
means• However, we calculate and test t in
essentially the same fashion
Independent Samples t-Test
21
21
xxsXX
t−
−=
2
22
1
21
21
ns
ns
XXt+
−=or
This test is calculated by dividing the mean difference between two groups by the “dispersion”or “variation” observed between the two groups
Independent Samples t-Test:Degrees of Freedom
• 1 df lost for each σ estimated by s using x-bar
• Since there are two independent groups in this analysis, we must estimate σ twice
• df = (n1 + n2) - 2
Independent Samples t-Test: Example
• Let’s return to the example used for the matched samples test
• As a competent researcher, you realize that simply showing a change over time is not enough to prove the efficacy of your treatment– People spontaneously change over time
• Show that an untreated control group does not change over the same period of time that your treatment group does change
Independent Samples t-Test: Example
SIB
ScoresTx Group
Ctrl Group
Time 1 Time 2
SIB
Scores
Tx
Tx
SIB
Scores
SIB
Scores
SIB
Scores
Time 3
= ?
Independent Samples t-Test: Example
• At time 1, the control and treatment SIB groups have equal SIB scores
• Administer the treatment for 2 weeks to Txgroup– The Control group receives no intervention
during these two weeks• Compare SIB scores of Tx and Control
group after 2 weeks• Provide Control group w/ intervention if
desired
Independent Samples t-Test: Example
• Research Hypothesis:– Your treatment for SIB will reduce SIB
scores in the Tx group after 2 weeks• H1: μt < μc
• Null Hypothesis – Your treatment for SIB will have no effect
• H0: μt = μc
• Evaluate the efficacy of your treatment
Independent Samples t-Test: Example
261791191074108Tx
12161513168119101312ControlTime 2 Data
Control Group
∑x 135
∑x2 1729
(∑x)2 18225
x-bar 12.27
s2 7.29
s 2.69
n 11
Tx Group
∑x 93
∑x2 941
(∑x)2 8649
x-bar 8.45
s2 15.47
s 3.93
n 11
Independent Samples t-Test: Example
• Select:• Rejection region
• α = .05• “Tail” or directionality
• We have evidence that the treatment probably works, so we make a one-tailed hypothesis here (scores for the Tx group will be lower than the Control group at time 2)
Independent Samples t-Test: Example
• Generate sampling distribution of the mean assuming H0 is true
• Independent Samples t-Test• Given our sampling distribution:
• Conduct the statistical test
Independent Samples t-Test: Example
1129.7
1147.15
27.1245.8
+
−=t
66.41.182.3+
−=t
07.282.3−
=t
2
22
1
21
21
ns
ns
XXt+
−=
44.182.3−
=t
65.2−=t tobt(20) = -2.65
Evaluating Statistical Significance of the t-Test
• First note:– α = .05– Tail or directionality: one-tailed– t-Value = -2.65– Degrees of freedom (df)
• For the Independent Samples t-Test– (n1 + n2) - 2 – (11+11)-2 – 22 - 2 = 20
p. 747 in Howell Text
1) Find row for TAIL
2) In the ROW for the correct tail, find α
3) Find df
4) Track ROW of dfacross to COLUMN of α
The numerical value you obtain is called tcrit
tcrit(20) = 1.725
Evaluating Statistical Significance of the t-Test
0
tobt = -2.65
Because tobt falls within the rejection region identified by tcrit we reject H0
tcrit = - 1.725
Independent Samples t-Test: One Complication
• There is a slight problem with the form of the equation we used…– ONLY can be applied
to groups with equal sample sizes
– A major limitation in real-world research
2
22
1
21
21
ns
ns
XXt+
−=
Pooled Variance Estimate
• This equation permits tests with different sample sizes
• Generates an estimate of the total variance between groups weighted by the size of each group– Therefore, larger samples have a greater
impact on the variance – Vice-versa for small samples
Pooled Variance Estimate
2)1()1(
21
222
2112
−+−+−
=nn
snsnsp
Using the Pooled Variance Estimate
2
22
1
21
21
ns
ns
XXt+
−= 2
2
1
221
ns
ns
XXtpp +
−=
21
2
21
11nn
s
XXt
p +
−=
Using the Pooled Variance Estimate: Example
261791191074108Tx
No Data121615131611ControlTime 2 Data
Control Group
∑x 83
∑x2 1171
(∑x)2 6889
x-bar 13.83
s2 4.57
s 2.14
n 6
Tx Group
∑x 93
∑x2 941
(∑x)2 8649
x-bar 8.45
s2 15.47
s 3.93
n 11
Using the Pooled Variance Estimate: Example
261157.4)16(47.15)111(2
−+−+−
=ps2
)1()1(
21
222
2112
−+−+−
=nn
snsnsp
1585.227.1542 +
=ps15
57.4)5(47.15)10(2 +=ps
1555.1772 =ps 84.112 =ps
Using the Pooled Variance Estimate: Example
21
2
21
11nn
s
XXt
p +
−=
)61
111(84.11
83.1345.8
+
−=t
)0909.1667(.84.1138.5+
−=t
tobt(15) = -3.07
)2576(.84.1138.5−
=t
05.338.5−
=t 75.138.5−
=t 07.3−=t
Evaluating Statistical Significance of the t-Test
• First note:– α = .05– Tail or directionality: one-tailed– t-Value = -3.99– Degrees of freedom (df)
• For the Independent Samples t-Test– (n1 + n2) - 2 – (11+6)-2 – 17 - 2 = 15
p. 747 in Howell Text
1) Find row for TAIL
2) In the ROW for the correct tail, find α
3) Find df
4) Track ROW of dfacross to COLUMN of α
The numerical value you obtain is called tcrit
tcrit(15) = 1.753
Evaluating Statistical Significance of the t-Test
0
tobt = -3.99
Because tobt falls within the rejection region identified by tcrit we reject H0
tcrit = - 1.753
Effect Size of The Independent Samples t-Test
σμμ 21 −=d or
psXX
d 21 −=
d = .20 -- Small effect
d = .50 -- Medium effect
d = .80 -- Large effect
Effect Size of The Independent Samples t-Test
8.45 13.833.44
d −=
5.383.44
d −=
psXX
d 21 −=
An effect size exceeding the convention for a large effect
1.56d = −
Statistical Tests We Have Learned
1. z-Test• 1 group• 1 set of data• μ & σ known
2. One-Sample t-Test• 1 group• 1 set of data• μ known• Estimate σ with s using
x-bar
3. Matched Samples t-Test
• 1 group• 2 sets of data• μ & σ unknown• Estimate σD with sD
using D-bar
4. Independent Samples t-Test
• 2 groups• 2 sets of data• μ & σ unknown• Estimate σ twice with s
using x-bar
Choosing the Best Test
Choosing the Best Test
• Flow-chart available on the website:– http://www.personal.kent.edu/~marmey
• Also refer to the diagram on p. 11 of your Howell text
• Try the review problems on the website for an example of the types of questions I might ask on an exam!