35845 Topic Group AssignmentNumber of Pages 1 (Double Spaced.docx

35845 Topic: Group Assignment
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Author: Jackson, S. L. (2017). Statistics plain and simple, (4th
ed.). Boston, MA: Cengage Learning.
Basic Probability Concepts
The Rules of Probability
Probability and the Standard Normal Distribution

Review of Key Terms
Module Exercises
Critical Thinking Check Answers
Module 8: Hypothesis Testing and Inferential Statistics
Null and Alternative Hypotheses
Two-Tailed and One-Tailed Hypothesis Tests
Type I and Type II Errors in Hypothesis Testing
Probability, Statistical Significance, and Errors
Using Inferential Statistics
Review of Key Terms
Module Exercises
Critical Thinking Check Answers
Chapter 4 Summary and Review
In this chapter you will be introduced to the concepts of
probability and hypothesis testing. Probability is the study of
likelihood and uncertainty. Most decisions that we make are
probabilistic in nature. Thus, probability plays a critical role in
most professions and in our everyday decisions. We will discuss
basic probability concepts along with how to compute
probabilities and the use of the standard normal curve in making
probabilistic decisions.
probability The study of likelihood and uncertainty; the number
of ways a particular outcome can occur, divided by the total
number of outcomes.
Hypothesis testing is the process of determining whether a
hypothesis is supported by the results of a research project. Our
introduction to hypothesis testing will include a discussion of
the null and alternative hypotheses, Type I and Type II errors,
and one- and two-tailed tests of hypotheses as well as an
introduction to statistical significance and probability as they
relate to inferential statistics.
hypothesis testing The process of determining whether a
hypothesis is supported by the results of a research study.
MODULE 7

Probability
Learning Objectives
•Understand how probability is used in everyday life.
•Know how to compute a probability.
•Understand and be able to apply the multiplication rule.
•Understand and be able to apply the addition rule.
•Understand the relationship between the standard normal curve
and probability.
In order to better understand the nature of probabilistic
decisions, consider the following court case of The People v.
Collins, 1968. In this case, the robbery victim was unable to
identify his assailant. All that the victim could recall was that
the assailant was female with a blonde pony tail. In addition, he
remembered that she fled the scene in a yellow convertible that
was driven by an African American male who had a full beard.
The suspect in the case fit the description given by the victim,
so the question was “Could the jury be sure, beyond a
reasonable doubt, that the woman on trial was the robber?” The
evidence against her was as follows: She was blonde and often
wore her hair in a pony tail; her codefendant friend was an
African American male with a moustache, beard, and a yellow
convertible. The attorney for the defense stressed the fact that
the victim could not identify this woman as the woman who
robbed him, and that therefore there should be reasonable doubt
on the part of the jury.
The prosecutor, on the other hand, called an expert in
probability theory who testified to the following: The
probability of all of the above conditions (being
blonde and often having a pony tail and having an African
American male friend and his having a full beard, and his
owning a yellow convertible) co-occurring when these
characteristics are independent was 1 in 12 million. The expert
further testified that the combination of characteristics was so
unusual that the jury could in fact be certain “beyond a
reasonable doubt” that the woman was the robber. The jury
returned a verdict of “guilty” (Arkes & Hammond, 1986;

Halpern, 1996).
As can be seen in the previous example, the legal system
operates on probability and recognizes that we can never be
absolutely certain when deciding whether an individual is
guilty. Thus, the standard of “beyond a reasonable doubt” was
established and jurors base their decisions on probability,
whether they realize it or not. Most decisions that we make on a
daily basis are, in fact, based on probabilities. Diagnoses made
by doctors, verdicts produced by juries, decisions made by
business executives regarding expansion and what products to
carry, decisions regarding whether individuals are admitted to
colleges, and most everyday decisions all involve using
probability. In addition, all games of chance (for example,
cards, horse racing, the stock market) involve probability.
If you think about it, there is very little in life that is certain.
Therefore, most of our decisions are probabilistic and having a
better understanding of probability will help you with those
decisions. In addition, because probability also plays an
important role in science, that is another important reason for us
to have an understanding of it. As we will see in later modules,
the laws of probability are critical in the interpretation of
research findings.
Basic Probability Concepts
Probability refers to the number of ways a particular outcome
(event) can occur divided by the total number of outcomes
(events). (Please note that the words outcome and event will be
used interchangeably in this module.) Probabilities are often
presented or expressed as proportions. Proportions vary between
0.0 and 1.0, where a probability of 0.0 means the event certainly
will not occur and a probability of 1.0 means that the event is
certain to occur. Thus, any probability between 0.0 and 1.0
represents an event with some degree of uncertainty to it. How
much uncertainty depends on the exact probability with which
we are dealing. For example, a probability close to 0.0
represents an event that is almost certain not to occur, and a

probability close to 1.0 represents an event that is almost
certain to occur. On the other hand, a probability of .50
represents maximum uncertainty. In addition, keep in mind that
probabilities tell us about the likelihood of events in the long
run, not the short run.
Let's start with a simplistic example of probability. What is the
probability of getting a “head” when tossing a coin? In this
example, we have to consider how many ways there are to get a
“head” on a coin toss (there is only one way, the coin lands
heads up) and how many possible outcomes there are (there are
two possible outcomes, either a “head” or a “tail”). So, the
probability of a “head” in a coin toss is:
p(head)=NumberofwaystogetaheadNumberofpossibleoutcomes=
12=.50p(head)=Number of ways to get a headNumber of possible
outcomes=12=.50
This means that in the long run, we can expect a coin to land
heads up 50% of the time.
Let's consider some other examples. How likely would it be for
an individual to roll a 2 in one roll of a die? Once again, let's
put this into basic probability terms. There is only one way to
roll a 2, the die lands with the 2 side up. How many possible
outcomes are there in a single roll of a die? There are six
possible outcomes (any number between 1 and 6 could appear
on the die). Hence, the probability of rolling a 2 on a single roll
of a die would be 1/6, or about .17. Representing this in a
formula as we did for the previous example:
p(2)=Numberofwaystogeta2Numberofpossibleoutcomes=16=.17
p(2)=Number of ways to get a 2Number of possible outcomes=16
=.17
Let's make it a little more difficult. What is the probability of
rolling an odd number in a single roll of a die? Well, there are
three odd numbers on any single die (1, 3, and 5). Thus, there
are three ways that an odd number can occur. Once again, how
many possible outcomes are there in a single roll of a die? Six
(any number between 1 and 6). Therefore, the probability of
rolling an odd number on a single roll is 3/6, or .50.

