use logic of hypothesis testing to answer the 4 research questions in the first file.

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

with a mean of  = 51.9 and a standard deviation of x = 10.2. Is there

evidence that the drug improves memory scores? Set  = .05.

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
is normally distributed with a mean of  = 79.1 and a standard deviation

of  = 6.1. In your sample of 22 music-listening students, the mean

was X = 76.15. Is there evidence to conclude that listening to music is
detrimental to exam performance? Set  = .05.

4. A common reading achievement test for fifth grade students has a

nationwide mean of  = 70.0. A teacher would like to know if 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
the students differ from the national average, setting  = .01.

b. Would this test be significant if you had set  = .05? Explain

why you reached a different conclusion depending on what alpha

level you used.

1

Null

Hypothesis

Significance

Testing

Hypothesis Testing
 Method to decide if an observed result is unlikely to have

occurred by chance

 Assumes data were obtained using a random sampling
procedure
 Probability associated with test may be wrong if sample is

not random

 In most real studies, p < .05 probably is not really p < .05!

Hypotheses
 Research Hypothesis

 Outcome you expect if your theory is true

 Statistical Hypothesis

 A research hypothesis stated in terms of the distribution
under investigation.

 Two types:

 Null Hypothesis (H0)

 Alternative Hypothesis (H1)

Null Hypothesis (H0)
 States that the population parameter is some particular

value

 Makes a prediction opposite of the research

hypothesis

 Rejecting H0 supports the research hypothesis

Alternative Hypothesis (H1)
 States that the population parameter is some alternative

range of values

 The opposite of H0
 Makes a prediction consistent with the research

hypothesis

Examples of Null and Alternate
Hypotheses

 H0 :  = 0 (treatment had no effect)

 Ha :   0 (treatment did have an effect)

 H0 :   100 (no effect or negative effect)

 Ha :   100 (positive effect)

 H0 : r = 0 (no linear relationship)

 Ha : r  0 (linear relationship is not zero)

2

Nondirectional Hypothesis
 Two-tailed

 Null will be rejected if test statistic is much higher OR
lower than prediction

Directional Hypothesis
 One-tailed

 Decide ahead of time: Null will be rejected if test statistic
is much higher than prediction
 (or much lower, depending on your prediction!)

 Use only if willing to ignore an extreme value in opposite
direction from what is expected

How to test a null Hypothesis

 Assume that H0 is true, determine sampling
distribution

 Draw a random sample from the population

 What is the probability that this sample could have
been drawn if H0 was actually true?

Declaring

Statistical Significance

 If the observed results are very unlikely to have occurred
just by chance if H0 was actually true…

 You conclude that H0 is probably NOT true – you REJECT
the null

Statistical Significance

 H0 is rejected

 The observed effect is greater than we would expect due
to chance if the null were true

Practical Significance

 The observed difference has conceptual or practical
meaning

 statistically significant may not be practically meaningful
(and vice-versa)

3

Test Statistic

 Numeric summary of how far an observed estimate is
from the parameter specified in H0

 Common ones are z, t, c2, F

Critical Value

 Value of a test statistic that cuts off the desired alpha
level (region of rejection)

 Usually a = .05

 Look this up in a table

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)

 Probability is associated with your decision

 Probability is NOT associated with the truth of the
universe

 Within each possible reality, decision probabilities
sum to 1.0

An analogy with the legal system
 US legal system is based on the premise “innocent until

proven guilty”

 Jury tests the null hypothesis of “didn’t commit the
crime”

 Rejecting the null = “guilty”

 Failing to reject the null = “not guilty”
 (notice that we don’t say innocent)

 What is a?

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)

Probabilityand 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 . The variability of the sampling 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:

1) The mean of the sampling distribution of means will be . Any given sample mean will

not be exactly the same as , but in the long run, they will average out to be exactly 

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

based on a sampling distribution of means with  = 100,  = 15 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




 which is equivalent to X

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 with  = 100 and  = 15. But

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


   the probability of getting a Z of .27 or greater is .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

.025 in the tail beyond it? Look in the table, it’s Z = 1.96.

We will refer to the area beyond 1.96 (that is, less than –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

Exampleof 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

is  = 100 with a standard deviation of  = 15. So the research 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

3.Select 

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.

H0:  = 100 HA:   100

In this case,  refers to the mean of the population that our 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.

2.Select statistical procedure

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 population standard deviation () is

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.

3.Select 

In many cases, I will just tell you what  level to use. But you should understand why a

researcher might choose one  level compared to another. Remember that  tells us the

probability of making a Type I error, or rejecting a null that should not have been rejected. Most

of the time, researchers in the social sciences use an  of .05, 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

may want to choose a more conservative  level, such as .01. For this example, let’s be boring

and go with  = .05.

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







where 0 is the mean in the null hypothesis

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

have  = .05, and our test was non-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-score is  1.96.

.025 .025

-1.96 +1.96

6.Compare (4) and (5)

7.If (4) exceeds (5), reject H0

8.Otherwise, fail to reject H0

OK, so our zobt was 2.80 and our zcrit was  1.96. So what do 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:

H0:   100 HA:   100

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 for  = .05, one-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:

H0:  = 100 HA:   100 and zcrit = 1.96.

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 population was  = 15, because we

know that IQ tests have been developed over many years. We were able to use the z-test because

 was known.

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 will use the sx instead of x. Also,

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:

H0:   18 HA:   18

2.Select statistical procedure

this time, we need to use a one-sample t-test

3.Select 

We’ll be boring again and go with  = .05.

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

value for tcrit. With df = 29 and  = .05, tcrit will be -1.699.

6.Compare (4) and (5)

7.If Zobt or tobt exceeds Zcrit or tcrit, reject H0

8.Otherwise, fail to 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.