Welcome to PSYC 2321:

Analysis of Behavioural
Data

CHAPTER 11
INSTRUCTOR: NICOLE JENNI
T-test comes in 3 forms
• Compare one group mean to a known
population μ when you don’t know σ
Single sample (Ch 9)
• Compare one group mean to some other null
value of interest (e.g., scale midpoint)

• Compare mean change in repeated measures

experiment
Paired Samples (Ch 10) • Compare mean differences when participants
are linked (e.g., parent and child, twin siblings)

• Compare difference between two

Independent Samples (Ch 11) group means in independent groups
experiment
 Chapter 11
Analyzing Paired Analyzing Independent
Samples Data Groups Data
A B C D E

=
Independent-Samples t Tests
Analyzing Paired Samples Data
A B C D E
Used to compare two means in a
between-groups design
=
Provides a situation in which each
participant is assigned to only one
Analyzing Independent Groups Data
condition
 Ways to express the null…
Conceptually speaking… Is the Two-tailed Null Hypothesis
difference between two group H0: |μ1 – μ2|= 0
means significantly different
or
from zero?
H0: μ1 = μ2
Corresponding Research Hypothesis
We tend to say… Are two
independent group means H1: |μ1 – μ2| ≠ 0
significantly different from each or
other? H1: μ1 ≠ μ2
 Distribution of differences between means
This graph represents the
beginning of a distribution of
differences between means.

It includes only 30 differences,

whereas the actual distribution
would include all possible
differences.
 Pooled variance that
Variance of one
incorporates two
sample
sample variances

Pooled variance estimate

Survey example: do introverts and extraverts differ in
how determined they are to do well in stats?
Descriptive Statistics on the DV Determined
GROUP “1” GROUP “2”
INTROVERTS EXTRAVERTS

n1 = 70 people n2 = 73 people

Mean Determined = 3.6357 Mean Determined = 3.5616

Standard Deviation = 0.92439 Standard Deviation = 0.89732

After I entered the data in SPSS, I

asked for an “Independent
Samples t Test”. This is the first
table in the output. Introvert
→ Extravert
 Analyzing Independent Groups Data
FAMILIAR SET OF ANALYSES, ADAPTED FOR
TWO-GROUP CONTEXT
𝑀1 − 𝑀2
𝑡𝑜𝑏𝑡 = 2 𝑛1 − 1 2 𝑛2 − 1 2
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠1 + 𝑠2
𝑁−2 𝑁−2
Standard error of the difference
between group means
2 2
𝑀1 −𝑀2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
Cohen’s 𝑑 = 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = +
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑛1 𝑛2
Confidence Interval around the
difference between means REMEMBER: I have combined steps c, d, and e on pages 254-255 of the
𝜇1 − 𝜇2 textbook into one equation.
 Finding your t-critical
Q3. What’s the closest
critical value of t?
(assume two-tail)
Degrees of Freedom
A. ±1.645
dfx = nx -1
dfy. = = ny -1 B. ±1.658
dftotal = dfx + dfy -OR- C. ±1.960
dftotal = N-2 D. ±1.980
choose a more conservative value if the exact df isn’t listed.
 2
Independent Groups t-test: calculate 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
2
𝑀1 − 𝑀2 𝑠𝑖𝑛𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.92439 𝑠𝑖𝑛𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.854497
𝑡𝑜𝑏𝑡 = 2
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑒𝑥𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.89732 𝑠𝑒𝑥𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.805183

2 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 What is the pooled
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = + variance estimate?
𝑛1 𝑛2
A. 0.82931
2 𝑛𝑋 − 1 2 𝑛𝑌 − 1 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌 B. 0.91067
𝑁−2 𝑁−2
C. 0.91057 A
Introvert D. 0.15234
Extravert
 2
Independent Groups t-test: calculate 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
𝑀1 − 𝑀2 𝑠𝑖𝑛𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.92439 𝑠 2
𝑖𝑛𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.854497
𝑡𝑜𝑏𝑡 =
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑒𝑥𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 2
0.89732 𝑠𝑒𝑥𝑡𝑟𝑎𝑣𝑒𝑟𝑡 = 0.805183

2 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = + 2 70 − 1 73 − 1
𝑛1 𝑛2 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.854497 + 0.805183
143 − 2 143 − 2
2
2 𝑛𝑋 − 1 2 𝑛𝑌 − 1 2 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.48936 0.854497 + 0.510638 0.805183
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌
𝑁−2 𝑁−2 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.418158 + 0.411157
2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.829315
 Independent Groups t-test: calculate 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
2
𝑀1 − 𝑀2 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.829315
𝑡𝑜𝑏𝑡 =
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 What is the pooled
2 2
variance estimate?
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = + A. 0.011361
𝑛1 𝑛2