Represented as a formula this would be:
p(oddnumber)=NumberofwaystogetanoddnumberNumberofpossi
bleoutcomes=36=.50p(odd number)=Number of ways to get an o
dd numberNumber of possible outcomes=36=.50
What if I asked you what the probability of rolling a single-
digit number is in a single roll of a die? A die has six numbers
on it, and each is a single-digit number. Thus, there are six
ways to get a single-digit number. How many possible outcomes
are there in a single roll of a die? Once again, six. Hence, the
probability of rolling a single-digit number is 6/6, or 1.0. If
someone asked you to place a bet on this occurring, you could
not lose on this bet! Once again, as a formula this would be:
p(single-digitnumber)=Numberofwaystogetasingle-
digitnumberNumberofpossibleoutcomse=66=1.0p(single-
digit number)=Number of ways to get a single-
digit numberNumber of possible outcomse=66=1.0
Now that you have a basic idea of where probabilities come
from, let's talk a little bit more about how we use probabilities.
Keep in mind that probabilities tell us something about what
will happen in the long run. Therefore, when we think about
using some of the probabilities that we just calculated, we have
to think about using them in the long run. For example, we
determined that the probability of rolling a 2 on a single roll of
a die was .17. This means that over many rolls of the die, it will
land with the 2 side up about 17% of the time. We cannot
predict what will happen on any single roll of the die, but over
many rolls of the die, we will roll a 2 with a probability of .17.
This means that with a very large number of trials, we can
predict with great accuracy what proportion of the rolls will end
up as 2s. However, we cannot predict which particular rolls will
yield a 2. So when we think about using probabilities, we need
to think about using them for predictions in the long run, not
the short run.
1.What is the probability of pulling a king from a standard (52-
card) deck of playing cards?

2.What is the probability of pulling a spade from a standard
deck of playing cards?
3.What is the probability of rolling an even number on a single
roll of a die?
4.Imagine that you have a bag that contains 4 black poker chips
and 7 red poker chips. What is the probability of pulling a black
poker chip from the bag?
The Rules of Probability
Often we are concerned with the probability of two or more
events occurring and not just the probability of a single event
occurring. For example, what is the probability of rolling at
least one 4 in two rolls of a die, or what is the probability of
getting two tails in two flips of a coin?
FIGURE 7.1 Tree diagram of possible coin toss outcomes
Let's use the coin-toss example to determine the probability of
two tails occurring in two flips of a coin. Based on what we
discussed in the previous section, we know that the probability
of a tail on one flip of a coin is 1/2, or 50. The same is true for
the second toss, the probability of a tail is 1/2, or .50. However,
let's think about the possible outcomes for two tosses of a coin.
One outcome is a head on the first toss and a head on the second
toss (HH). The other outcomes would be a head followed by a
tail (HT), a tail followed by a head (TH), and a tail followed by
a tail (TT). These four possible outcomes are illustrated in the
tree diagram in Figure 7.1.
Notice that the probability of two tails or any one of the other
three possible outcomes is 1/4, or .25. But how are these
probabilities calculated? The general rule that we apply here is
known as the multiplication rule, or the and rule. When the
events are independent and we want to know the probability of
one event “and” another event, we use this rule.
The multiplication rule says that the probability of a series of
outcomes occurring on successive trials is the product of their
individual probabilities, when the events are independent (do
not impact one another). Thus, when using the multiplication
rule, we multiply the probability of the first event by the

probability of the second event. Therefore, for the present
problem, the probability of a tail in the first toss is 1/2, or
.50, and the probability of a tail in the second toss is 1/2, or .50.
When we multiply these two probabilities, we have .50 × .50 =
.25. This should make some sense to you because the
probability of both events occurring should be less than that of
either event alone. We can represent the problem as follows:
multiplication rule A probability rule stating that the probability
of a series of outcomes occurring on successive trials is the
product of their individual probabilities, when the sequence of
outcomes is independent.
p(tailonfirsttossandtailonsecondtoss)=p(tailonfirsttoss)×p(tailon
secondtoss)p(tail on first toss and tail on second toss)=p(tail on f
irst toss) ×p(tail on second toss)
Let's try another example. Assuming that the probabilities of
having a girl and having a boy are both .50 for single-child
births, what is the probability that a couple planning a family of
three children would have the children in the following order:
girl, girl, boy?
You can see in the tree diagram in Figure 7.2 that the
probability of girl, girl, boy is .125. Let's use the and rule to
double-check this probability. The probability of a girl as the
first child is 1/2, or .50. The same is true for the probability of
a girl as the second child (.50) and the probability of a boy as
the third child (.50). In order to determine the probability of
this sequence of births, we multiply: .50 × .50 × .50 = .125.
FIGURE 7.2 Tree diagram of possible birth orders
In addition to being able to calculate probabilities based on a
series of independent events (as in the preceding examples), we
can also calculate the probability of one event or another event
occurring on a single trial when the events are mutually
exclusive. Mutually exclusive means that only one of the events
can occur on a single trial. For example, a coin toss can either
be heads or tails on a given trial, but not both. When dealing
with mutually exclusive events, we apply what is known as
the addition rule, which states that the probability of one

outcome or the other outcome occurring on a particular trial is
the sum of their individual probabilities. In other words, we
are adding the two probabilities together. Thus, the probability
of having either a girl or a boy when giving birth would be:
addition rule A probability rule stating that the probability of
one outcome or another outcome occurring on a particular trial
is the sum of their individual probabilities, when the outcomes
are mutually exclusive.
p(girlorboy)=p(girl)+p(boy)=.50+.50=1.00p(girl or boy) = p(girl
) + p(boy) = .50 + .50=1.00
This is sometimes referred to as the or rule because we are
determining the probability of one event or the other event.
Let's try another problem using the or rule. What is the
probability of drawing either a club or a heart when drawing
one card from a deck of cards? The probability of drawing a
club is 13/52, or .25. The same holds for drawing a heart (13/52
= .25). Thus, the probability of drawing either a club or a heart
card on a single draw would be .25 + .25 = .50.
p(cluborheart)=p(club)+p(heart)=.25+.25=.50p(club or heart) =
p(club) + p(heart) = .25 + .25 = .50
THE RULES OF PROBABILITY
Rule
Explanation
Example
The Multiplication Rule
The probability of a series of independent outcomes occurring
on successive trials is the product of their individual
probabilities. This is also known as the and rule because we
want to know the probability of one event and another event.
In order to determine the probability of one coin toss of a head
followed by (and) another coin toss of a head, we multiply the
probability of each individual event: 50 × .50 = .25
The Addition Rule
The probability of one outcome or another outcome occurring
on a particular trial is the sum of their individual probabilities

