Example of a one-sample Z-test The previous lecture in P.docx

Example of a one-sample Z-test
The previous lecture in Powerpoint explained the general
procedure for conducting a hypothesis
test. Now let’s go through an example of a hypothesis test to
see how it actually works.
Say you’ve developed a new drug that you think might influence
people’s cognitive ability,
although you aren’t really sure what it will do. (Yes, this is a
really bad study!) You give the
drug to a sample of N = 49 people and then test their IQ. Say
their IQ turns out to be X = 106.
Based on decades of test development, we know that in the
general U.S. population, the mean IQ
question that we want to address
is whether the people who take the “IQ Pill” will have IQ scores
that are different from the
general population (suggesting that the pill had some effect), or
whether they basically look the

same as everybody else in the country (suggesting that the pill
didn’t do anything).
Let’s test this question by using the formal steps of hypothesis
testing:
1.Generate H0 and HA
2.Select statistical procedure
4.Calculate observed statistic for your data
5.Determine critical statistic
6.Compare (4) and (5)
7.If (4) exceeds (5), reject H0
8.Otherwise, fail to reject H0
1. Generate H0 and HA
So what are H0 and HA? We don’t know whether the drug will
make people’s IQs get higher or
lower, so we need to use a 2-tailed or non-directional
hypothesis test. Since the mean in the
general population is 100, we will test whether the mean in our
test group is equal to 100.

treatment group represents. This is a
theoretical population of people who have taken our IQ pill.
Obviously, this population doesn’t
exist in reality, because the only people who have ever taken
our pill are the 49 people who were
in our study. But IN THEORY, a whole lot of other people
could also take this drug, and IN
THEORY, they should respond to it in the same way that our 49
participants respond. So we are
making an inference from our 49 participants to that very large
group of people who in theory
could also have taken this drug.
At this point, I’m just going to tell you that the statistical
procedure that you will use is
something called the Z-test. This is a test that is appropriate for
testing whether one sample is
different from some specified value when the value of the

known. Over the next few days, we will be introduced to some
other statistical procedures that
are appropriate for different kinds of situations than this, and
you will need to choose which one
is the best to use.
should understand why a
probability of making a Type I error, or rejecting a null that
should not have been rejected. Most
meaning that there is a 5% chance
of making a Type I error. If this type of mistake has
particularly bad consequences in your study
(e.g., telling people that a very expensive medication is
effective when in fact it is not), then you
For this example, let’s be boring
and go wit

4.Calculate observed statistic for your data
Now we need to compute what is called a test statistic for our
data. For a z-test, the procedure is
to compare the mean of our sample to the mean that we would
have expected if the null
hypotheses was true, divided by the standard error of the mean.
The general formula looks like
this:
Zobt =
0
X
X
N
Using the numbers in our example, we get:
106 100 6

2.80
15 2.14
49
obt
Z
5.Determine critical statistic
Of course, we don’t know whether our test is significant until
we have something to compare our
obtained z to. So we need to compute the critical value of z.
Then, if our observed z exceeds the
critical value, we will be able to reject the null. How do we get
zcrit? Remember, we wanted to
-directional (because we
didn’t know whether the pill would
make people more intelligent or less intelligent). So we need to
find the z-score that will put 5
percent of the z-distribution in the tails beyond that score.
That z-

.025 .025
-1.96 +1.96
7.If (4) exceeds (5), reject H0
we conclude? 2.80 is larger than
+1.96, so we can reject the null. Looking at the picture again,
you can see that 2.80 is in the tail
beyond 1.96, so it is in the region of rejection.

