Exercise 14Understanding Simple Linear RegressionStatistical Tec.docx

Exercise 14
Understanding Simple Linear Regression
Statistical Technique in Review
In nursing practice, the ability to predict future events or
outcomes is crucial, and researchers calculate and report linear
regression results as a basis for making these predictions.
Linear regression
provides a means to estimate or predict the value of a
dependent variable based on the value of one or more
independent variables. The regression equation is a
mathematical expression of a causal proposition emerging from
a theoretical framework. The linkage between the theoretical
statement and the equation is made prior to data collection and
analysis. Linear regression is a statistical method of estimating
the expected value of one variable,
y
, given the value of another variable,
x
. The focus of this exercise is
simple linear regression
, which involves the use of one independent variable,
x
, to predict one dependent variable,
y
.
The regression line developed from simple linear regression is
usually plotted on a graph, with the horizontal axis representing
x
(the independent or predictor variable) and the vertical axis
representing the
y
(the dependent or predicted variable; see
Figure 14-1
). The value represented by the letter
a

is referred to as the
y
intercept, or the point where the regression line crosses or
intercepts the
y
-axis. At this point on the regression line,
x
= 0. The value represented by the letter
b
is referred to as the slope, or the coefficient of
x
. The slope determines the direction and angle of the regression
line within the graph. The slope expresses the extent to which
y
changes for every one-unit change in
x
. The score on variable
y
(dependent variable) is predicted from the subject's known
score on variable
x
(independent variable). The predicted score or estimate is
referred to as
Ŷ
(expressed as
y
-hat) (
Cohen, 1988
;
Grove, Burns, & Gray, 2013
;
Zar, 2010
).
FIGURE 14-1

GRAPH OF A SIMPLE LINEAR REGRESSION LINE
140
Simple linear regression is an effort to explain the dynamics
within a scatterplot (see
Exercise 11
) by drawing a straight line through the plotted scores. No
single regression line can be used to predict, with complete
accuracy, every
y
value from every
x
value. However, the purpose of the regression equation is to
develop the line to allow the highest degree of prediction
possible, the
line of best fit
. The procedure for developing the line of best fit is the
method of least squares
. If the data were perfectly correlated, all data points would fall
along the straight line or line of best fit. However, not all data
points fall on the line of best fit in studies, but the line of best
fit provides the best equation for the values of
y
to be predicted by locating the intersection of points on the line
for any given value of
x
.
The algebraic equation for the regression line of best fit is
y
=
bx
+
a
, where:
y
=
dependent

variable
(
outcome
)
x
=
independent
variable
(
predictor
)
b
=
s
l
o
p
e
o
f
t
h
e
l
i
n
e
(
beta
,
or
what
the
increase

in
value
is
along
the
x
-axis
for
every
unit
of
increase
in
the
y
value
)
,
also
called
the
regression
coefficient
.
a
=
y
−
i
n
t
e
r
c

e
p
t
(
the
point
where
the
regression
line
intersects
the
y
-axis
)
,
also
called
the
regression
constant
(
Zar
,
2
0
1
0
)
.
In
Figure 14-2
, the
x

-axis represents Gestational Age in weeks and the
y
-axis represents Birth Weight in grams. As gestational age
increases from 20 weeks to 34 weeks, birth weight also
increases. In other words, the slope of the line is positive. This
line of best fit can be used to predict the birth weight
(dependent variable) for an infant based on his or her
gestational age in weeks (independent variable).
Figure 14-2
is an example of a line of best fit that was not developed from
research data. In addition, the
x
-axis was started at 22 weeks rather than 0, which is the usual
start in a regression figure. Using the formula
y
=
bx
+
a
, the birth weight of a baby born at 28 weeks of gestation is
calculated below.
Formula
:
y
=
b
x
+
a
In
this
example
,
a
=

500
,
b
=
20
,
and
x
=
28
weeks
y
=
20
(
28
)
+
500
=
560
+
500
=
1
,
060
grams
FIGURE 14-2
EXAMPLE LINE OF BEST FIT FOR GESTATIONAL AGE
AND BIRTH WEIGHT
141