when the two outcomes are mutually exclusive. This is also
known as the or rule because we want to know the probability of
one event or another event.
In order to determine the probability of tossing a head or a tail
on a single coin toss, we sum the probability of each individual
event: 50 +.50 = 1.0
1.Which rule, the multiplication rule or the addition rule, would
be applied in each of the following situations?
a.What is the probability of a couple having a girl as their first
child followed by a boy as their second child?
b.What is the probability of pulling a spade or a diamond from a
standard deck of cards on a single trial?
c.What is the probability of pulling a spade (and then putting it
back in the deck) followed by pulling a diamond from a
standard deck of cards?
d.What is the probability of pulling a jack or a queen from a
standard deck of cards on a single trial?
2.Determine the probability for each of the examples in exercise
1.
Probability and the Standard Normal distribution
As you might remember from Chapter 3, z scores can be used to
determine proportions under the standard normal curve. In that
chapter, we used z scores and the area under the standard
normal curve to determine percentile ranks. We will now
use z scores and the area under the standard normal curve
(Table A.1) to determine probabilities. As you might remember,
the standard normal curve has a mean of 0 and a standard
deviation of 1. In addition, as discussed in Chapter 3, the
standard normal curve is symmetrical and bell-shaped and the
mean, median, and mode are all the same. Take a look at Figure
7.3, which represents the area under the standard normal curve
in terms of standard deviations. We looked at this figure in the
previous module (Figure 6.1), and based on this figure, we see
that approximately 68% of the observations in the distribution
fall between −1.0 and + 1.0 standard deviations from the mean.

This approximate percentage holds for all data that are normally
distributed. Notice also that approximately 13.5% of the
observations fall between −1.0 and −2.0 and another 13.5%
between +1.0 and +2.0, and that approximately 2% of the
observations fall between − 2.0 and − 3.0 and another 2%
between + 2.0 and + 3.0. Only .13% of the scores are beyond
a z score of either ±3.0. If you sum the percentages in Figure
7.3, you will have 100%—all of the area under the curve,
representing everybody in the distribution. If you sum half of
the curve, you will have 50%—half of the distribution.
FIGURE 7.3 area under the standard normal curve
We can use the areas under the standard normal curve to
determine the probability that an observation falls within a
certain area under the curve. Let's use a distribution that is
normal to illustrate what we mean. Intelligence test scores are
normally distributed with a mean of 100 and a standard
deviation of 15. We could use the standard normal curve to
determine the probability of randomly selecting someone from
the population who had an intelligence score as high or higher
than a certain amount. For example, if a school psychologist
wanted to know the probability of selecting a student from the
general population who had an intelligence test score of 119 or
higher, we could use the area under the standard normal curve
to determine this. First we have to convert the intelligence test
score to a z score. As you might remember, the formula for
a z score is:
z=X−μσz=X−μσ
where X represents the individual's score on the intelligence
test, Âµ represents the population mean, and σ represents the
population standard deviation. Using this formula, we can
calculate the individual's z score as follows:
z=X−μσ=119−10015=1915=+1.27z=X−μσ=119−10015=1915=+1
.27
FIGURE 7.4 Standard normal curve with z = +1.27 indicated
Thus, we know that this individual's z score falls +1.27 standard
deviations above the mean. As in Chapter 3, it is helpful to

represent this on a figure where the z score of + 1.27 is
indicated. This is illustrated in Figure 7.4.
Now, in order to determine the probability of selecting a student
with an intelligence test score of 119 or higher, we need to turn
to Table A.1 in Appendix A. We begin by looking up a z score
of 1.27 and find that for this score, a proportion of .39797 of
the scores fall between the score and the mean of the
distribution and a proportion of .10203 of the scores fall beyond
the score. Referring to Figure 7.4, we see that the proportion of
the curve in which we are interested is the area beyond the
score, or .10203. This means that the probability of randomly
selecting a student with an intelligence test score of 119 or
higher is .10203, or just slightly higher than 10%. We can
represent this problem in standard probability format as
follows:
p(X≥119)=.10203p(X≥119)=.10203
Let's try a couple more probability problems using the
intelligence test score distribution. First, what is the probability
of the school psychologist randomly selecting a student with an
intelligence test score of 85 or higher? Secondly, what is the
probability of the school psychologist selecting a student with
an intelligence test score of 70 or lower?
Let's begin with the first problem. We need to convert the
intelligence test score to a z score and then consult Table A.1.
z=X−μσ=85−10015=−1515=−1.0z=X−μσ=85−10015=−1515=−1.
0
When we consult Table A.1, we find that for a z score of −1.0,
15866 of the scores fall below this score and .34134 of the
scores fall between this score and the mean of the distribution.
This z score is illustrated in Figure 7.5 along with the area in
which we are interested—the probability of a student with an
intelligence test score of 85 or higher being selected.
FIGURE 7.5 Standard normal curve with z = −1.0 indicated
In order to determine the probability of selecting a student with
an intelligence test score this high or higher, we take the area
between the mean and the z score (.34134) and add the .50 from

the other half of the distribution to it. Hence, the probability of
selecting a student with an intelligence test score of 85 or
higher is .84134, or approximately 84%. You should see that the
probability of this happening is fairly high because when we
look at Figure 7.5 we are talking about a large proportion of
people who fit this description. This can be represented as
follows:
P(X≥85)=.84134P(X≥85)=.84134
Let's work the second problem, the probability of selecting a
student with an intelligence test score of 70 or lower. Once
again we begin by converting this score into a z score.
z=X−μσ=70−10015=−3015=−2.0z=X−μσ=70−10015=−3015=−2.
0
Next, we represent this on a figure with the z score indicated
along with the area in which we are interested (anyone with this
score or a lower score). This is illustrated in Figure 7.6.
Consulting Table A.1, we find that for a z score of −2.0, .02275
of the scores are below the score (beyond it) and .47725 of the
scores are between the score and the mean of the distribution.
We are interested in the probability of selecting a student with
an intelligence test score of 70 or lower. Can you figure out
what that would be? If you answered .02275, you are correct.
Therefore, there is slightly more than a 2% chance of selecting
a student with an intelligence test score this low or lower—a
fairly low probability event. This can be represented as follows:
p(X≤70)=.02275p(X≤70)=.02275
FIGURE 7.6 Standard normal curve with z = −2.0 indicated
Let's apply what we have learned about using the standard
normal curve to calculate probabilities together with the
addition rule from earlier in the module to determine the
probability of selecting a child whose intelligence test score is
70 or lower or 119 or higher. We have already determined
the z scores for each of these intelligence test scores in our
previous problems. The intelligence test score of 70 converts to
a z score of −2.0 and the intelligence test score of 119 converts
to a z score of + 1.27. Moreover, we have already determined