Thus, our conclusion is that we can reject H0. But it is not
enough to stop there. Read any
psychology journal and you will find that they NEVER actually
say “we rejected the null
hypothesis.” Rejecting the null isn’t that interesting. What is
interesting is what you can
conclude based on your rejected null. In this case, our
conclusion is that there is evidence that
the “IQ Pill” that you developed has an effect on people’s IQ
scores.
Would it be appropriate to say that there is evidence that people
who took the IQ pill are
smarter? The group mean was 106, which is obviously higher
than the national average of 100,
right?
This is a matter of some disagreement among statisticians.
Technically, our null hypothesis was
set up so that we would have also been able to reject the null if
people had scored six points
LOWER on the IQ test. Say the sample had scored 94 on the IQ
test instead of 106. Then we
would get:

94 100 6
2.80
15 2.14
49
obt
Z
-2.80 is beyond –1.96, so we would also be able to reject the
null in this case. So with a non-
directional or two-tailed null hypotheses, you would be willing
to reject the null hypothesis in
either direction. So technically you should not interpret the
direction of the effect, just that there
was a difference.
However, in the real world, most researchers DO interpret the
direction of the effect even if they
used a two-tailed hypothesis. But if you really wanted to
predict a particular direction, you
should start things off a little bit differently…

.025 .025
-1.96 +1.96 2.80
Using a directional hypothesis test
What if we actually did think from the very beginning that
taking this drug would make people
smarter? In that case, we would have set up a directional null
hypothesis that would test the
theory that people who take the pill will have IQ scores that are
higher than 100. Our hypotheses
would look like this:
Notice that the null hypothesis is that the mean is “less than or
equal to” 100. This is because we
would only reject the null if people’s scores were GREATER
than 100. If it turned out that
people who took the pill got scores of exactly 100 or even lower
than 100, then clearly the drug
is not having the effect we expected it to, and we would not be
able to reject the null. We will
only be able to reject the null if it turns out that people’s IQ
scores after taking the drug are

GREATER than 100.
How else will this change our hypothesis test? Well, now we
will only have a region of rejection
in the upper tail of the distribution, rather than having regions
of rejection in both tails of the
distribution. That means that if we want to keep a = .05, we
will end up putting all of that .05
probability in the upper tail. What is the z-score that puts .05
in tail beyond it? That value is
actually not on your table, but it’s right between two values that
are on the table. The exact z-
score that will put .05 in the tail beyond is z = 1.645. So zcrit
-tailed, is 1.645.
Using a one-tailed test does not change the value of zobt, so
that will still be 2.80. So can we
reject the null? Sure. 2.80 is greater than 1.645, so we can
reject the null. This time, we are able
to conclude that taking the IQ pill is making people smarter.
Example of a non-significant result
What if the IQ pill had a very small effect in our sample? Say

that we set up the 2-tailed, non-
directional null hypothesis that we used for the first example,
except that this time the mean of
the sample was only X = 98. How will this change our test?
Well, we’re back to the original
hypotheses with:
But this time our zobt will also be affected:
98 100 2
.93
15 2.14
49
obt
Z
What can we conclude in this case? Our zobt does not exceed
zcrit , so we cannot reject the null.
The conclusion in this case is that there is no evidence that the
IQ pill has an effect on people’s
IQ scores.

Notice that this is not the same as saying that the IQ pill does
not affect people’s IQs. It is still
possible that it does affect IQ scores, but this particular study
did not provide sufficient evidence
to say that it does. It’s like when OJ Simpson was acquitted of
murdering his wife. We don’t
know for sure that he was innocent of the crime, but we do
know that the evidence that was
presented at the trial was not sufficient to convince the jury
beyond a reasonable doubt that he
actually committed the crime. There was not enough evidence
to prove he was guilty, so it was
concluded that he was not guilty. See how this is not the same
thing as proving that he was
innocent. Maybe OJ should have taken one of the IQ pills that
you developed!
One-sample t-test
In the example with the IQ pill, we used a z-test to determine
whether a group mean is

statistically different from a known population parameter.
With the IQ test, we knew that the standard deviation of the
know that IQ tests have been developed over many years. We
were able to use the z-test because
But most of the time, we are testing things for which we do not
know the standard deviation of
the population. If we don’t know the population standard
deviation, we must estimate it using a
sample. Remember the formula for the standard deviation when
we are estimating a population
value by using a sample:
2
( )
1
X
X X
s
N

