Hypothesis Testing - Personal.kent.edu

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Hypothesis Testing
Comparing One Sample to its
Population
Hypothesis Testing w/ One
Sample

If the population mean (μ) and standard
deviation (σ) are known:


Testing if our sample mean ( X) is significantly
different from our sampling distribution of the
mean
Similar to testing if how different an individual
score is from other scores in the sample

What is this test called?
Hypothesis Testing w/ One
Sample

z-score formula [for an individual score (x)] =
z

X 

z-score formula [for means ( X )] =
z
X 

N
Hypothesis Testing w/ One
Sample


Testing score versus
standard deviation for
an distribution of scores
Testing mean versus
standard deviation for
distribution of sample
means

I.e. standard error
z
z
X 

X 

N
Hypothesis Testing w/ One
Sample

Two implications of this formula:

1. Because we are dividing by N (actually √N),
with the same data (same sample & population
mean and σ), but larger sample size, our p-value
will be smaller (i.e. more likely to be significant)

All statistical tests that produce p-values will be
sensitive to sample size – i.e. with enough people
anything is significant at p < .05
Hypothesis Testing w/ One
Sample

Two implications of this formula:

2. If you recall, this formula was derived from the
formula for the normal distribution



This means that your data must be normally
distributed to use this test validly
However, this test is robust to violations of this
assumption – i.e. you can violate it, if you have (a) a
large enough sample or (b) your population data is
normally distributed
Why?
Hypothesis Testing w/ One
Sample

The Central Limit Theorem:
Given a population with mean μ and variance
σ2, the sampling distribution of the mean (the
distribution of sample means) will have a
mean equal to μ (i.e., μ = μ) and a variance
(σ2) equal to σ2/N (and standard deviation, σ
= σ/√N). The distribution will approach the
normal distribution as N, the sample size,
increases.
X
X
X
Hypothesis Testing w/ One
Sample

Example #1:

You want to test the hypothesis that the current
crop of Kent State freshman are more depressed
than Kent State undergraduates in general.




What is your sample and what is your population?
What is your Ho and your H1?
Are you using a one- or two-tailed test?
Assuming that for current Kent State freshman, their
mean depression score is 15, while the mean for all
previous Kent State undergrads (N = 100,000) is 10,
and their standard deviation is 5
Hypothesis Testing w/ One
Sample
15  10
z
= 5/.0158 = 316.46
5
100,000



Look up value in Table E10 with value in “Smaller
Portion”
p < 0.0000
Since this is less than .05 (or .025 if we were using a
two-tailed test), we could conclude that the current
batch of freshman is significantly more depressed
than previous undergrads

Also notice the effect that our large N had on our p-value
Hypothesis Testing w/ One
Sample

Example #2:

You want to test the hypothesis that a treatment conducted
at the ward you are working on in a hospital is more
beneficial than the average treatment used in other wards
of the hospital
 Get into groups of 2 or more
 State Ho and H1
 State if you’re using a one- or two-tailed test and why.
 Given that the mean wellness score of the people on your
ward is 87, and that the mean for the entire hospital (N =
20) is 76 and the standard deviation is 15, is your ward
significantly better? State the p-value that supports your
claim.
Hypothesis Testing w/ One
Sample

Example #2:
87  76
z
= 11/3.36 = 3.27
15
20
Smaller Portion = p < .0006
Hypothesis Testing w/ One
Sample




Most often, however, we don’t know the μ and σ,
because this is what we’re trying to estimate with
our sample in the first place
The formula for the t statistic accomplishes this by
substituting s2 for σ2 in the formula for the z statistic
Because of this substitution, we have a different
statistic, which requires that we use a different table
than E.10
Don’t worry too much about why it’s different (you
can skip pg. 274 in your book if you want, you won’t
be tested on it)
Hypothesis Testing w/ One
Sample

Testing mean versus
standard deviation for
distribution of sample
means


I.e. standard error
Testing mean versus
standard deviation for
sample
z
X 

N
X 
t
s
N
Hypothesis Testing w/ One
Sample


After computing our t statistic, we need to
compare it with the t-table (called the
Student’s T-Table; E.6, pg. 519)
First, we will need to become familiar with the
concept of degrees of freedom or df


df = N – 1
This represents the number of individual subjects
data points that are free to vary, if you know the
mean or s already
Hypothesis Testing w/ One
Sample

For example:


If we already know that a particular set of data has a mean
of 5, and 10 scores in total (n = 10)
Once we have nine of those scores, we can calculate the
tenth, however, if we have eight scores we do not know
what the other two scores could be

We can solve x + 5 = 10, but not x + y = 10, because
in the latter we have more than one unknown (x and y)


x and y could be 5 and 5, 8 and 2, 4 and 6, 7 and 3, etc.
Therefore, nine scores are free to vary, then the tenth is
fixed
Hypothesis Testing w/ One
Sample
Two-Tailed Significance Level
df
10
15
20
25
30
100
.10
1.812
1.753
1.725
1.708
1.697
1.660
.05
2.228
2.131
2.086
2.060
2.042
1.984
.02
2.764
2.602
2.528
2.485
2.457
2.364
.01
3.169
2.947
2.845
2.787
2.750
2.626
Hypothesis Testing w/ One
Sample



df: we know = N – 1
Two-Tailed vs. One-Tailed Tests and Significance
Level (α): we determine in advance
@ p = .05 for a Two-Tailed Test (with 10 df), the
Critical Value of T, or the value of t equal to a
“Smaller Value” of 2.5% (if we were using the ztable), is 2.228