that the probability of selecting a student with a score of 70 or
lower is .02275 and that the probability of selecting a student
with an intelligence test score of 119 or higher is .10203. Thus,
applying the addition rule, the probability of selecting a student
with a score that is 70 or lower or 119 or higher would be the
sum of these two probabilities, or .02275 + .10203. These two
probabilities sum to .12478, or just about 12.5%. This can be
represented as follows:
=p(X≤70orX≥119)=p(X≤70)+p(X≥119)=p(X≤70orX≥119)=(.022
75)+(.10203)=p(X≤70orX≥119)=(.02275)+(.10203)=p(X≤70 or
X≥119)= p(X≤70)+p(X≥119)=p(X≤70 or X≥119)= (.02275)+(.10
203)=p(X≤70 or X≥119)=(.02275)+(.10203)
Now let's turn to using the multiplication rule, discussed earlier
in the module, with the area under the standard normal curve
(Table A.1). In this case, we want to determine the probability
of selecting two students who fit different descriptions. For
example, what is the probability of selecting one student with
an intelligence test score equal to or below 80, followed by
another student with an intelligence test score equal to or above
125? Once again, we begin by converting the scores to z scores.
z=X−μσ=80−10015=−2015=−1.33z=X−μσ=80−10015=−2015=−
1.33
z=X−μσ=125−10015=2515=+1.67z=X−μσ=125−10015=2515=+1
.67
Consequently, the intelligence test scores convert to z scores of
−1.33 and +1.67, respectively. Next we use Table A.1 to
determine the probability of each of these events.
Consulting Table A.1, we find that the probability of selecting a
student with a score of 80 or lower is .09175 and the probability
of selecting a student with a score of 125 or higher is .04745.
These z scores and proportions are represented in Figure 7.7.
We now apply the multiplication rule to determine the
probability of selecting the first person followed by the second
person. Thus, we multiply the first probability by the second
probability, or .09175 × .04745 = .00435. This can be
represented as follows:

=p(X≤80andX≥125)=p(X≤80)+p(X≥125)=p(X≤80andX≥125)=(.
09175)+(.04745)=p(X≤80andX≥119)=.00435=p(X≤80 and X≥12
5)= p(X≤80)+p(X≥125)=p(X≤80 and X≥125)= (.09175)+(.04745
)=p(X≤80 and X≥119)= .00435
FIGURE 7.7 Standard normal curve with z = −1.33 and z =
+1.67 indicated
Thus, the probability of the first event followed by the second
event has a very low probability of less than 1%.
1.If SAT scores are normally distributed with a mean of 1,000
and a standard deviation of 200, what is the probability of a
student scoring 1,100 or higher on the SAT?
2.For this hypothetical SAT distribution, what is the probability
of a student scoring 910 or lower on the SAT?
of a student scoring 910 or lower or 1,100 or higher?
of selecting a student who scored 910 or lower followed by a
student who scored 1,100 or higher on the SAT?
REVIEW OF KEY TERMS
addition rule (p. 117)
hypothesis testing (p. 111)
multiplication rule (p. 116)
probability (p. 111)
MODULE EXERCISES
(Answers to odd-numbered questions appear in Appendix B.)
1.Imagine that I have a jar that contains 50 blue marbles and 20
red marbles.
a.What is the probability of selecting a red marble from the jar?
b.What is the probability of selecting a blue marble from the
jar?
c.What is the probability of selecting either a red or a blue
marble from the jar?
d.What is the probability of selecting a red marble (with
replacement) followed by a blue marble?
2.What is the probability of a couple having children in the

following birth order: boy, boy, boy, boy?
3.What is the probability of selecting either a 2 or a 4 (of any
suit) from a standard deck of cards?
4.If height is normally distributed with a mean of 68 inches and
a standard deviation of 5 inches, what is the probability of
selecting someone who is 70 inches or taller?
5.For the distribution described in exercise 4, what is the
probability of selecting someone who is 64 inches or shorter?
probability of selecting someone who is 70 inches or taller or 64
inches or shorter?
probability of selecting someone who is 70 inches or taller
followed by someone who is 64 inches or shorter?
CRITICAL THINKING CHECK ANSWERS
Critical Thinking Check 7.1
1.452=.077452=.077
2.1352=.251352=.25
3.36=.5036=.50
4.411=.36411=.36
1.
(a)Multiplication rule
(b)Addition rule
(c)Multiplication rule
(d)Addition rule
2.
(a).50 × .50 = .25
(b)(1352)+(1352)=.25+.25=.50(1352)+(1352)=.25+.25=.50
(c)(1352)+(1352)=.25+.25=.0625(1352)+(1352)=.25+.25=.0625
(d)(452)+(452)=.077+0.77=.154(452)+(452)=.077+0.77=.154
1.z = +1.50, p = .30854
2.z = −.45, p = .32634
3..32634 +.30854 = .63488
4..32634 × .30854 = .101

MODULE 8
Hypothesis Testing and Inferential Statistics
Learning Objectives
•Differentiate null and alternative hypotheses.
•Differentiate one- and two-tailed hypothesis tests.
•Explain how Type I and Type II errors are related to hypothesis
testing.
•Explain what statistical significance means.
•Explain the difference between a parametric test and a
nonparametric test.
Research is usually designed to answer a specific question, for
example, “Do science majors score higher on tests of
intelligence than students in the general population?” The
process of determining whether this statement is supported by
the results of the research project is referred to as hypothesis
testing.
Suppose a researcher wants to examine the relationship between
type of after-school program attended by a child and
intelligence level. The researcher is interested in whether
students who attend an after-school program that is
academically oriented (math, writing, computer use) score
differently on an intelligence test than students who do not
attend such programs. The researcher will form a hypothesis.
The hypothesis might be that children in academic after-school
programs will have different IQ scores than children in the
general population. Because most intelligence tests are
standardized with a mean score (μ) of 100 and a standard
deviation (σ) of 15, the students in the academic after-school
program would have to score higher or lower than 100 for the
hypothesis to be supported.
Null and Alternative Hypotheses
Most of the time, researchers are interested in demonstrating the
truth of some statement. In other words, they are interested in
supporting their hypothesis. It is impossible statistically,