The symbol for this is a lower-case s, to indicate that it’s an
estimate.
We had to divide by N-1 in the formula instead of N because if
we didn’t then our estimate
would be biased. If you use the formula where you just divide
by N instead of dividing my N-1,
on average the standard deviation that you compute will tend to
underestimate the true
population value. To adjust for this bias, we divide by N-1.
Doing this will make the standard
deviation estimate a little bigger, adjusting for the bias.
Notice that the bigger your N is, the less it matters that you
have to divide by N-1. The
difference between 5 and 5 - 1 = 4 is pretty big. But the
difference between 50000 and 50000 - 1
= 49999 is practically nothing at all. This is consistent with the
idea that bigger samples will
give us more accurate estimates of the true population standard

deviation.
Now if we want to do a null hypothesis test, the formal steps for
doing it will be the same as
before, except that the formula for the obtained test statistic
we will call this a t-statistic rather than a z-statistic. More on
that in a minute:
tobt = 0
X
X
s
N
Notice that in the formula on the right side, the standard error
of the estimate is still just X
s
N
.

We don’t need to divide by N-1 for this part, because we
already did that when we computed sx.
The null hypothesis test will proceed in the same way that we
did it before, except that now we
can no longer use the z-table to look up probabilities. When we
use a t-statistic instead of a z-
statistic, we must look up probabilities in a t-table, rather than
in a z-table.
Huh? Well, it turns out that the sampling distribution of the
mean is not quite normal when you
use an estimated standard deviation, particularly when your
estimate is based on a small sample.
The probabilities of the z-distribution (e.g., that .05 is beyond
1.645 in the upper tail) are not
quite accurate when you are using an estimated standard
deviation.
Instead of looking up zcrit using the z-table, we need to look up
our critical value using the t-
distribution. The t-distribution is pretty similar to the z-
distribution except that it is a little bit
flatter in the middle, and has more area in the tails.

Also, the t-distribution differs depending on how big your
sample is. More specifically, it varies
depending on how many degrees of freedom you have. For the
t-test of a mean, degrees of
freedom is N-1. Why? Because the t-test uses an estimated
standard deviation. And when you
estimate a standard deviation, you have to divide by N-1.
2
( )
1
X
X X
s
N
When we look up tcrit using the t-table, we will use a different
line on the table depending on how

many degrees of freedom we have. Let’s try an example:
A researcher wanted to know whether smoking cigarettes
reduces olfactory sensitivity (makes
your sense of smell worse). On a test of olfactory sensitivity,
the mean is known to be 18 where
higher scores mean better sensitivity, so the researcher wants to
see whether people who smoke
have olfactory sensitivity scores that are lower than 18. The
researcher collects data from a
sample of 30 smokers and finds that they have a mean score of
X =17.2 and a standard deviation
of sx =1.52. Let’s go through the formal steps of hypothesis
testing:
1.Generate H0 and HA
This is a one-tailed test where we think that scores will be
lower, so the hypotheses are:

this time, we need to use a one-sample t-test
4.Calculate observed Z or t for your data
17.2 18.0 .8
2.88
1.52 .27
30
obt
t
5.Determine critical Z or t
Now we need to break out the t-table. How many degrees of
freedom do we have? 30 – 1 = 29.
Since this is a one-tailed test, we look in the t-table column for
the one-tailed test. And since our
alternate hypothesis has a “less than” symbol, that means that
we will need to use a negative
-1.699.

