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Hypothesis Testing:
Inferential statistics
These will help us to decide if we should:
1) believe that the relationship we found in our sample
data is the same as the relationship we would find if we
tested the entire population
OR
2) believe that the relationship we found in our sample
data is a coincidence produced by sampling error
Inferential statistics
-Univariate statistical analysis: tests hypotheses involving
only one variable
-Bivariate statistical analysis: tests hypotheses involving
two variables
-Multivariate statistical analysis: tests hypotheses and
models involving multiple variables
Specific Steps in Hypothesis
Testing
1.
2.
3.
4.
Specify the hypothesis.
Select an appropriate statistical test.
Two-tailed or One-tailed test.
Specify a decision rule (alpha = ???).
Compute the critical value.
5. Calculate the value of the test statistic
and perform the test.
6. State the conclusion.
Hypotheses and statistical tests
In general, when you make a testable hypothesis, you
specify the relationship you expect to find between your
IV and the DV.
If you specify the exact direction of the relationship (i.e.,
longer math tests will increase test anxiety), then you will
perform a 1-tailed test.
At other times, you may not know or predict a specific
result direction but rather just that performance will
change (ie. longer math tests will affect test anxiety),
then you will perform a 2-tailed test.
Hypothesis
• NULL Hypothesis: world view or Status
quo.
• Alternative Hypothesis: Researcher’s
theory
– -H0 The average age of a large class is 25.
– -H1 The average age of a large class is
different than 25 (two tail)
– - H1 The average age of a large class is less
than 25.
– Other examples?
Hypothesis Testing about Means
# Groups
One
Two
Purpose
Test
Comment
Sample and
Population
comparison
Sample and
Population
comparison
Comparing two
sample means
Z-test
If  is known;
and large
samples
t-test
If  is
unknown
Z-test
If  is known;
and large
samples
Comparing two
sample means
t-test
If  is
unknown
Hypothesis testing about
means
# Groups
Three or More
Purpose
Test
Comparing multiple
sample means
F
Comment
ANOVA
framework
The One-sample Experiment
Let’s say you know the value of a particular characteristic
in the population (this is uncommon)
- i.e., Computer Industry Satisfaction is normal
(mean=100, SD=15)
It turns out that we have one CS score for a company X(X
= 84)
This is a pretty high score. It’s lower than the industry
average, but it is “within range.” Based on this one score
can I say that X’s score is significantly, different than
industry score?
Statistical Hypotheses (1-tailed)
H1: X has CS lower than industry
H2: X has an equal to or greater CS than industry
Given: i = 100
Alternative hypothesis (H1): x < i
Null hypothesis (H0): x  i
Statistical Hypotheses (2-tailed)
H1: X has different CS than industry
H2: X has an equal CS to industry
Given: i = 100
Alternative hypothesis (H1): x ≠ I
Null hypothesis (H0): x = i
Logic of Statistical Hypothesis Testing
We measured CS on a sample of companies (for the
moment N=1) and find that mean CS for the sample of
companies is less than the industry mean of 100. Does
this mean that X has lower CS?
Or , the possible explanations are:
1) X indeed has lower CS
2) X has the same CS, but sampling error
produced the smaller mean CS score.
We could run everyone in the population (not possible), or use
inferential statistics to choose between these two alternatives.
Logic of Statistical Hypothesis Testing
The logic goes like this:
Inferential statistics will tell us how likely it would be to
get a sample mean like the one we measured given
that the null hypothesis (H0) is true.
If it is really UNlikely that we would get a mean score
like we did, drawing a sample from the population of
the x = 100), then we conclude that our sample did
not come from that population, but from a different one.
We reject the null hypothesis.
Logic of Statistical Hypothesis Testing
All statistical tests use this logic:
- determine the probability that sampling
error (random chance) has produced the
sample data from a population described by
the null hypothesis (H0).
Let’s look at the z-test to see how this works.
Assumptions of the z-test
1. We have randomly selected one sample (for the
moment N=1)
2. The dependent variable is at least approximately
normally distributed in the population and involves
an interval or ratio scale
3. We know the mean of the population of raw scores
under some other condition of the independent
variable
4. We know the true standard deviation of the
population xdescribed by the null hypothesis
Before you compute the test statistic
1. Choose your alpha () level, the criterion you
will use to determine whether to accept or
reject H0, .05 is typical in
psychology/mkt/mgm
2. Identify the region of rejection (1- or 2-tailed)
3. Determine the critical value for your statistic
- for z-test, the critical value is labeled zcrit
A sampling distribution for H0 showing the region of
rejection for  = .05 in a 2-tailed z-test.
2-tailed regions
A sampling distribution for H0 showing the region of
rejection for  = .05 in a 1-tailed z-test.
1-tailed region, above mean
A sampling distribution for H0 showing the region of
rejection for  = .05 in a 1-tailed z-test where a
decrease in the mean is predicted.
1-tailed region, below mean
Finding zcrit (2-tailed)
To find zcrit, we will use
our z-table to find the zscore that gives us the
appropriate proportion
of scores in the region
of rejection (in the tails
of the distribution).
Finding zcrit (1-tailed)
To find zcrit, we will use
our z-table to find the zscore that gives us the
appropriate proportion
of scores in the region
of rejection (in the tails
of the distribution).
Computing zobt for the test
You calculate the z-score for your sample mean
in a similar way as we did for a single score
and it is labeled zobt.
t obt 
X 
SEM:
X
zobt = 84 - 100 = 16
15
15
SX
sX 
N
= -1.07
What does it mean?
Interpreting zobt relative to zcrit
You then compare your zobt value to the zcrit
2-tail
value. If zobt is beyond zcrit (more into the tail
of the distribution), then we will say that we
1-tail
“reject the null hypothesis”.
Note: in this case we fail to reject the null
hypothesis. This does not mean the null is
true. Indeed, in this case it is almost certainly
not true.
Don’t forget this it is really important!!!
We then report that our results were “not
significant” or “nonsignificant” and report it like
this: z = +1.07, p > .05
Now suppose that we have a sample
of 3 companies CS scores.
3  1.73
SO,
X
15
X 