however, to demonstrate that something is true. In fact,
statistical techniques are much better at demonstrating that
something is not true. This presents a dilemma for researchers.
They want to support their hypotheses, but the techniques
available to them are better for showing that something is false.
What are they to do? The logical route is to propose exactly the
opposite of what they want to demonstrate to be true, then
disprove or falsify that hypothesis. What is left (the initial
hypothesis) must then be true (Kranzler, 2007).
Let's use our sample hypothesis to demonstrate what we mean.
We want to show that children who attend academic after-
school programs have different IQ scores from those who do
not. We understand that statistics cannot demonstrate the truth
of this statement. We therefore construct what is known as
a null hypothesis (symbol H0). Whatever the research topic, the
null hypothesis always predicts that there is no difference
between the groups being compared. This is typically what the
researcher does not expect to find. Think about the meaning of
null—nothing or zero. The null hypothesis means you have
found nothing—no difference between the groups.
null hypothesis The hypothesis predicting that no difference
exists between the groups being compared.
For the sample study, the null hypothesis would be that children
who attend academic after-school programs are of the same
intelligence level as other children. Remember, we said that
statistics allow us to disprove or falsify a hypothesis. Therefore,
if the null hypothesis is not supported, our original
hypothesis—that children who attend academic after-school
programs have different IQs from other children—is all that is
left. In statistical notation, the null hypothesis for this study
would be:
H0: μ0 = μacademic program = μgeneral population
The purpose of the study, then, is to decide whether H0 is
probably true or probably false.
The hypothesis that the researcher wants to support is known as
the alternative hypothesis (Ha), or the research hypothesis (H1).

The statistical notation for Ha is:
Ha:μ0≠μ1,orμacademicprogram≠μgeneralpopulationHa:μ0≠μ1, o
r μacademic program≠ μgeneral population
alternative hypothesis (research hypothesis) The hypothesis that
the researcher wants to support, predicting that a significant
difference exists between the groups being compared.
When we use inferential statistics, we are trying to reject H0,
which means that Ha is supported.
Two-Tailed and One-Tailed Hypothesis Tests
The manner in which the previous alternative hypothesis (Ha)
was stated reflects what is known statistically as a two-tailed
hypothesis, or a nondirectional hypothesis—an alternative
hypothesis in which the researcher expects to find differences
between the groups but is unsure what the differences will be.
In this case, the researcher would predict a difference in IQ
scores between children in academic after-school programs and
those in the general population, but the direction of the
difference would not be predicted. Those in academic programs
would be expected to have either higher or lower IQs but not
the same IQs as the general population of children. The
statistical notation for a two-tailed test is as follows:
H0:μ1≤μ1,orμacademicprogram=μgeneralpopulationHa:μ0≤μ1,o
rμacademicprogram≠μgeneralpopulationH0:μ1≤μ1, or μacademi
c program=μgeneral populationHa:μ0≤μ1, or μacademic program
≠μgeneral population
two-tailed hypothesis (nondirectional hypothesis) An alternative
hypothesis in which the researcher predicts that the groups
being compared differ but does not predict the direction of the
difference.
The alternative to a two-tailed or nondirectional test is a one-
tailed hypothesis, or a directional hypothesis—an alternative
hypothesis in which the researcher predicts the direction of the
expected difference between the groups. In our example, the
researcher would predict the direction of the difference—
namely, that children in academic after-school programs will be
more intelligent than children in the general population. Thus,

the alternative hypothesis would be written as follows:
Ha:μ0>μ1;orμacademicprogram>μgeneralpopulationHa:μ0>μ1; o
r μacademic program > μgeneral population
one-tailed hypothesis (directional hypothesis) An alternative
hypothesis in which the researcher predicts the direction of the
expected difference between the groups.
When we use a directional alternative hypothesis, the null
hypothesis is also, in some sense, directional. If the alternative
hypothesis is that children in academic after-school programs
will have higher intelligence test scores, then the null
hypothesis is that being in academic after-school programs
either will have no effect on intelligence test scores or will
decrease intelligence test scores. Thus, the null hypothesis for
the one-tailed directional test might more appropriately be
written as follows:
H0:μ0≤μ1;orμacademicprogram≤μgeneralpopulationH0: μ0≤μ1;
or μacademic program ≤ μgeneral population
In other words, if the alternative hypothesis for a one-tailed test
is μ0 > μ1, then the null hypothesis is μ0 ≤ μ1, and to reject H0,
the children in academic after-school programs have to have
intelligence test scores higher than those in the general
population. In our example, the one-tailed hypothesis makes
more sense.
Assume that the researcher has selected a random sample of
children from academic after-school programs to compare their
IQs with the IQs of children in the general population (as noted
previously, we know that the mean IQ for the population is
100). If we collected data and found that the mean intelligence
level of the children in academic after-school programs is
“significantly” (a term that will be discussed shortly) higher
than the mean intelligence level for the general population, we
could reject the null hypothesis. Remember that the null
hypothesis states that no difference exists between the sample
and the population. Thus, the researcher concludes that the null
hypothesis—that there is no difference or that children
performed more poorly—is not supported. When the null

hypothesis is rejected, the alternative hypothesis—that those in
academic programs have higher IQ scores than those in the
general population—is supported. We can say that the evidence
suggests that the sample of children in academic after-school
programs represents a specific population that scores higher on
the IQ test than the general population.
If, on the other hand, the mean IQ scores of the children in
academic after-school programs are not significantly different
from the population mean score, then the researcher has failed
to reject the null hypothesis and, by default, has failed to
support the alternative hypothesis. In this case, the alternative
hypothesis—that the children in academic programs have higher
IQs than the general population—is not supported.
Type I and Type II Errors in Hypothesis Testing
Any time we make a decision using statistics, there are four
possible outcomes (see Table 8.1). Two of the outcomes
represent correct decisions, whereas two represent errors. Let's
use our example to illustrate these possibilities.
TABLE 8-1The four possible outcomes in statistical decision
making
THE TRUTH (UNKNOWN TO THE RESEARCHER)
THE RESEARCHER’S DECISION
H0 is true
H0 is false
Reject H0 (say it is false)
Type I error
Correct decision
Fail to reject H0 (say it is true)
Correct decision
Type II error
If we reject the null hypothesis (the hypothesis stating that there
is no IQ difference between groups), we may be correct in our
decision, or we may be incorrect. If our decision to reject H0 is
correct, that means there truly is a difference in IQ between
children in academic after-school programs and the general