7.If Zobt or tobt exceeds Zcrit or tcrit, reject H0
Our obtained t of –2.88 is farther in the tail of the distribution
than our critical t of –1.699, so we
can reject the null. Our conclusion is that smoking does reduce
olfactory sensitivity.
CMGT/400 v7
Security Risk Mitigation Plan Template
CMGT/400 v7
Page 2 of 2Security Risk Mitigation Plan Template
Instructions: Replace the information in brackets [ ] with
information relevant to your project.
A Risk Management Analyst identifies and analyzes potential
issues that could negatively impact a business in order to help
the business avoid or mitigate those risks.
Take on the role of Risk Management Analyst for the
organization you chose in Week 1. Research the following
information about your chosen organization. Create a Security
Risk Mitigation Plan using this template.[Organization Name]
Security Policies and Controls
[Response]

Password Policies
[Response]
Administrator Roles and Responsibilities
[Response]
User Roles and Responsibilities
[Response]
Authentic Strategy
[Response]
Intrusion Detection and Monitoring Strategy
[Response]
Virus Detection Strategies and Protection
[Response]
Auditing Policies and Procedures
[Response]
Education Plan
Develop an education plan for employees on security protocols
and appropriate use.
[Response]
Risk Response
Include: Avoidance, Transference, Mitigation, and Acceptance.
[Response]
Change Management/Version Control
[Response]
Acceptable Use of Organization Assets and Data
[Response]

Employee Policies
Explain the separations of duties and training.
[Response]
Incident Response
Document incident types and category definitions, roles and
responsibilities, reporting requirements and escalation, and
cyber-incident response teams.
[Response]
Incident Response Process
Discuss the incident response process including: preparation,
identification, containment, eradication, recovery, and lessons
learned.
[Response]
Copyright© 2018 by University of Phoenix. All rights reserved.
Copyright© 2018 by University of Phoenix. All rights reserved.
Probability and Decision Making
Most of the statistics that we have talked about so far involve
describing distributions of samples
or describing relationships within samples. These are examples
of descriptive statistics. But
most of the time, we are actually interested in using the data
from our sample to make an
inference to a population. In that case, we would be doing
inferential statistics.

Inferential statistics are based on the principles of probability.
In order to use a sample to make
an inference about a population, we need to consider the
probability of different events occurring
in the population based on what we observe in our sample.
Before we can do that, we should
start with a (very) brief review of some key terms in probability
The probability of an event occurring [p(A)] is equal to the
relative frequency of the event in the
long run. For example, pass completion average = # passes
completed divided by # passes
attempted. In the 2019 season, Russell Wilson attempted 516
passes, 341 of which were caught.
Thus, we predict that he has a 341/516 = .66 probability of
completing the next pass he throws.
The limits of probability are 0 to 1. Probability of an event
occurring plus probability of an
event not occurring equals 1.0. P(A) + P(not-A) = 1.0.
The rules of probability only apply to random events.
Remember that a random sample
(sometimes called a “probability sample”) requires that all

elements or individuals within the
population have an equal probability of being selected for the
sample.
Probability Distributions
Empirical probability distribution = measured probability. This
is a distribution based on
observation of actual events. Pass completion average is an
example of this. Another example is
in Consumer Reports magazine where they report repair rates of
various automobiles. Car
models with lower repair rates are expected to be less likely to
need repairs in the future.
Theoretical probability distribution = based on theory. This is a
distribution based on
assumptions about the probability of events occurring. It is
NOT based on guessing! Theoretical
probability distributions can be created for events for which we
have very accurate knowledge
about the probability of certain events occurring. For example,
you know that a fair coin has .5
probability of coming up heads. You don’t need to toss a coin a
thousand times to figure this

out, you just compute it by this formula:
# of outcomes that satisfy the event
P(event) =
# of possible outcomes
So for the coin example,
1
.5
2
heads
heads tails
P(rolling a six on a 6-sided die) =
1
.17
6