Once we calculate our t-score, we compare it with our
rejection region and our critical t to determine if we reject
H0 or fail to reject it
Hypothesis Testing w/ One
Sample



p = .025; z = 1.96; t = 2.228
Our rejection region is the area in red – it is all in one tail because we’re
using a one-tailed test (in this example), and is in the lower half of the
distribution because we’re interested in scores below the population mean
(again, in this example)
If we were interested in scores above the population mean, we would move
our rejection region to the other tail
Hypothesis Testing w/ One
Sample

Factors that influence the z and t statistics:





The difference between the sample mean and population
mean – greater differences = greater t and z values
The magnitude of s (or s2) – since we’re dividing by s,
smaller values of s result in larger values of t or z [i.e. we
want to decrease variability in our sample (error)]
The sample size – the bigger the bigger t and z
The significance level (α) – the smaller the α, the higher the
critical t to reject Ho – although raising α also raises our
Type I Error, so we probably won’t want to do this without
good reason
Whether the test is one- or two-tailed – two-tailed tests split
α into two tails of p< .025, instead of one tail at p < .05
Hypothesis Testing w/ One
Sample

Example #1:

You’ve administered a therapy for people with anorexia that
will supposedly assist them in gaining weight. The following
data are amount of weight gained in pounds over your 16
session therapy for 29 participants. Does this represent a
significantly increased degree of weight gain compared to
the average weight gained without treatment (-.45 lbs.)?



What are Ho and H1?
Will you be using a one- or two-tailed test?
Why?
Based on this, what is your df? What is your
critical t?
Hypothesis Testing w/ One
Sample

Example #1:
1.7
-9.1
.7
2.1
-.1
-1.4
-.7
1.4
-3.5
-.3
14.9
-3.7
3.5
-.8
17.1
2.4
-7.6
12.6
1.6
1.9
11.7
3.9
6.1
.1
1.1
15.4
-4.0
-.7
20.9
Hypothesis Testing w/ One
Sample

Example #1:

Critical t (df = 28; two-tailed test at .05) = 2.048
Sample Mean = 87.2/29 = 3.0069
s2 = (1757.8 – [(87.2)2/29])/ 28 = 53.41
s = 7.3085
t = (3.0069 - -.45)/(7.3085/√29) = 2.5472, p < .05

t > Critical t and in our rejection region, which is above the
population mean (since we’re only interested in people
gaining weight), therefore we reject Ho and conclude that
our treatment is more effective than no treatment at all
Hypothesis Testing w/ One
Sample

Example #2:

A family therapist states that parents talk to their
teenagers an average of 27 minutes per week.
Surprised by that claim, a psychologist decided to
collect some data on the amount of time parents
spend in conversation with their teenage children.
For the n = 12 parents, the study revealed that
following times (in minutes) devoted to
conversation in a week:
Hypothesis Testing w/ One
Sample

Example #2:

Do the psychologists findings differ significantly from the
therapist’s claim? If so, is the family expert’s claim an
overestimate or underestimate of actual time spent talking
to children? Use the .05 level of significance with two tails.
29
22
19
25
27
28
21
22
24
26
30
22
Hypothesis Testing w/ One
Sample

Example #2:




H0 = μ = 27
Critical t (df = 11) = ±2.201
Mean = 24.58, s2 = 12.24, (s/√n) = 1.00, and t(11)
= -2.41
Reject H0, and conclude that the data are
significantly different (less time) from the
therapist’s claim
Hypothesis Testing w/ One
Sample

Often, if we’re reporting the results of our
experiments to the public, or the results of an
assessment (psychological or otherwise) to a
client, we want to emphasize to them that our
measurements are made with error, or that
our samples include sampling error

We can do this by including intervals around the
scores we report, indicating that the “true” score
measured without error lies in this interval
Hypothesis Testing w/ One
Sample

This is what is known as a Confidence
Interval


In keeping with the p < .05 tradition, we are often
looking for the 95% confidence interval, or the
scores that 95% of our distribution lie, but we can
do this for any interval
They are calculated just like for z-scores, where
we plug the t values into the formula and work
backwards
Hypothesis Testing w/ One
Sample

For a 95% CI:


Given your df (we’ll assume df = 9 for this
example) and type of test (assume a two-tailed
test for now), look up your critical values of t from
Table E.6 (t = ± 2.262)
Plug into your formula with your Sample Mean
and s (which we’ll assume are 1.463 and .341,
respectively), and solve for μ
Hypothesis Testing w/ One
Sample

For a 95% CI:
± 2.262 = (1.463 – μ)/(.341/√10)
μ = ±2.262(.108) + 1.463 = ±.244 + 1.462
.244 + 1.463 = 1.707
-.244 + 1.463 = 1.219
 CI.95 = 1.219 ≤ μ ≤ 1.707
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