population of children. However, our decision could be
incorrect. The result may have been due to chance. Even though
we observed a significant difference in IQ between the children
in our study and the general population, the result might have
been a fluke—maybe the children in our sample just happened
to guess correctly on a lot of the questions. In this case, we
have made what is known as a Type I error—we
rejected H0 when in reality we should have failed to reject it (it
is true that there really is no IQ difference between the sample
and population). Type I errors can be thought of as false
alarms—we said there was a difference, but in reality there is
no difference.
Type I error An error in hypothesis testing in which the null
hypothesis is rejected when it is true.
What if our decision is to not reject H0, meaning we conclude
that there is no difference in IQ between the children in the
academic after-school program and children in the general
population? This decision could be correct, meaning that in
reality there is no IQ difference between the sample and the
population. However, it could also be incorrect. In this case, we
would be making a Type II error—saying there is no difference
between groups when in reality there is a difference. Somehow
we have missed the difference that really exists and have failed
to reject the null hypothesis when it is false. All of these
possibilities are summarized in Table 8.1.
Probability, Statistical Significance, and Errors
Type II error An error in hypothesis testing in which there is a
failure to reject the null hypothesis when it is false.
Suppose we actually did the study on IQ level and academic
after-school programs. In addition, suppose we found that there
was a difference between the IQ levels of children in academic
after-school programs and children in the general population
(those in the academic programs scored higher). Lastly, suppose
that this difference is statistically significant at the .05 (or the
5%) level (also known as the .05 alpha level). To say that a
result has statistical significance at the .05 level means that a

difference as big as or bigger than what we observed between
the sample and the population could have occurred by chance
only 5 times or fewer out of 100. In other words, the likelihood
that this result is due to chance is small. If the result is not due
to chance, then it is most likely due to a true or real difference
between the groups. If our result were statistically significant,
we would reject the null hypothesis and conclude that we have
observed a significant difference in IQ scores between the
sample and the population.
statistical significance An observed difference between two
descriptive statistics (such as means) that is unlikely to have
occurred by chance.
Remember, however, that when we reject the null hypothesis,
we could be correct in our decision, or we could be making a
Type I error. Maybe the null hypothesis is true, and this is one
of those 5 or fewer times out of 100 when the observed
differences between the sample and the population did occur by
chance. This means that when we adopt the .05 level of
significance (the .05 alpha level), as often as 5 times out of 100
we could make a Type I error. The .05 level, then, is the
probability of making a Type I error (for this reason, it is also
referred to as a p value, which means probability value—the
probability of a Type I error). In the social and behavioral
sciences, alpha is typically set at .05 (as opposed to .01 or .08
or anything else). This means that researchers in these areas are
willing to accept up to a 5% risk of making a Type I error.
What if you want to reduce your risk of making a Type I error
and decide to use the .01 alpha level, reducing the risk of a
Type I error to 1 out of 100 times? This seems simple enough:
Simply reduce alpha to .01, and you have reduced your chance
of making a Type I error. By doing this, however, you have now
increased your chance of making a Type II error. Do you see
why? If I reduce my risk of making a false alarm—saying a
difference is there when it really is not—I increase my risk of
missing a difference that really is there. When we reduce the
alpha level, we have insisted on more stringent conditions for

accepting our research hypothesis, making it more likely that
we could miss a significant difference when it is present. We
will return to Type I and Type II errors in the next module when
we cover statistical power and discuss alternative ways of
addressing this problem.
Which type of error, Type I or Type II, do you think is
considered more serious by researchers? Most researchers
consider a Type I error more serious. They would rather miss a
result (Type II error) than conclude that there is a meaningful
difference when there really is not (Type I error). What about in
other arenas—for example, in the courtroom? A jury could make
a correct decision in a case (find guilty when truly guilty, or
find innocent when truly innocent). They could also make either
a Type I error (say guilty when innocent) or Type II error (say
innocent when guilty). Which is more serious here? Most people
believe that a Type I error is worse in this situation also. How
about in the medical profession? Imagine a doctor attempting to
determine whether or not a patient has cancer. Here again, the
doctor could make one of the two correct decisions or could
make one of the two types of errors. What would the Type I
error be? This would be saying that cancer is present when in
fact it is not. What about the Type II error? This would be
saying that there is no cancer when in fact there is. In this
situation, most people would consider a Type II error to be
more serious.
HYPOTHESIS TESTING
Concept
Description
Example
Null Hypothesis
The hypothesis stating that the independent variable has no
effect and that there will be no difference between the two
groups
H0: μ0 = μ1 (two-tailed)
H0: μ0 ≤ μ1 (one-tailed)

H0: μ0 ≥ μ1 (one-tailed)
Alternative Hypothesis or Research Hypothesis
The hypothesis stating that the independent variable has an
effect and that there will be a difference between the two groups
Ha: μ0 ≠ μ1 (two-tailed)
Ha: μ0 > μ1 (one-tailed)
Ha: μ0 < μ1 (one-tailed)
Two-Tailed or Nondirectional Test
An alternative hypothesis stating that a difference is expected
between the groups, but there is no prediction as to which group
will perform better or worse
The mean of the sample will be different from or unequal to the
mean of the general population
One-Tailed or Directional Test
An alternative hypothesis stating that a difference is expected
between the groups, and it is expected to occur in a specific
direction
The mean of the sample will be greater than the mean of the
population, or the mean of the sample will be less than the mean
of the population
Type I Error
The error of rejecting H0 when we should have failed to reject
it
This error in hypothesis testing is equivalent to a “false alarm,”
saying that there is a difference when in reality there is no
difference between the groups
Type II Error
The error of failing to reject H0 when we should have rejected
it
This error in hypothesis testing is equivalent to a “miss,” saying
that there is not a difference between the groups when in reality
there is difference between the groups
Statistical Significance
When the probability of a Type I error is low (.05 or less)
The difference between the groups is so large that we conclude
it is due to something other than chance