Independent events are when the occurrence of one event does
not influence the probability of
another event. Examples of this are coin tosses, dice rolls, and
slot machines. (Mistaken beliefs
about “hot dice” or someone being “due” for a jackpot on a
machine that hasn’t paid out in a
while are referred to as the Gambler’s Fallacy.)
Dependent events are when the occurrence of one event does
influence the probability of another
event. Card games like Blackjack and poker are based on
dependent events, because once
certain cards have been dealt, they cannot be dealt again in that
cycle.
Sampling with replacement is when the selected sample is
returned to the population before the
next sample is drawn.
Sampling without replacement is when the selected sample is
not returned to the population
before subsequent samples are drawn. The probability of events
occurring changes with the new
samples. For example, say you have a raffle with three prizes.
First prize winner is not eligible

for the other two prizes, second winner is not eligible for the
third prize.
If they sell 100 tickets:
P(1st prize) = 1/100
P(2nd prize) = 1/99
P(3rd prize) = 1/98
Probability and the Standard Normal Curve
The standard normal curve is a theoretical probability
distribution. It specifies the theoretical
probability of having certain values within a distribution. We
can use the normal curve to
determine probabilities associated with events that are
approximately normally distributed.
Say IQ scores are normally distributed with a mean of 100 and a
standard deviation of 15. (this
is pretty much true.)
What’s the probability of drawing one person at random from
the population who has an IQ of at

least 110 (or higher)? To answer this, we need to compute a z-
score for that person:
110 100
.67
15
X
X
X
Z
Looking in the Z-table, we see that the probability of having a
Z-score of .67 or greater (area in
the tail above) is p = .2514. So there is about a 25% chance of
randomly grabbing a person with
an IQ of 110 or greater from the population with a mean of 100
and standard deviation of 15.
What we just did refers to determining probability of single

observations. But often we want to
know probabilities associated with means.
What’s the probability of drawing a random sample of 16 people
who have a mean IQ of at least
110? You might guess that this probability will be smaller than
the probability of getting just
one person with an IQ of at least 110. Of course, some of these
16 people could have IQs of less
than 110, but then some would need to have IQs of greater than
110 to balance it out, so that the
group mean is at least 110.
In order to answer this question, you need to remember what we
talked about back before Test 1,
sampling distributions.
A sampling distribution of the mean is a theoretical distribution.
It is based on what the
distribution of means would look like if you took an infinite
number of samples of size N from
the population. We never actually bother to do that (who has

time?) but we know that if we did,
in theory, the distribution would have some specific properties.
Remember that a sampling distribution has a Mean and
variability. The Mean of the sampling
distribution is equal to
distribution is smaller than the
variability of the population.
The variability of the sampling distribution is called the
Standard Error. In this case, because we
are calculating estimates of the Mean, the variability is the
Standard Error of the Mean.
Note that this is different than the Standard Deviation of the
sample or the population. (It is also
different than the Standard error of the estimate that we learned
about with regression. I wish
these terms didn’t all sound so similar. But the difference
between them all is very important, so
pay attention and be careful!).
Standard Error of the Mean is the variability of the sampling
distribution of Means. It is the

standard deviation of the sampling distribution of means.
Because of the Central Limit Theorem, we know the following
things:
Any given sample mean will
2) If you took an infinite number of samples of size N, the
standard error of the mean (i.e.,
the standard Deviation of the sampling distribution of means)
would be:
X
X
N
What that means is that the variability of the sampling
distribution is smaller when your
samples are bigger. Bigger samples mean you are more likely
to get a good (accurate)

estimate of the true population Mean.
3) As the size of the sample increases, the shape of the sampling
distribution of the mean
will approach normal. What’s really amazing is that this is true
even if the shape of the
original distribution is not normal.
Because of the central limit theorem, we can use a single
sample of size N to estimate properties
of the sampling distribution (rather than actually needing to
take an infinite number of samples).
OK, so now that you know about the sampling distribution of
means, we can get back to our
earlier question about how to determine the probability of
getting a sample of N = 16 people who
have an average IQ of at least 110.
We’re going to create a Z-score like we did before, except that
now we will create a Z-score
and N = 10. Instead of

comparing one person’s score (X) to the sample mean, we will
be comparing the sample mean to
the mean of the population. Instead of dividing by the standard
deviation of X, we will divide by
the standard error of the estimate. The general formula for the
Z-score will be:
X
X
X
Z
X
X
N

