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analysis check for Demonstration 8.1 on pages 227 – 228, (36 points):We reviewed the basic hypothesis procedure for a z-score test in Chapter 8 in class on 3/3/20 and 3/10/20; and most of the later chapter statistical test designs have also been applied to this basic hypothesis testing procedure.Complete all 6 items below, and write all work as directed; Be sure to give each item number with an answer that has only one or two words or numerical values as directed below for Demo 8.1. (If more than one answer is submitted for an item, the item will be scored as an incorrect answer.) If you are not sure of an answer, check the information on the reference pages provided or other chapter pages, which can be helpful. Correctly formatted examples: Final Project 1. special training, 2. scores, 3. 65, and so on.Section A: State some items below that are relevant to the null and alternative hypotheses.Choose: special training (in reading skills) OR scores (on a standardized test)(Reference page: 14)1. The Demo 8.1 independent variable is: __________2. The Demo 8.1 dependent variable is: __________Section B: Write the values indicated below that are a part of the formulas and calculations for the z statistic for the hypothesis test. (Reference pages: 65, 94)3. The Demo 8.1 numerical value of the population mean is: __________4. The Demo 8.1 numerical value of the population standard deviation is: __________Section C: State information related to the decision and the conclusion statement regarding the null hypothesis.5. Would the Demo. 8.1 final obtained z statistic value of 1.67 be in the middle of the distribution on page 228 or in one of the tails of the distribution? Choose/write A or B: __________ (Reference page: 205)A. middleB. tails6. According to Step 4 decision and conclusion information on page 228:Does the Demo 8.1 study indicate a statistically significant effect? (Reference pages: 206, 213 – 214)Choose Conclusion statement A or B below. Write A or B: __________A. There is a statistically significant effect.B. There is not a significant effect.

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DEMONSTRATION 8.1

227

MONSTRATION 8.1

HYPOTHESIS TEST WITH Z

STEP 1

A researcher begins with a known population in this case, scores on a standardized test

that are normally distributed with p = 65 and a = 15. The researcher suspects that spe-

cial training in reading skills will produce a change in the scores for the individuals in the

population. Because it is not feasible to administer the treatment (the special training) to

everyone in the population, a sample of n = 25 individuals is selected, and the treatment

is given to this sample. Following treatment, the average score for this sample is M = 70.

Is there evidence that the training has an effect on test scores?

State the hypothesis and select an alpha level.

The researcher would like to know what would happen if the special training were

given to the entire population. Although this population is unknown, it is possible to make

hypotheses about the treated population. The null hypothesis says that the treatment has

no effect, so the treated population is identical to the original untreated population. In

symbols,

Ho: u = 65 (After special training, the mean is still 65.)

The alternative hypothesis states that the treatment does have an effect that causes a

change in the population mean. In symbols,

H: 4 65

(After special training, the mean is different from 65.)

At this time, you also select the alpha level. Traditionally, a is set at .05 or .01. If there

is particular concern about a Type I error, or if a researcher desires to present overwhelm-

ing evidence for a treatment effect, a smaller alpha level can be used (such as a = .001).

For this demonstration, we will set alpha to .05. Thus, we are taking a 5% risk of commit-

ting a Type I error.

STEP 2

Locate the critical region.

We begin by looking at all the possible outcomes that could be obtained if the null

hypothesis is true. This is the distribution of sample means for n = 25. The distribution

will be normal with an expected value of u = 65 (from Ho) and a standard error of

o

15

OM=

–

Vn

15

= 3

5

V25

With a = .05, we want to identify the most unlikely 5% of this distribution. Therefore,

we divide the alpha level between the two tails, with 2.5% or p = 0.0250 in each tail.

According to the unit normal table, the boundaries for the critical region are z-scores of

+1.96 and -1.96. The distribution and the critical region are shown in Figure 8.14.

STEP 3

Obtain the sample data, and compute the test statistic.

For this demonstration, the researcher obtained a sample mean of M = 70. This sample

mean corresponds to a z-score of

Z=

M-M _ 70 – 65

OM

3

= +1.67

228

CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

09

1

FIGURE 8.14

The critical region for

Demonstration 8.1 consists

of the extreme tails with

boundaries of x = -1.96 and

z = +1.96.

B!

Om=3

+

OM

D

M

5

65

17

1

0

-1.96

fe

1.96

50

1)

9.

STEP 4

Make a decision about Ho, and state the conclusion.

b.

sample

mean of M = 70 is not an extreme or unusual value to be obtained from a population with

= 65. Therefore, our statistical decision is to fail to reject Ho. Our conclusion for the

study is that the data do not provide sufficient evidence that the special training changes

test scores.

1. A

st

le

an

DEMONSTRATION 8.2

EFFECT SIZE USING COHEN’S d

P

P

9

We will compute Cohen’s d using the research situation and the data from Demonstration

8.1. Again, the original population mean was u = 65 and, after treatment (special train-

ing), the sample mean was M = 70. Thus, there is a 5-point mean difference. Using the

population standard deviation, o = 15, we obtain an effect size of

2

Cohen’s d =

mean difference

standard deviation

5

15

0.33

According to Cohen’s evaluation standards (see Table 8.2), this is a medium treatment

effect.