1.A researcher hypothesizes that children from the South weigh
less (because they spend more time outside) than the national
average. Identify H0 and Ha. Is this a one- or two-tailed test?
2.A researcher collects data on children's weights from a
random sample of children in the South and concludes that
children from the South weigh less than the national average.
The researcher, however, did not realize that the sample
included many children who were small for their age and that in
reality there is no difference in weight between children in the
South and the national average. What type of error was made?
3.If a researcher decides to use the .10 level rather than using
the conventional .05 level of significance, what type of error is
more likely to be made? Why? If the .01 level is used, what type
of error is more likely? Why?
Using Inferential Statistics
Now that we have an understanding of the concept of hypothesis
testing, we can begin to discuss how hypothesis testing is used.
The simplest type of study involves only one group and is
known as a single-group design. The single-group design lacks a
comparison group—there is not a control group of any sort. We
can, however, compare the performance of the group (the
sample) to the performance of the population (assuming that
population data are available).
single-group design A research study in which there is only one
group of participants.
Earlier in the module, we illustrated hypothesis testing using a
single-group design—comparing the IQ scores of children in
academic after-school programs (the sample) to the IQ scores of
children in the general population. The null and alternative
hypotheses for this study were:
H0:μ0≤μ1,orμacademicprogram≤μgeneralpopulationHa:μ0≤μ1,o
rμacademicprogram>μgeneralpopulationH0:μ0≤μ1, or μacademi
c program ≤μgeneral population Ha:μ0≤μ1, or μacademic progra
m >μgeneral population
To compare the performance of the sample to that of the

population, we need to know the population mean (μ) and the
population standard deviation (σ). We know that for IQ tests, μ
= 100 and σ = 15. We also need to decide who will be in the
sample. Random selection will increase our chances of getting a
representative sample of children enrolled in academic after-
school programs. How many children do we need in the sample?
We will see in later modules that the larger the sample, the
greater the power of the study. We will also see that one of the
assumptions of the statistical procedure we will be using to test
our hypothesis is a sample size of 30 or more.
Once we have chosen our sample, we need to collect the data.
To collect IQ score data, we could either administer an
intelligence test to the children or look at their academic files to
see whether they had already taken such a test.
Once the data are collected, we can begin to analyze them
using inferential statistics—procedures for drawing conclusions
about a population based on data collected from a sample.
Inferential statistics involve the use of procedures for drawing
conclusions based on the scores collected in a research study
and going beyond them to make inferences about a population.
In the following chapter (Chapter 5) we will describe two
inferential statistical tests—the z test and t test. Both of these
are parametric tests—tests that require us to make certain
assumptions about estimates of population characteristics, or
parameters. These assumptions typically involve knowing the
mean (μ) and standard deviation (σ) of the population and that
the population distribution is normal. Parametric tests are
generally used with interval or ratio data. The alternative to a
parametric test is a nonparametric test; that is, it does not
involve the use of any population parameters. In other words, μ
and σ are not needed, and the underlying distribution does not
have to be normal. Nonparametric tests are most often used with
ordinal or nominal data and will be discussed more fully
in Chapter 10.
inferential statistics Procedures for drawing conclusions about a
population based on data collected from a sample.

parametric test A statistical test that involves making
assumptions about estimates of population characteristics, or
parameters.
nonparametric test A statistical test that does not involve the
use of any population parameters—μ and s are not needed, and
the underlying distribution does not have to be normal.
SINGLE-SAMPLE RESEARCH AND INFERENTIAL
STATISTICS
Concept
Description
Examples
Parametric Inferential Statistics
Inferential statistical procedures that require certain
assumptions about the parameters of the population represented
by the sample data, such as knowing μand s and that the
distribution is normal Most often used with interval or ratio
data
z test
t test
(discussed in Chapter 5)
Nonparametric Inferential Statistics
Inferential procedures that do not require assumptions about the
parameters of the population represented by the sample data; μ
and σ are not needed, and the underlying distribution does not
have to be normal Most often used with ordinal or nominal data
Chi-square tests Wilcoxon tests (discussed in Chapter 10)
1.How do inferential statistics differ from descriptive statistics?
2.How does single-sample research involve the use of
hypothesis testing? In other words, in a single-group design,
what hypothesis is tested?
REVIEW OF KEY TERMS
alternative hypothesis (research hypothesis) (p. 126)
inferential statistics (p. 131)
nonparametric test (p. 132)

null hypothesis (p. 126)
one-tailed hypothesis (directional hypothesis) (p. 126)
parametric test (p. 131)
single-group design (p. 131)
statistical significance (p. 128)
two-tailed hypothesis (nondirectional hypothesis) (p. 126)
Type I error (p. 128)
Type II error (p. 128)
MODULE EXERCISES
(Answers to odd-numbered questions appear in Appendix B.)
1.The admissions counselors at Brainy University believe that
the freshman class they have just recruited is the brightest yet.
If they wanted to test this belief (that the freshmen are brighter
than the other classes), what would the null and alternative
hypotheses be? Is this a one- or two-tailed hypothesis test?
2.To test the hypothesis in exercise 1, the admissions counselors
select a random sample of freshmen and compare their scores on
the SAT to those of the population of upper-classmen. They find
that the freshmen do in fact have a higher mean SAT score.
However, what they are unaware of is that the sample of
freshmen was not representative of all freshmen at Brainy
University. In fact, the sample over-represented those with high
scores and under-represented those with low scores. What type
of error (Type I or Type II) did the counselors make?
3.A researcher believes that family size has increased in the last
decade in comparison to the previous decade—that is, people
are now having more children than they were before. What
would the null and alternative hypotheses be in a study designed
to assess this? Is this a one- or two-tailed hypothesis test?
4.What are the appropriate H0 and Ha for each of the following
research studies? In addition, note whether the hypothesis test is
one- or two-tailed.
a.A study in which researchers want to test whether there is a
difference in spatial ability between left- and right-handed
people
b.A study in which researchers want to test whether nurses who

work 8-hour shifts deliver higher-quality work than those who
work 12-hour shifts
c.A study in which researchers want to determine whether crate-
training puppies is superior to training without a crate
5.Assume that each of the following conclusions represents an
error in hypothesis testing. Indicate whether each of the
statements is a Type I or II error.
a.Based on the data, the null hypothesis was rejected.
b.There was no significant difference in quality of work
between nurses who work 8- and 12-hour shifts.
c.There was a significant difference between right- and left-
handers in their ability to perform a spatial task.
d.The researcher failed to reject the null hypothesis based on
these data.
6.Explain the difference between parametric and nonparametric
statistics.
CRITICAL THINKING CHECK ANSWERS
1.H0:μSouthernchildren≥μchildreningeneralHa:μSouthernchildr
en<μgeneralpopulationH0: μSouthern children≥μchildren in gene
ralHa:μSouthern children<μgeneral population
This is a one-tailed test.
2.The researcher concluded that there was a difference when in
reality there was no difference between the sample and the
population. This is a Type I error.
3.With the .10 level of significance, the researcher is willing to
accept a higher probability that the result may be due to chance.
Therefore, a Type I error is more likely to be made than if the
researcher used the more traditional .05 level of significance.
With a .01 level of significance, the researcher is willing to
accept only a .01 probability that the result may be due to
chance. In this case, a true result is more likely to be missed,
meaning that a Type II error is more likely.
1.Inferential statistics allow researchers to make inferences
about a population based on sample data. Descriptive statistics