B. 0.02321
2 𝑛𝑋 − 1 2 𝑛𝑌 − 1 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌
𝑁−2 𝑁−2 C. 0.15342

D. 0.15234 D
Introvert

Extravert
 Independent Groups t-test: calculate 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
2
𝑀1 − 𝑀2 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.829315
𝑡𝑜𝑏𝑡 =
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = +
2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑛1 𝑛2
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = +
𝑛1 𝑛2
0.829315 0.829315
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = +
𝑛𝑋 − 1 2 𝑛𝑌 − 1 2 70 73
2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌
𝑁−2 𝑁−2
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0.11847 + 0.01136

𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0.023208
Introvert
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0.15234
Extravert
Independent Groups t-test: calculate t-obtained
𝑀1 − 𝑀2
𝑡𝑜𝑏𝑡 =
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒

3.6357 − 3.5616
𝑡𝑜𝑏𝑡 =
0.1523

𝑡𝑜𝑏𝑡 = 0.4864

Introvert
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0.15234
Extravert
Null Hypothesis Sampling
Distribution of the Difference
Between Group Means, at N-2
degrees of freedom

tcrit(141) = -1.980 tcrit(141) = 1.980

.025 .025

All possible
-5 -4
values of t

If our sample mean difference = 0, μ1 – μ2 = 0

then tobtained = 0

When we reject the null hypothesis with the t sampling distribution, we’re saying we think we drew
our sample from a population that has a non-zero t (i.e., a difference between group means).
Null Hypothesis Sampling
Distribution of the Difference
𝑡𝑜𝑏𝑡 = 0.4864
Between Group Means, at N-2
degrees of freedom Is our difference
between means
tcrit(141) = -1.980 tcrit(141) = 1.980
significantly
different from zero?
A. Yes
B. No
.025 .025

-5 -4

μ1 – μ2 = 0
B.
tobtained is between the critical values, not outside either of them. It
looks like these two groups were drawn from the null population
(no difference). p > .05.
Computing Cohen’s D

Q7. What is the effect size?

𝑀1 −𝑀2
Cohen’s 𝑑 = 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 A. 0.0741
B. 0.08935
C. 0.08137
D. 0.4864

2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.82931
Introvert
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0.15234
Extravert
Computing Cohen’s D

𝑀1 −𝑀2
Cohen’s 𝑑 = 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.82931

3.6357 − 3.5616
𝑑=
0.910665 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.910665

𝑑 = 0.081
 Add a Confidence Interval to identify the range of plausible
values for the difference between means of whatever
population our sample belongs to.
2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 0.82931
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0.15234 Q8. What is the 95CI
tcritical = 1.98 A.[-0.22, 0.38]
B. [-1.56, 1.72]
Lower Boundary Upper Boundary
C.[-0.07, 0.23]
MDifflower = MDiff - tcritical(𝑠𝑑𝑖𝑓𝑓 ) MDiffupper = MDiff + tcritical(𝑠𝑑𝑖𝑓𝑓 ) D. [-0.15, 0.38]

-.30 -.20 -.10 0 .10 .20 .30 .40 .50 .60 .70 .80 .90

Difference Between Group Means

Add a Confidence Interval to identify the range of plausible
values for the difference between means of whatever
population our sample belongs to.

Lower Boundary Upper Boundary

MDifflower = MDiff - tcritical(𝑠𝑑𝑖𝑓𝑓 ) MDiffupper = MDiff + tcritical(𝑠𝑑𝑖𝑓𝑓 )

MDifflower = 0.074 – 1.98(0.1523) MDiffupper = 0.074 + 1.98(0.1523)
MDifflower = -0.22 MDiffupper = 0.38

-.30 -.20 -.10 0 .10 .20 .30 .40 .50 .60 .70 .80 .90

Difference Between Group Means

The Plan: Analyzing Our Data

t-test Confidence
• Means, SD, n per group Intervals Cohen’s d
• Difference between group • t-critical • Difference between
means • Difference between
• Pooled standard group means
group means Conclusions
• Standard Error of the • Pooled standard
deviation deviation
difference between
• Standard Error of the means
difference between
means
 Every statistical test relies on some
assumptions
ASSUMPTIONS OF THE Z TEST/SINGLE SAMPLE T-
TEST/ PAIRED SAMPLES T TEST
•DV measured using a scale variable (so can
calculate means) What happens if
•Population(s) normally distributed or N ≥ 30 we violate these
• N = n1 + n2 assumptions?
•Participants randomly selected from
population(s)
• Careful generalizing
Every statistical test relies on some
assumptions
ASSUMPTIONS OF THE INDEPENDENT SAMPLES T-TEST

