In this assignment students will demonstrate their understanding of the distribution of means doing all steps of hypothesis testing.

For each problem students will write out all steps of hypothesis testing including populations, hypotheses, cutoff scores, and all relevant calculations. Assignments will be typed and uploaded in a word document to blackboard.

PSY 3200

Unit #4 AS: Hypothesis Testing for Distribution of Means

In this assignment students will demonstrate their understanding of the distribution of means doing all steps of hypothesis testing.
For each problem students will write out all steps of hypothesis testing including populations, hypotheses, cutoff scores, and all relevant calculations. Assignments will be typed and uploaded in a word document to blackboard.

1.

A nationwide survey in 1995 revealed that U.S. grade-school children spend an average of µ = 8.4 hours per week doing homework. The distribution is normal with σ = 3.2. Last year, a sample of n = 100 grade-school children was given the same survey. For this sample, the mean number of homework hours was 7.1. Has there been a significant change in the homework habits of grade-school children? Test with α = .05.

2.

On the basis of her newly developed technique, a student believes she can reduce the amount of time schizophrenics spend in an institution. As director of training at a nearby institution, you agree to let her try her method on 20 schizophrenics, randomly sampled from your institution. The mean duration that schizophrenics stay at your institution is 85 weeks, with a standard deviation of 15 weeks. The scores are normally distributed. The results of the experiment show that patients treated by the student stay at the institution a mean duration of 78 weeks. What do you conclude about the student’s technique? Use α = .05.

3.

A psychologist has developed a standardized test for measuring the vocabulary skills of 4-year-old children. The scores on the test form a normal distribution with μ = 60 and σ = 10. A researcher would like to use this test to investigate the idea that children who grow up with no siblings develop vocabulary skills at a different rate than children in large families. A sample of n = 25 children is obtained, and the mean test score for this sample is 63. On the basis of this sample, can the researcher conclude that vocabulary skills for children with no siblings are significantly different from those of the general population? Test at the .01 level of significance.

4.

The average age for licensed drivers in a county is 42.6, with a standard deviation of 12, and the distribution is approximately normal. A county police officer was interested in whether the average age of those receiving speeding tickets is less that the average age of the population who has a license. She obtained a sample of 16 drivers with speeding tickets. The average age for this sample was 34.4. Do all the steps of hypothesis testing using the 0.01 significance level.

Directions

• In the following slide is a chart with 9 rows, each row is a practice problem

• In each problem you are given:

• an X value

• a Mean

• a Standard Deviation

• an Alpha level

• Whether it is a 1 or 2 tailed test

• (if it is one tailed you are told the direction of high/positive, or low/negative)

• For each of these problems you will practice steps 3, 4, and 5 of hypothesis testing:

• Detrmine the cutoff score

• Solve for your Z score

• Decide whether or not to reject your null hypothesis

Study Population
Sample

Score
Alpha Level Tails of Test

M SD X

A 100 10 80 .05 1 (low)

B 100 20 80 .01 2

C 74.3 4.8 80 .05 2

D 78 0.8 80 .01 1 (high)

E 90 4.7 80 .05 2

F 78 0.8 80 .01 2

G 76 2.3 80 .05 1 (high)

H 74 2.6 80 .01 2

I 76 2.3 80 .05 1 (low)

Answers

• A

• Step 3:

• Z cutoff = -1.64

• Step 4:

• Z = -2

• Step 5:

• Reject

Answers

• B

• Step 3:

• Z cutoff = +/-2.57

• Step 4:

• Z = -1

• Step 5:

• Fail to reject

Answers

• C

• Step 3:

• Z cutoff = +/-1.96

• Step 4:

• Z = 1.19

• Step 5:
• Fail to reject

Answers

• D

• Step 3:

• Z cutoff = 2.33

• Step 4:

• Z = 2.5

• Step 5:
• Reject

Answers

• E

• Step 3:
• Z cutoff = +/-1.96
• Step 4:

• Z = -2.13

• Step 5:
• Reject

Answers

• F

• Step 3:

• Z cutoff = +/-2.57

• Step 4:
• Z = 2.5
• Step 5:

• Fail to Reject

Answers

• G

• Step 3:

• Z cutoff = 1.64

• Step 4:

• Z = 1.74

• Step 5:
• Reject

Answers

• H

• Step 3:
• Z cutoff = +/-2.57
• Step 4:

• Z = 2.31

• Step 5:
• Fail to Reject

Answers

• I

• Step 3:

• Z cutoff = -1.64

• Step 4:
• Z = 1.74
• Step 5:
• Fail to Reject

Hypothesis Testing and the

Distribution of Means

PSY3200 Unit 4

Defining Hypothesis Testing

In this unit we will be taking the concept of the normal distribution and finally connecting it with an experiment. We
will do this in a process known as hypothesis testing. Hypothesis testing is a procedure used to see if a
hypothesis supports a particular theory; it uses the normal curve, probability, and sampling (Aron, Coups, & Aron,
2013). The steps for hypothesis testing are as follows: Identify your populations and state your hypotheses;
identify the characteristics of the comparison distribution; determine the cutoff score at which you would reject
your null hypothesis; calculate your test score; compare your test score to the cutoff score and decide whether
you would reject your null hypothesis or not (Aron et al., 2013).

