Students t -Distribution The t -Distribution, t -Tests, & - - PowerPoint PPT Presentation

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

• f 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 /

2

σ

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 μ will tend to be similar to one another… – The relationships between…

• σx

2 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 ) ( x x z − =

15 ) 100 83 ( − = z 15 ) 17 (− = z 3 3 . 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

5 15 ) 100 122 ( − = z 24 . 2 15 ) 22 ( = z 70 . 6 ) 22 ( = z

n x z σ μ) ( − =

Note: this equation is a modification of the

• riginal 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

100 15 ) 100 122 ( − = z 10 15 ) 22 ( = z 5 . 1 ) 22 ( = z

67 . 14 = z

2 15 ) 100 122 ( − = z 41 . 1 15 ) 22 ( = z 64 . 10 ) 22 ( = z

07 . 2 = z

1 15 ) 100 122 ( − = z 1 15 ) 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 x z σ μ) ( − =

σ μ) ( − = x z

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

• f s

Sampling distribution

• f 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

• ne degree of freedom calculating s using the sample

mean and not the population mean.

One-Sample t-Test

x

s x t ) ( μ − = n s x t

x

) ( μ − = n s x t

x 2

) ( μ − =

• r
• r

z-Test vs. One-Sample t-Test

n x z σ μ) ( − =

n s x t

x

) ( 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

n s x t

x

) ( μ − =

24 82 ) 569 643 ( − = t 73 . 16 ) 74 ( = t

90 . 4 82 ) 74 ( = t

42 . 4 = t

μVerbal = 569 x-bar = 643 s = 82 n = 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 df across to COLUMN

• f α

The numerical value you

• btain 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 = 2.069 tcrit = - 2.069 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
• f 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

n s D t

D

− =

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

5 4 2 7 4 4 1 3 4 4 5 D 25 16 4 49 16 16 1 9 16 16 25 D2 2 6 17 9 11 9 10 7 4 10 8 Time 2 7 10 19 16 15 13 11 10 8 14 13 Time 1

∑D = 43 D-bar = 3.91 ∑D2 = 193 (∑D)2 = 1849

Calculating the Matched Samples t-Test

) 1 ( ) (

2 2 2

− − = ∑

∑

n n x x s

) 1 11 ( 11 43 193

2 2

− − = s

) 10 ( 11 1849 193

2

− = s

) 10 ( 09 . 168 193

2

− = s

) 10 ( 91 . 24

2 =

s 49 . 2

2 =

s

49 . 2 = s

58 . 1 = s

Calculating the Matched Samples t-Test

n s D t

D

− = 11 58 . 1 91 . 3 − = t

32 . 3 58 . 1 91 . 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 df across to COLUMN

• f α

The numerical value you

• btain is called tcrit

tcrit(10) = 2.228

Evaluating Statistical Significance of the t-Test

tcrit = 2.228 tcrit = - 2.228 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
• n 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

2 1

2 1 x x

s X X t

−

− =

2 2 2 1 2 1 2 1

n s n s X X t + − =

• r

This test is calculated by dividing the mean difference between two groups by the “dispersion”

• r “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 Scores Tx 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 Tx

group

– 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

2 6 17 9 11 9 10 7 4 10 8 Tx 12 16 15 13 16 8 11 9 10 13 12 Control

Time 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

11 29 . 7 11 47 . 15 27 . 12 45 . 8 + − = t

66 . 41 . 1 82 . 3 + − = t

07 . 2 82 . 3 − = t

2 2 2 1 2 1 2 1

n s n s X X t + − =

44 . 1 82 . 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 df across to COLUMN

• f α

The numerical value you

• btain is called tcrit

tcrit(20) = 1.725

Evaluating Statistical Significance of the t-Test

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

• f the equation we

used…

– ONLY can be applied to groups with equal sample sizes – A major limitation in real-world research

2 2 2 1 2 1 2 1

n s n s X X t + − =

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 (

2 1 2 2 2 2 1 1 2

− + − + − = n n s n s n sp

Using the Pooled Variance Estimate

2 2 2 1 2 1 2 1

n s n s X X t + − =

2 2 1 2 2 1

n s n s X X t

p p +

− =

2 1 2 2 1

1 1 n n s X X t

p

+ − =

Using the Pooled Variance Estimate: Example

2 6 17 9 11 9 10 7 4 10 8 Tx No Data 12 16 15 13 16 11 Control

Time 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

2 6 11 57 . 4 ) 1 6 ( 47 . 15 ) 1 11 (

2

− + − + − =

p

s 2 ) 1 ( ) 1 (

2 1 2 2 2 2 1 1 2

− + − + − = n n s n s n sp

15 85 . 22 7 . 154

2

+ =

p

s

15 57 . 4 ) 5 ( 47 . 15 ) 10 (

2

+ =

p

s

15 55 . 177

2 = p

s

84 . 11

2 = p

s

Using the Pooled Variance Estimate: Example

2 1 2 2 1

1 1 n n s X X t

p

+ − =

) 6 1 11 1 ( 84 . 11 83 . 13 45 . 8 + − = t ) 0909 . 1667 (. 84 . 11 38 . 5 + − = t

tobt(15) = -3.07

) 2576 (. 84 . 11 38 . 5 − = t

05 . 3 38 . 5 − = t

75 . 1 38 . 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 df across to COLUMN

• f α

The numerical value you

• btain is called tcrit

tcrit(15) = 1.753

Evaluating Statistical Significance of the t-Test

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

σ μ μ

2 1 −

= d

• r

p

s X X d

2 1 −

=

d = .20 -- Small effect d = .50 -- Medium effect d = .80 -- Large effect

Effect Size of The Independent Samples t-Test

8.45 13.83 3.44 d − =

5.38 3.44 d − =

p

s X X d

2 1 −

=

An effect size exceeding the convention for a large effect

1.56 d = −

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!