•DV measured using a scale variable (so can

calculate means)
•Population(s) normally distributed or N ≥ 30
•Subjects are randomly selected from
population(s)
• Careful generalizing
•Homogeneity of Variance (HOV)
Homogeneity of Variance (HOV) Assumption
HOMOSCEDASTIC: populations that have the same variance
HETEROSCEDASTIC: populations that have different variances

Because we are ‘pooling’ our variance Independent Samples T-Tests are

estimate from the two different Robust to violations of HOV ONLY if
samples… sample sizes are perfectly equal (n1 = n2
We are assuming that the samples all
come from populations with the same
variances
▪In other words, we want to check that
these variances are not significantly
different
Testing our Homogeneity of Variance (HOV)
Assumption with SPSS

Levene’s Test for Variances

Levene’s test is testing the H0: that the If we want to check that our group
variances are the same variances are the SAME, we are looking
for a Levene’s test result that is:
Ie. That there is NO difference
between the variances in the A) Significant
population
B) Not Significant

B
Independent samples t-test: Caffeine and
Reaction Times Study
oDoes caffeine improve reaction times?
oGroup 1 gets real Caffeine, Group 2 gets placebo

oEverybody plays a reaction time task

oDependent variable: reaction time in milliseconds (ms)
 We want to conduct a one-tailed test
oHo: People who consume caffeine will have
Fill in the blanks:
slower or equal reaction times compared to
people who consume placebo A. Not different / Different
Ho: μCaffeine ≥ μPlacebo B. Different / Not different
C. Slower or equal / Faster
oH1: People who consume caffeine will have D. Faster or equal / Slower
faster reaction times compared to people who
consume a placebo
H1: μCaffeine < μPlacebo C
 We want to conduct a one-tailed test
oHo: People who consume caffeine will have
Fill in the blanks:
slower or equal reaction times compared to
people who consume placebo A. ≥ / <
Ho: μCaffeine ≥ μPlacebo B. ≤ / >
C. = / ≠
oH1: People who consume caffeine will have D. ≠ / ≠
faster reaction times compared to people who
consume a placebo
A
H1: μCaffeine < μPlacebo
We want to conduct a one-tailed test
oHo: People who consume caffeine will have ***NOTE***
slower or equal reaction times compared to “Slower” reaction
time means a
people who consume placebo larger millisecond
value
Ho: μCaffeine ≥ μPlacebo
oH1: People who consume caffeine will have
***NOTE***
faster reaction times compared to people who “Faster” reaction
consume a placebo time means a
smaller millisecond
H1: μCaffeine < μPlacebo value
 Alpha one tailed = .05
Caffeine = 7 people
What is t-critical? Placebo = 6 people

Ho: μCaffeine≥μPlacebo A. + 2.201

H1: μCaffeine<μPlacebo B. – 2.201
C. + 1.796
D. – 1.796

D
Data collected
Caffeine (ms) Placebo (ms)
40 45
45 60
55 55
35 50
40 55
45 50
40

Caffeine
Placebo
 2
calculate 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
Caffeine (ms) Placebo (ms) 𝑋ത1 − 𝑋ത2 What is the pooled
𝑡𝑜𝑏𝑡 =
40 45 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 variance estimate?
45 60 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 A. 34.5779
55 55 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = +
𝑛1 𝑛2
35 50 B. 34.4871
40 55
45 50 2 𝑛𝑋 − 1 2 𝑛𝑌 − 1 2 C. 5.85388
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌
40 𝑁−2 𝑁−2
D. 5.8803
A
2 𝑛𝑋 − 1 2 𝑛A𝑌 − 1 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌
Caffeine 𝑁−2 𝑁−2
Placebo
 Independent Groups t-test: calculate 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
2
𝑀1 − 𝑀2 𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 34.5779
𝑡𝑜𝑏𝑡 =
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 What is the standard error
2 2
estimate?
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = + A. 10.7027
𝑛1 𝑛2

B. 3.5607
2 𝑛𝑋 − 1 2 𝑛𝑌 − 1 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 = 𝑠𝑋 + 𝑠𝑌
𝑁−2 𝑁−2 C. 3.2715

D. 12.6786 C
Caffeine
Placebo
Independent Groups t-test: calculate t-obtained
𝑀1 − 𝑀
𝑡𝑜𝑏𝑡 =
𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒

42.8571 − 52.5000
𝑡𝑜𝑏𝑡 =
3.2715

𝑡𝑜𝑏𝑡 = -2.948

𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 3.2715
Caffeine
Placebo
Null Hypothesis Sampling
Distribution of the Difference
Between Group Means, at N-2 𝑡𝑜𝑏𝑡 = -2.948
degrees of freedom Is our difference between
means significantly different
tcrit(11) = -1.796 from zero?

Make sure to check the

direction of the effect!