simply describe a data set.
2.Single-sample research allows researchers to compare sample
data to population data. The hypothesis tested is whether the
sample performs similarly to the population or whether the
sample differs significantly from the population and, thus,
represents a different population.
CHAPTER FOUR SUMMARY AND REVIEW
Probability and Hypothesis Testing
CHAPTER SUMMARY
This chapter consisted of an introduction to probability and
hypothesis testing. There was a discussion of how to calculate
basic probabilities, how to use the multiplication and addition
rules, and lastly how to use the area under the normal curve to
calculate probabilities. With respect to hypothesis testing, there
was a discussion of the null and alternative hypotheses, one-
and two-tailed hypothesis tests, and Type I and Type II errors in
hypothesis testing. In addition, the concept of statistical
significance was defined. The most simplistic use of hypothesis
testing—a single-group design—in which the performance of a
sample is compared to the general population was presented to
illustrate the use of inferential statistics in hypothesis testing.
CHAPTER 4 REVIEW EXERCISES
(Answers to exercises appear in Appendix B.)
Fill-in Self-Test
Answer the following questions. If you have trouble answering
any of the questions, restudy the relevant material before going
on to the multiple-choice self-test.
1._______________ is the study of likelihood and uncertainty.
2.The rule that says that the probability of a series of outcomes
occurring on successive trials is the product of their individual
probabilities is the _______________ rule.
3.The rule that says that the probability of one outcome or the
other outcome occurring on a particular trial is the sum of their
individual probabilities is the _______________ rule.

4.The hypothesis predicting that no difference exists between
the groups being compared is the _______________.
5.An alternative hypothesis in which the researcher predicts the
direction of the expected difference between the groups is a
_______________.
6.An error in hypothesis testing in which the null hypothesis is
rejected when it is true is a _______________.
7.When an observed difference, say between two means, is
unlikely to have occurred by chance, we say that the result has
_______________.
8._______________ tests are statistical tests that do not involve
the use of any population parameters.
Multiple-Choice Self-Test
Select the single best answer for each of the following
questions. If you have trouble answering any of the questions,
restudy the relevant material.
1.The study of likelihood and uncertainty is to
_______________ and the process of determining whether a
hypothesis is supported by the results of a research project is to
_______________.
a.hypothesis testing; probability
b.hypothesis testing; inferential statistics
c.probability; hypothesis testing
d.inferential statistics; probability
2.The and rule is to _______________ and the or rule is to
_______________.
a.multiplication rule; addition rule
b.addition rule; multiplication rule
c.multiplication rule; multiplication rule
d.addition rule; addition rule
3.The probability of rolling a 5 on one roll of a standard die is
a..25
b..50
c..20
d..17
4.The probability of rolling either a 2 or a 5 on one roll of a

standard die is
a..34
b..25
c..50
d..03
5.The probability of rolling a 2 followed by a 6 on a standard
die is:
a..34
b..25
c..50
d..03
6.If a psychology exam is normally distributed with a mean of
75 and a standard deviation of 5, what is the probability of
someone scoring 80 or higher on the exam?
a..15866
b..34134
c..84134
d..30598
7.If a psychology exam is normally distributed with a mean of
75 and a standard deviation of 5, what is the probability of
someone scoring 65 or higher on the exam?
a..02275
b..47725
c..52275
d..97725
8.Inferential statistics allow us to infer something about the
_______________ based on the _______________.
a.sample; population
b.population; sample
c.sample; sample
d.population; population
9.The hypothesis predicting that differences exist between the
groups being compared is the _______________ hypothesis.
a.null
b.alternative
c.one-tailed

d.two-tailed
10.Null hypothesis is to alternative hypothesis as
_______________ is to a. effect; no effect
b.Type I error; Type II error
c.no effect; effect
d.None of the alternatives is correct.
11.One-tailed hypothesis is to directional hypothesis as
_______________ hypothesis is to _______________
hypothesis.
a.null; alternative
b.alternative; null
c.two-tailed; nondirectional
d.two-tailed; one-tailed
12.When using a one-tailed hypothesis, the researcher predicts
a.the direction of the expected difference between the groups.
b.that the groups being compared will differ in some way.
c.nothing.
d.only one thing.
13.In a study of the effects of caffeine on driving performance,
researchers predict that those in the group that is given more
caffeine will exhibit worse driving performance. The
researchers are using a _______________ hypothesis.
a.two-tailed
b.directional
c.one-tailed
d.both directional and one-tailed
14.In a recent study, researchers concluded that caffeine
significantly increased anxiety levels. What the researchers
were unaware of, however, was that several of the participants
in the no-caffeine group were also taking antianxiety
medications. The researchers’ conclusion is a(n)
_______________ error.
a.Type II
b.Type I
c.null hypothesis
d.alternative hypothesis

15.When alpha is .05, this means that
a.the probability of a Type II error is .95.
b.the probability of a Type II error is .05.
c.the probability of a Type I error is .95.
d.the probability of a Type I error is .05.
Introduction to Hypothesis Testing Table
PSYCH/625 Version 5
1
Introduction to Hypothesis Testing Table
Identify whether each research scenario listed in the left column
of the top of the matrix, would be a one- or two-tailed test and
then justify your decision. For each research scenario in the
bottom of the matrix, write what the null and alternative
hypotheses would be.
Research Scenario
One-tail or two-tail test
Justification
A local dermatologist claims that 30% of mole biopsies are
unnecessary. Last month at his clinic, 210 out of 634 had benign
biopsy results. Is there enough evidence to reject the
dermatologist’s claim?
A local dermatologist claims that more than 30% of mole
biopsies are unnecessary. Last month at his clinic, 210 out of
634 had benign biopsy results. Is there enough evidence to
reject the dermatologist’s claim?
A local dermatologist claims that less than 30% of mole
biopsies are unnecessary. Last month at his clinic, 210 out of

634 had benign biopsy results. Is there enough evidence to
reject the dermatologist’s claim?
Research Scenario
Null hypothesis:
Alternative hypothesis:
A pharmaceutical company believes their weight loss drug is
more effective than any program on the market. They study two
groups of people: Group A is given the drug and Group B is
given a placebo. They then compare weight loss over the course
of 3 months.
A college professor believes class attendance is different on
sunny versus cloudy days.
Researchers hypothesize that rats fed donuts for breakfast will
perform worse on a task than those fed a high-protein breakfast.
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