&

PROBLEMS

1. In the Z-score formula as it is used in a hypothesis test,

a. Explain what is measured by M – u in the numerator.

b. Explain what is measured by the standard error in

the denominator.

2. The value of the 2-score that is obtained for a hypothe-

sis test is influenced by several factors. Some factors

influence the size of the numerator of the Z-score and

other factors influence the size of the standard error in

the denominator. For each of the following, indicate

whether the factor influences the numerator or denom-

inator of the Z-score and determine whether the effect

would be to increase the value of z (farther from zero)

or decrease the value of z (closer to zero). In each

case, assume that all other components of the 2-score

remain constant.

a. Increase the sample size.

b. Increase the population standard deviation.

c. Increase the difference between the sample

mean and the value of u specified in the null

hypothesis.

14

CHAPTER 1 INTRODUCTION TO STATISTICS

an experiment uses only 10-year-old children as particpants (holding age constant), then

the researcher can be certain that one group is not noticeably older than another.

DEFINITION

In the experimental method, one variable is manipulated, while another vari-

able is observed and measured. To establish a cause-and-effect relationship

between the two variables, an experiment attempts to control all other variables

to prevent them from influencing the results.

Terminology in the experimental method Specific names are used for the two vari.

ables that are studied by the experimental method. The variable that is manipulated by

the experimenter is called the independent variable. It can be identified as the treatment

conditions to which participants are assigned. For the example in Figure 1.6, tempera-

ture is the independent variable. The variable that is observed to assess a possible ef-

fect of the manipulation is the dependent variable.

DEFINITIONS

The independent variable is the variable that is manipulated by the researcher.

In behavioral research, the independent variable usually consists of the two (or

more) treatment conditions to which subjects are exposed. The independent

variable consists of the antecedent conditions that were manipulated prior to

observing the dependent variable.

The dependent variable is the one that is observed in order to assess the effect

of the treatment.

In psychological research, the dependent variable is typically a measurement or score

obtained for each participant. For the temperature experiment (Figure 1.6), the dependent

variable is the number of words recalled on the memory test. Differences between groups

in performance on the dependent variable suggest that the manipulation had an effect. That

is, changes in the dependent variable depend on changes in the independent variable.

An experimental study evaluates the relationship between two variables by manipu-

lating one variable (the independent variable) and measuring one variable (the depen-

dent variable). Note that in an experiment only one variable is actually measured. You

should realize that this is very different from a correlational study, in which both vari-

ables are measured and the data consist of two separate scores for each individual.

Often an experiment will include a condition in which the subjects do not receive

any treatment. The scores from these subjects are then compared with scores from sub-

jects who do receive the treatment. The goal of this type of study is to demonstrate that

the treatment has an effect by showing that the scores in the treatment condition are

substantially different from the scores in the no-treatment condition. In this kind of re-

search, the no-treatment condition is called the control condition, and the treatment

condition is called the experimental condition.

DEFINITIONS

Individuals in a control condition do not receive the experimental treatment.

Instead, they either receive no treatment or they receive a neutral, placebo treat-

ment. The purpose of a control condition is to provide a baseline for compari-

son with the experimental condition.

Individuals in the experimental condition do receive the experimental treatment.

Note that the independent variable always consists of at least two values. (Something

must have at least two different values before you can say that it is “variable.”) For the

temperature experiment (Figure 1.6), the independent variable is temperature (using

65

SECTION 3.3 THE MEDIAN

original values has been multiplied by 3 (to change from yards to feet), and the result-

ing values total EX = 150, with M = 30 feet. Multiplying each score by 3 has also

caused the mean to be multiplied by 3. You should realize, however, that although the

numerical values for the individual scores and the sample mean have been multiplied,

the actual measurements are not changed.

Original Measurement

in Yards

Conversion to Feet

(Multiply by 3)

TABLE 3.3

Measurement of five pieces of

10

rope.

9

12

8

11

30

27

36

24

33

ΣΧ = 50

M = 10 yards

ΣΧ = 150

M = 30 feet

1. a. Compute the mean for the following sample of scores:

LEARNING CHECK

4

4.

6, 1, 8, 0, 5

10

5.

b. Add 4 points to each score, and then compute the mean.

c. Multiply each the original scores by 5, and then compute the mean.

2. A population has a mean of u = 80.

a. If 6 points were added to every score, what would be the value for the new mean?

b. If every score were multiplied by 2, what would be the value for the new mean?

3. A sample of n = 5 scores has a mean of M = 8. If you added a new score of

a

X = 2 to the sample, what value would you obtain for the new sample mean?

ANSWERS

1. a. M = 20 = 4 b. M = 40 = 8 c. M = 190 = 20

2. a. The new mean would be 86. b. The new mean would be 160.

3. The original sample has n = 5 and X = 40. With the new score, the sample has n = 6 and

X = 42. The new mean is 2 = 7.

3.3

THE MEDIAN

The second measure of central tendency we will consider is called the median. The me-

dian is the score that divides a distribution exactly in half. If the scores are listed in

order from smallest to largest, the median is the midpoint of the list.