For our example,
110 100 10
2.67
15 3.75
16
Z
What’s the probability of getting a Z-score of 2.67 or greater?
Look in the table, p = .0038. It is
a LOT less likely that we would get a sample of 16 people with
a mean IQ of 110 than it is that
we would get one single person with an IQ of 110 (p = .2514).
Deciding whether a sample represents a population
So, if the probability of getting 16 people with a mean IQ of at
least 110 just by chance is only
.0038, that might make you start thinking that maybe there is
something other than just chance
operating here. Maybe those 16 people weren’t actually
randomly sampled from a population

with a mean of 100 and a standard deviation of 15. Maybe
those 16 people do not represent the
general population of people in the U.S. Maybe they actually
represent some other population,
such a population of college students who have higher than
average IQs.
We get suspicious about the representativeness of the sample
because the probability of
obtaining a sample with those characteristics (mean IQ of 110)
is very unlikely if those people
were really just randomly sampled from the general population
what exactly do we mean by very unlikely? How unlikely does
an event have to be before we
start getting suspicious?
Suppose that the sample of 16 people we drew only had a mean
IQ of 101. Would we be
suspicious that they were not really representative of the
general population? Let’s see:

101 100 1
.27
15 3.75
16
Z
.3936.
If there’s a 39% chance that we could have found 16 people
with a mean IQ of 101, it doesn’t
seem so strange that it could have happened just by chance.
So what probability do we want to use as our cutoff for “too
unlikely?” Well, the conventional
level in social sciences is usually p = .05 or less (that is, p <
.05).
Often, it is easier to think about this in the opposite way, by
asking what is the Z-score that will
give us exactly 5% probability in the tails of the distribution?
This is referred to as the critical
value.

One important question is whether we just want our 5% to be in
one tail of the distribution or
whether we want it to be divided up between the top and bottom
end of the distribution. In this
case, we probably would have been just as surprised if we
randomly sampled a group of 16
people and found that they had an average IQ of 90 (10 points
below the mean instead of 10
points above the mean). So we would think things are very
unlikely if they happened to be much
higher than average as well as much lower than average. So we
will split our 5% up between the
two tails of the distribution.
If we have 5% split up between the two ends of the distribution,
that means we have 2.5% or
.025 in each tail of the distribution. Let’s find the critical
value. What is the Z-score that has

–1.96
or greater than +1.96) as the
region of rejection. If we end up getting a sample that gives us
a Z-score that is in the region of
rejection, we will conclude that it is too unlikely that this could
have happened just by random
chance. There must be something else going on, such as the
group we sampled not actually
being representative of the general population.
And this leads us right into the topic of hypothesis testing,
which we’ll discuss next.
.025 .025
-1.96 +1.96
1
Null
Hypothesis
Significance

Testing
Hypothesis Testing
de if an observed result is unlikely to have
occurred by chance
procedure
not random
eally p < .05!
Hypotheses
under investigation.
Hypothesis (H1)
Null Hypothesis (H0)
value

Alternative Hypothesis (H1)
he population parameter is some alternative
range of values
hypothesis
Examples of Null and Alternate
Hypotheses
did have an effect)
2
Nondirectional Hypothesis
-tailed

ed if test statistic is much higher OR
lower than prediction
Directional Hypothesis
-tailed
is much higher than prediction
y if willing to ignore an extreme value in opposite
direction from what is expected
How to test a null Hypothesis
distribution
sample could have
been drawn if H0 was actually true?
Declaring Statistical Significance
just by chance if H0 was actually true…
– you REJECT
the null
Statistical Significance

to chance if the null were true
Practical Significance
meaning
ant may not be practically meaningful
(and vice-versa)
3
Test Statistic
from the parameter specified in H0
Critical Value
the desired alpha
level (region of rejection)
Hypothesis Testing: Formal Steps
1. Generate H0 and H1
2. Select statistical procedure (z, t, etc)
3. Select a