.05
A. Yes
All possible
values of t
-5 -4 B. No

If our sample mean difference = 0, μ1 – μ2 = 0

then tobtained = 0

Caffeine
Placebo
 SPSS Output

Levene’s Test is testing the assumption that our two Did we violate our HOV assumption?
samples have equal variances in the population (HOV) p=0.78

Here we can see this test is NOT SIGNIFICANT (p=.78) A. YES, we violated this assumption
B. NO, we did not
 HOV Assumption
• When we design an experiment, we generally hypothesize that our
manipulation (ie caffeine) will cause some mean difference between our
groups

• We do NOT expect this manipulation to change the variability in our

experimental groups

• Ie. There should still be the same ‘spread’ or variability in the caffeine and
non caffeine group

• IF caffeine did happen to have an effect on group variance, a significant

Levene’s test will tell us, and will apply a correction to our test (row 2)
Writing a good conclusion
Participants who consumed caffeine (M= 42.86 ms, s=6.36
ms) performed the reaction time task significantly faster
than the placebo group (M= 52.50, s = 5.24) t(11)= -2.95,
p<.05.

A good conclusion ALWAYS states your group(s), the descriptive statistics

(means and standard deviations) and your test results. If you have it, you
should also report Cohen’s D and your confidence intervals.
 Where we’ve been…
Normal Single sample
To compare sampling z test
distribution
sample mean to a distribution of the
(when know μ (CI around mean,
population mean mean (𝜎𝑀 )
and σ) effect size)
 Where we’ve been…
Normal Single sample
To compare sampling z test
distribution
sample mean to a distribution of the
(when know μ (CI around mean,
population mean mean (𝜎𝑀 )
and σ) effect size)

To compare sample t distribution Single sample

sampling t test
mean to a (when don’t know
distribution of the
population mean
mean (𝑠𝑀 )
σ) (CI around mean,
or particular score df = N-1 effect size)
 Where we’ve been…
Normal Single sample
To compare sampling z test
distribution
sample mean to a distribution of the
(when know μ (CI around mean,
population mean mean (𝜎𝑀 )
and σ) effect size)

To compare sample t distribution Single sample

sampling t test
mean to a (when don’t know
distribution of the
population mean
mean (𝑠𝑀 )
σ) (CI around mean,
or particular score df = N-1 effect size)

Paired Samples
sampling distribution (repeated measures) t-
To compare means of t distribution
of the mean difference test
two related groups df = N-1
(𝑠𝑋ത ) (CI around mean
difference, effect size)
 Where we’ve been…
Normal Single sample
To compare sampling z test
distribution
sample mean to a distribution of the
(when know μ (CI around mean,
population mean mean (𝜎𝑀 )
and σ) effect size)

To compare sample t distribution Single sample

sampling t test
mean to a (when don’t know
distribution of the
population mean
mean (𝑠𝑀 )
σ) (CI around mean,
or particular score df = N-1 effect size)

Paired Samples
sampling distribution (repeated measures) t-
To compare means of t distribution
of the mean difference test
two related groups df = N-1
(𝑠𝑋ത ) (CI around mean
difference, effect size)

sampling Independent
To compare means distribution of the groups t-test
of two t distribution
difference between (CI around
independent two means df = N-2
groups difference between
(𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 ) means, effect size)
 Where z-distribution is the normal
curve, and t’s reference t-distribution
Comparison Distributions as their associated dfs

Ch 7: Z-test Ch 9 single sample t-test

Sampling Distribution of the mean
• Centered around population mean Sampling Distribution of the mean
• Error = SEM • Centered around population mean
• Error = SEM

𝜎 𝑠
𝜎𝑀 = 𝑠𝑀 =
𝑁 𝑁

Sampling Distribution differences

Ch 9: Paired Ch 10: Independent between means

samples t-test
Sampling Distribution of mean samples t-test • Centered around 0
difference • Error = standard error of difference
• Centered around 0 between means
• Error = standard error of mean 2 2
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 𝑠𝑝𝑜𝑜𝑙𝑒𝑑
difference s 𝑠𝑑𝑖𝑓𝑓 = +
D
𝑠𝑀 = 𝑛1 𝑛2
𝑁
Learn to dissociate your symbols
Standard deviation = s
Variance =𝑠 2
Standard deviation of ‘difference scores’ = 𝑠𝐷
2
Pooled variance = 𝑠𝑝𝑜𝑜𝑙𝑒𝑑

Standard error of the mean: 𝑠𝑀

Standard error of the differences between means: 𝑠𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
Which test is best?
IS OUR MEAN DIFFERENT FROM A SPECIFIC ARE THESE TWO SAMPLE MEANS DIFFERENT
POPULATION MEAN? FROM EACH OTHER?

Do we know the population standard Are the data correlated or independent?

deviation? (σ)

Yes No Correlated Independent

Single sample Paired samples Independent

Z test samples t-test
t-test t-test