DEFINITION

The median is the score that divides a distribution in half so that 50% of the

individuals in the distribution have scores at or below the median.

Earlier, when we introduced the mean, specific symbols and notation were used to

identify the mean and to differentiate a sample mean and a population mean. For the

median, however, there are no symbols or notation. Instead, the median is simply iden-

94

Slo

CHAPTER 4 VARIABILITY

CRAP

SS = Σχ2 – (ΣΧ)?

M

$

X

1

= 38 – (8)

4

1

0

6

1

0

36

1

64

38 –

4

= 38 – 16

ΣΧ = 8.

ΣΧ2 = 38

= 22

2

Note that the two formulas produce exactly the same value for SS. Although the for

mulas look different, they are in fact equivalent. The definitional formula provides the

most direct representation of the concept of SS, however, this formula can be awkward

to use, especially if the mean includes a fraction or decimal value. If you have a small

group of scores and the mean is a whole number, then the definitional formula is fine,

otherwise use the computational formula.

ROK

THE

3

VARIANCE AND STANDARD

DEVIATION

With the definition and calculation of SS behind you, the equations for variance and

standard deviation become relatively simple. Remember that variance is defined as the

mean squared deviation. The mean is the sum divided by N, so the equation for the pop-

ulation variance is

In the same way that sum of

squares, or SS, is used to refer to

the sum of squared deviations,

the term mean square, or MS, is

often used to refer to variance,

which is the mean squared

deviation

SS

variance =

N

Standard deviation is the square root of variance, so the equation for the population

standard deviation is

standard deviation =

SS

VN

There is one final bit of notation before we work completely through an example

computing SS, variance, and standard deviation. Like the mean (u), variance and stan-

dard deviation are parameters of a population and will be identified by Greek letters.

To identify the standard deviation, we use the Greek letter sigma (the Greek letters,

standing for standard deviation). The capital letter sigma (2) has been used already, 50

we now use the lowercase sigma, o, as the symbol for the population standard devia-

tion. To emphasize the relationship between standard deviation and variance, we use of

as the symbol for population variance (standard deviation is the square root of the vari-

ance). Thus,

population standard deviation = o =

VO

SS

VN

(4.3)

population variance =

o?

SS

N

(4.4)

Using the definitional formula for SS, the complete calculation of population variance

can be expressed as

population variance = 02

Σ(X – μ)

N

(4.5)

SECTION 8.3 | AN EXAMPLE OF A HYPOTHESIS TEST

205

FIGURE

8,7

Population

according to the

null hypothesis

Distribution of M

for n = 16

according to Ho

Normal

u =18

04

OM=1.7

Locating the critical region

as a three-step process. You

begin with the population of

scores that is predicted by the

null hypothesis. Then, you

construct the distribution of

sample means for the sample

size that is being used. The

distribution of sample means

corresponds to all the possi-

ble outcomes that could be

obtained if He is true.

Finally, you use z-scores to

separate the extreme out-

comes (as defined by the

alpha level) from the high-

probability outcomes. The

extreme values determine the

critical

region

u = 18

Decision criteria

for the

hypothesis test

Reject Ho

Middle 95%

High probability

values if Ho is true.

Fail to reject Ho

Reject Ho

Z=-1.96

Z=1.96

distribution, z-scores of z = +1.96 separate the middle 95% from the extreme

5% (a proportion of 0.0250 in each tail). Thus, we have identified the sample

means that, according to the null hypothesis, have a high probability of occur-

rence, and the sample means that are very unlikely to occur. It is the unlikely sam-

ple means, those with z-score values beyond +1.96, that form the critical region

for the test.

STEP 3

Collect the data, and compute the test statistic. At this point, we would select our

sample of n = 16 pups whose mothers had received alcohol during pregnancy. The

birth weight would be recorded for each pup and the sample mean computed. As

noted, we obtained a sample mean of M = 15 grams. The sample mean is then con-

verted to a z-score, which is our test statistic.

М – р

z =

15 – 18

1

-3

= -3.00

1

OM

STEP 4

Make a decision. The z-score computed in step 3 has a value of -3.00, which is be-

yond the boundary of -1.96. Therefore, the sample mean is located in the critical

region. This is a very unlikely outcome if the null hypothesis is true, so our decision

is to reject the null hypothesis. In addition to this statistical decision concerning the

null hypothesis, it is customary to state a conclusion about the results of the research

study. For this example, we conclude that prenatal exposure to alcohol does have a

significant effect on birth weight.

206

CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

IN THE LITERATURE

REPORTING THE RESULTS OF THE STATISTICAL TEST

FIGUR

Samp

critics

have

alpha

Host

Sam

falli

pra

,

tests. When you are reading a scientific journal, for example, you typically will be

A special jargon and notational system are used in published reports of hypothesis

told explicitly that the researcher evaluated the data using a z-score as a test statistics

with an alpha level of .05. Nor will you be told that “the null hypothesis is rejected.

Instead, you will see a statement such as

The treatment with alcohol had a significant effect on the birth weight of newborn ra

2 = 3.00, p

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