 8.66
N 1.73

and

zobs 
X 
X
16

 1.85
8.66
Interpreting zobt relative to zcrit
You then compare your zobt value to the zcrit
2-tail
value. If zobt is beyond zcrit (more into the tail
of the distribution), then we will say that we
1-tail
“reject the null hypothesis”.
As a logical consequence, if we have rejected
the null hypothesis then we have supported
the alternative hypothesis. BUT we have not
PROVED anything.
We then report that the difference was
“statistically significant” and report it like this:
z = -1.86, p < .05
Errors
Since our statistical procedures are based on
probabilities, there is a possibility that our data
turned out as they did from chance alone.
Type I: rejecting the null hypothesis when it is
actually true, the probability of making this error is
equal to , our chosen criterion.
Type II: accepting the null hypothesis when it is
actually false, the probability of making this error
can be computed and it is labeled .
The One-sample t-test
We rarely know the population characteristics, particularly
the SD, and thus we rarely use the z-test.
If we know the population mean, but not the SD, we
must use the t-test rather than the z-test. The logic is
identical.
Assumptions of the t-test
1. We have randomly selected one sample
2. The dependent variable is at least approximately
normally distributed in the population and involves
an interval or ratio scale
3. We know the mean of the population of raw scores
under some other condition of the independent
variable
4. We do not know the true standard deviation of the
population described by the null hypothesis so we
will estimate it using our sample data.
Before you compute the test statistic
1. Choose your alpha () level, .05 is common
2. Identify the region of rejection (1- or 2-tailed)
3. Determine the critical value for your statistic
- for t-test, the critical value is labeled tcrit
Computing tobt for the test
You calculate tobt for your sample mean in a
similar way as we did for zobt.
sX
15
sX 

 8.66
N 1.73

tobs 
X 
X
16

 1.85
8.66
Compare tobt relative to tcrit
As with the z-test, for the t-test you will compare
tobt to tcrit. So how do you find tcrit? You look in
a table of the t-distribution.
The t-distribution contains all possible values of t
computed for random sample means selected
from the population described by the null
hypothesis (similar to the z-distribution).
One BIG difference, is that there are many t-distributions
for any population, one for every sample size. As N
increases, the t-distribution better approximates a
normal distribution, until N ~ 120.
Finding tcrit
Since there are different t-distributions for
different sample sizes, we must find tcrit from
the appropriate t-distribution.
The appropriate t-distribution will be the one that
matches our sample size, which we now call
degrees of freedom (df).
For the t-test, the degrees of freedom equals N-1,
When N-1 is more than 120, the t-distribution is
indistinguishable from a true normal distribution and
thus use df=.
Finding tcrit (2-tailed)
To find tcrit, use the t-table
to find the t for the df that
corresponds to your
sample size (N-1) for the
criterion () you have
chosen.
Finding tcrit (1-tailed)
To find tcrit, use the t-table
to find the t for the df that
corresponds to your
sample size (N-1) for the
criterion () you have
chosen.
Interpreting tobt relative to tcrit
If tobt is less than tcrit (away from the area of
rejection), then we will say that we “failed to
reject the null hypothesis”.
We then report that our results were “not
significant” or “nonsignificant” and report it like
this: t (2) = 1.86, p > .05
Modification of calculation when we compare
means of two different samples
What is the null?
The alternative?
Chi-Square Test Requirements
1. Quantitative data.
2. One or more categories.
3. Independent observations.
4. Adequate sample size (at least 10).
5. Simple random sample.
6. Data in frequency form.
7. All observations must be used
Think of types of research questions that can be answered
using this test
How do you know the expected frequencies?
1. You hypothesize that they are equal in each group (category)
2. Prior knowledge, census data
Color preference in a car dealership (univariate)
Category O
yellow
red
green
blue
white
E
35
50
30
10
25
(O - E)2
(O - E)
(O - E)2 /E
30
5
25
0.83
45
5
25
0.56
15
15
225
15
15
-5
25
1.67
45
-20
400
8.89
Chisquare=26.95
With df=5-1=4