Hypothesis Testing Steps

Step 1 requires us to do 2 parts. The first thing we will need to do for hypothesis testing is to define our
populations. To do this we always state (1) who is in this particular group, and (2) what we are measuring (Aron et
al., 2013). It is important for this not to state what we think will happen, that is for the hypothesis. Let’s take the
example hypothesis of “a new technique of studying will increase statistics scores.” In this scenario we have two
groups, one studying a new way (our experimental group), and one studying an old way (the comparison group).
For both groups we are measuring the same thing, their stats scores. So, for the first part of step 1 we will define
our populations as:

P1 = Stats scores for those who study the new way

P2 = Stats scores for those who study the old way

The other half of step 1 is to state our hypotheses, and we will have two of them: the research hypothesis and the
null hypothesis. The research hypothesis states that there will be a change or a difference (Aron et al., 2013). In
this case we are saying that the new way of studying will increase our grades, so our hypothesis will be:

H1 = P1 > P2

This translates to us saying: the stats scores for people who study the new way will be greater than those who
study the old way. The null hypothesis states there is no change or difference between the experimental and
comparison distributions, in other words, our independent variable will not affect our dependent variable (Aron et
al., 2013):

H0 = P1 = P2

This setup may seem unusual, but there is a reason for it. In the media today, we are always hearing things have
been proven to be true, whether it is a new product, a new study, a new way of doing something. If you think
about it, can you ever really prove something is true? If I say all cars are red and I go out to try and prove that, I
could spend 3 years of my life counting red cars; and even if I only found red cars, there is always that possibility
that somewhere a blue car exists, I just haven’t seen yet. I cannot prove a hypothesis true, but if I did see a blue
car, this would disprove my hypothesis. I cannot prove a hypothesis, but I can disprove one by finding
contradictory evidence. However, disproving a research hypothesis is not an efficient way to do research, so this
is where the rationale of the null hypothesis comes in. The null hypothesis states the experiment does not work
(the IV does not affect my DV), so we try to disprove that null hypothesis and reject it. If we can, it seems the
research supports my hypothesis (Aron et al., 2013). If we cannot, then I must accept the null hypothesis and the
fact the experiment did not work.

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Step 2 is to identify the characteristics of the comparison distribution (Aron et al., 2013). In order to see if the
experiment worked, we must compare it to those who did not receive the manipulation to check if there is a
difference. If I was doing a Z score: (X-M)/SD, I would find the X, the M, and the SD in this step. Any piece I’m
missing I would calculate. For our example we would calculate the M and SD of people who study the old way
and the X is the stat score for the person studying the new way.

Step 3 is to find the cutoff score for the comparison distribution. It is in this step where we ask: at what point is our
score different enough from the comparison distribution to say it was caused by our IV (Aron et al., 2013). Let’s
say the average score on our stats test was a 75. If you study the new way and get a 76, it is clearly higher, but
would you say it is different enough that it’s because of our new way of studying? Most likely you’d say that’s not
enough of a difference. If you get a 97 however, then you might be more inclined to say it was because of the
experiment. The question step 3 addresses is: at what point do we say it is enough? Because we are dealing with
a sample we know we can never be 100% certain, but if we can say we are 95% certain we have the right results,
and there is only a 5% chance the results we have are wrong, then that we can accept. The key to step 3 is to
find the Z scores that will give us that answer. Luckily the problems we did in the last unit will help us out. If we
look at the normal curve distribution, we know the mean is right in the middle (with a Z score of 0), 5% chance of
error is out in the tail, so we just have to figure out what Z score will give us 5% in the tail. That Z score is 1.64;
we call this the 0.05 significance level or α = 0.05. If we wanted to be more confident in our answer, we could go
for 99% accurate, or a 1% chance the results we got were wrong. In this case we’d need to reach a Z score of
2.33.

In both scenarios we had a positive Z score because we expected the scores to increase or go up. This is called
a directional hypothesis. In a directional hypothesis the researcher clearly indicated what direction she thought
the experiment would go in (Aron et al., 2013). In these types of problems, you would see key words like
increase, rise, or decrease, lower. There are also non-directional hypotheses in which the researcher indicates
something will change but does not specify which direction it will occur (Aron et al., 2013). These would use key
words such as effect, change, or difference. In these scenarios, you would need two other Z cutoff scores: +/-1.96
for the 0.05 significance level and +/-2.57 for the 0.01 significance level. Why do we need these two different
cutoff scores? We said using 1.64 allowed for just a 5% chance that the results we got were due to chance, but if
we were saying our hypothesis could go in either direction from the mean, then we would need acknowledge both
tails of the curve with that 5%. We take that 5% and split it evenly in both tails of the distribution, so the top and
bottom 2.5%. The same setup would be used for the 0.01 significance level.