4. Calculate observed statistic for your data
5. Determine critical value
6. Compare (4) and (5)
7. If (4) exceeds (5), reject H0
8. Otherwise, fail to reject H0
Truth of the Universe
H0 True (no effect) H0 False(effect exists)
Do not reject H0
(say there is no
effect)
Type I error
a
Correct
Decision
1 - b
Type II error
b
Correct
Decision
1 - a
Your Decision

Reject H0
(say the
effect exists)
universe
decision probabilities
sum to 1.0
An analogy with the legal system
proven guilty”
crime”
reject the null = “not guilty”
4
What really happened
H0 True (Didn’t do it) H0 False (did it)
Do not reject H0

(Not guilty)
Innocent person is
convicted
Type I error
Guilty person is
convicted
Correct decision
Guilty person
gets off
Type II error
Innocent person
goes home
Correct decision
Jury Decision
Reject H0
(guilty verdict)
Psychology 302, Winter 2020
Correlational Approaches to Research
Problem Set 4, due Wednesday, March 4th in class

1. For each of the following, determine whether the decision
reached by the
researcher in the first sentence is correct, given the information
in the
subsequent sentences. If the decision is incorrect, indicate what
type of
error was made.
a. Based on an initial test, a medical researcher concluded that
Serum A was not effective for treating a disease. However, 25
years later, many subsequent studies have found that Serum A
is effective, and it is now used regularly to treat the disease.
b. A researcher who studied literacy concluded that children
who
were raised by parents who read to them regularly learned to
read earlier than children whose parents did not read to them.
This finding has been consistently demonstrated in subsequent
studies over many years.
c. Researchers originally claimed that students who were
homeschooled performed worse in college compared to students
with public education. However, over many years, studies have
subsequently shown that home schooled children perform
equally well in college as public school students.
d. A researcher found that adults who followed a low fat diet
did
not lose any more weight compared to adults who were not
dieting. Several years of subsequent research have shown that

low fat diets are not effective compared to not dieting.
For problems 2-4, use the logic of hypothesis testing to answer
the research
question posed. Be sure to go through each of the formal steps.
Clearly
state your null and alternate hypotheses, your obtained and
critical
statistics, and whether you can reject the null. Be sure to
clearly state your
conclusions in words.
2. A researcher is testing the effectiveness of a new drug that is
intended to
improve learning and memory performance. A random sample
of 16 rats
are given the drug and then tested on a standard learning task.
The
mean of the sample is 56.5. In the general population of rats
(with no
drug), the average score on the standardized test is normally
distributed
Is there

3. You notice that a lot of students listen to music while
studying at the
library, and you suspect that this may be detrimental to their
learning.
You take a random sample of Intro Psychology students who
listen to
music while studying and you measure their scores on the Intro
Psych
final exam. In the population of ALL intro psych students, the
final exam
deviation
-listening students, the
mean
was X = 76.15. Is there evidence to conclude that listening to
music is
detrime
4. A common reading achievement test for fifth grade students
has a
students

who receive a new type of reading skills training are
significantly different
from the national average, but she doesn’t know whether they’re
likely to
be better or worse than average. She trains her class of N = 31
students
with the new technique and then gives them the test. The class
average
is X = 72.9 with a standard deviation of Sx= 7.5.
a. Using the formal logic of null hypothesis testing, test whether
Explain
why you reached a different conclusion depending on what
alpha
level you used.

Example of a one-sample Z-test The previous lecture in P.docx

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