In the end we are left with 4 cutoff scores for Z scores: For the α = 0.05 and a 1-tailed test: 1.64 or -1.64; for the α
= 0.01 and a 1-tailed test: 2.33 or -2.33; for the α = 0.05 and a 2-tailed test: +/-1.96; and for the α = 0.01 and a
2-tailed test: +/-2.57.

Step 4 is to solve the final equation. In this case we are solving for Z so we would follow the steps for solving for Z
= (X-M)/SD.

Step 5 we compare the Z scores from step 3 (the cutoff) to step 4 (your Z score) and we decide if we would reject
or fail to reject the null hypothesis. The rule is, if our score is further from the mean we would reject the null, if it is
not, we would fail to reject the null (or accept the null) (Aron et al., 2013). For example, if our Z score was 2, and
the cutoff Z score was +/-1.96, we would reject because it is further from the mean (which has a Z score of 0). If
we had a Z score of 1 and the same cutoff score (+/-1.96), then we would fail to reject. Some students do get
tripped up on the wording: reject sounds bad so many students think it means it’s wrong, but you have to
remember what you are rejecting: the null hypothesis, the one that says it didn’t work. In order to help understand
steps 3, 4, and 5 of hypothesis testing, a set of practice steps were added to an accompanying PowerPoint with
answers.

Distribution of Means

In our example we used the Z score formula of (X-M)/SD. This formula is used when you are comparing a single

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score to a sample, but we are almost never using just a single person for an experiment. We often compare a
sample to a population. We would take a group of people from a population, apply a manipulation and see how
they compare to the population they came from (Aron et al., 2013). This would update our Z score formula to: Z =
(M-µ)/σ. The mean of the sample minus the mean of the population, divided by the standard deviation of the
population. The problem with this is we would never actually know the mean of the population (it is impractical to
be able to get that).

To solve these problems, we will come up with a mean that will represent the population; and how we do that is by
repeatedly sampling the population (Aron et al., 2013). A single group to represent the population isn’t ideal as we
could have a group of outliers, so by getting multiple samples, we have a better chance of being a good
representation (Aron et al., 2013). We get a sample of 30 people from the population and get their mean; if do this
20 times and get 20 samples (each time we calculate their mean), we now have a set of scores made up of
means: a distribution of means. This distribution is a good substitute for the population because it has enough
numbers we can comfortably feel it represents the population. The more people from the population, the more like
the population our distribution will be (Aron et al., 2013). We will use this distribution as a comparison distribution,
and because we are saying the distribution is a representation of the population then the mean of that population
is equal to the mean of that distribution of means; so: µ=µm.

The same cannot be said about the variance however. The distribution of means is made up of numbers that are
means or averages, and if you recall the definition of mean was a balance point for the distribution. The variance,
which describes how the scores are deviating from the mean would be lower because it would be made up of less
extreme scores (since all the scores are means) (Aron et al., 2013). To calculate the variance of the distribution of
means we take the variance of the population and divide by the sample size: σ2m = (σ2 / n). The standard
deviation of the distribution of means (called the standard error) would be the square root of that: σm = (σ/√n).
The standard deviation of the population divided by the square root of the sample size.

The rest of the steps of hypothesis testing for the distribution of means is the same, except we use an updated
formula for Z scores: (M-µm)/σm. To review how we would do a hypothesis testing test for the distribution of
means: (1) define your populations and research and null hypotheses; (2) define the comparison distribution by
calculating µm and σm; (3) determine your cutoff score in which you would reject your null hypothesis; (4)
calculate your Z score; (5) decide whether to reject your null hypothesis.

Power

The final part of this unit will briefly discuss the concept of power. Power is the probability you will reject your null
hypothesis when you are supposed to (Aron et al., 2013). We know at the end of a hypothesis testing problem we
have 2 choices: reject or fail to reject your null hypothesis. These choices have nothing to do however with the
reality of what actually happened. It is entirely possible you did everything correct in your experiment but still got
the wrong answer. If your sample told you the scores were different you might reject your null hypothesis even
though the sample was entirely incorrect. We could make 2 possible errors: type I error and type II error. Type I
error is rejecting the null hypothesis when we are not supposed to; in other words, we said the experiment worked
when it did not (Aron et al., 2013). This may sound familiar, and that is because this is the basis of the α levels. α
is the probability of making a type I error. Type II error is failing to reject the null hypothesis when you were
supposed to; or the experiment did work but you said it did not (Aron et al., 2013). The problem in research is one
never knows if they actually made a type I or type II error. The key is to try and limit the possibilities of errors in
data such as confounding variables. The other key to avoid this is increase effect size. Effect is the amount 2
distributions do not overlap each other (Aron et al., 2013). The less they overlap, the more different they are,
which increases power.

(CSLO 1, CSLO 2, CSLO 5, CSLO 7, CSLO 9, CSLO 10, CSLO 11)

References

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Aron, A. Coups, E.J. & Aron, E. (2013) Statistics for Psychology (6th ed.) Chapter 3.

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