The regression line represents
y
for any given value of
x
. As you can see, some data points fall above the line, and some
fall below the line. If we substitute any
x
value in the regression equation and solve for
y
, we will obtain a
ŷ
that will be somewhat different from the actual values. The
distance between the
ŷ
and the actual value of
y
is called
residual
, and this represents the degree of error in the regression line.
The regression line or the line of best fit for the data points is
the unique line that will minimize error and yield the smallest
residual (
Zar, 2010
). The step-by-step process for calculating simple linear
regression in a study is presented in
Exercise 29
.
Research Article
Source
Flannigan, C., Bourke, T. W., Sproule, A., Stevenson, M., &
Terris, M. (2014). Are APLS formulae for estimating weight
appropriate for use in children admitted to PICU?
Resuscitation, 85
(7), 927–931.
Introduction
Medications and other therapies often necessitate knowing a

patient's weight. However, a child may be admitted to a
pediatric intensive care unit (PICU) without a known weight,
and instability and on-going resuscitation may prevent obtaining
this needed weight. Clinicians would benefit from a tool that
could accurately estimate a patient's weight when such
information is unavailable. Thus Flannigan et al. (2014)
conducted a retrospective observational study for the purpose of
determining “if the revised APLS UK [Advanced Paediatric Life
Support United Kingdom] formulae for estimating weight are
appropriate for use in the paediatric care population in the
United Kingdom” (
Flannigan et al., 2014
, p. 927). The sample included 10,081 children (5,622 males and
4,459 females), who ranged from term-corrected age to 15 years
of age, admitted to the PICU during a 5-year period. Because
this was a retrospective study, no geographic location, race, and
ethnicity data were collected for the sample. A paired samples
t-
test was used to compare mean sample weights with the APLS
UK formula weight. The “APLS UK formula ‘weight = (0.05 ×
age in months) + 4’ significantly overestimates the mean weight
of children under 1 year admitted to PICU by between 10%
[and] 25.4%” (
, p. 928). Therefore, the researchers concluded that the APLS
UK formulas were not appropriate for estimating the weight of
children admitted to the PICU.
Relevant Study Results
“Simple linear regression was used to produce novel formulae
for the prediction of the mean weight specifically for the PICU
population” (
Flannigan et al.
, 2014, p. 927). The three novel formulas are presented in
Figures 1
,
2

, and
3
, respectively. The new formulas calculations are more complex
than the APLS UK formulas. “Although a good estimate of
mean weight can be obtained by our newly derived formula,
reliance on mean weight alone will still result in significant
error as the weights of children admitted to PICU in each age
and sex [gender] group have a large standard
deviation . . . Therefore as soon as possible after admission a
weight should be obtained, e.g., using a weight bed” (
, p. 929).
FIGURE 1
Comparison of actual weight with weight calculated using APLS
formula “Weight in kg = (0.5 × age in months) + 4” and novel
formula “Weight in kg = (0.502 × age in months) + 3.161”
Resuscitation, 85
(7), p. 928.
FIGURE 2
formula “Weight in kg = (2 × age in years) + 8” and novel
formula “Weight in kg = (0.176 × age in months) + 7.241”
Resuscitation, 85
(7), p. 928.

FIGURE 3
formula “Weight in kg = (3 × age in years) + 7” and novel
formula “Weight in kg = (0.331 × age in months) − 6.868”
Resuscitation, 85
(7), p. 929.
144
Study Questions
1.
What are the variables on the
x
- and
y
-axes in
Figure 1
from the Flannigan et al. (2014) study?
2.
What is the name of the type of variable represented by
x
and
y
in
Figure 1
? Is
x
or
y

the score to be predicted?
3.
What is the purpose of simple linear regression analysis and the
regression equation?
4.
What is the point where the regression line meets the
y-
axis called? Is there more than one term for this point and what
is the value of
x
at that point?
5.
In the formula
y
=
bx
+
a
, is
a
or
b
the slope? What does the slope represent in regression
analysis?
6.
Using the values
a
= 3.161 and
b
= 0.502 with the novel formula in
Figure 1
, what is the predicted weight in kilograms for a child at 5

months of age? Show your calculations.
145
7.
What are the variables on the
x
-axis and the
y
-axis in
Figures 2
and
3
? Describe these variables and how they might be entered into
the regression novel formulas identified in
Figures 2
and
3
.
8.
Using the values
a
= 7.241 and
b
Figure 2
, what is the predicted weight in kilograms for a child at 4 years
of age? Show your calculations.
9.
Does
Figure 1
have a positive or negative slope? Provide a rationale for your
answer. Discuss the meaning of the slope of
Figure 1
.

10.
According to the study narrative, why are estimated child
weights important in a pediatric intensive care (PICU) setting?
What are the implications of these findings for practice?
146
Answers to Study Questions
1.
The
x
variable is age in months, and the
y
variable is weight in kilograms in
Figure 1
.
2.
x
is the independent or predictor variable.
y
is the dependent variable or the variable that is to be predicted
by the independent variable,
x
.
3.
Simple linear regression is conducted to estimate or predict the
values of one dependent variable based on the values of one
independent variable. Regression analysis is used to calculate a
line of best fit based on the relationship between the
independent variable
x
and the dependent variable
y
. The formula developed with regression analysis can be used to

predict the dependent variable (
y
) values based on values of the independent variable
x
.
4.
The point where the regression line meets the
y
-axis is called the
y
intercept and is also represented by
a
(see
Figure 14-1
).
a
is also called the regression constant. At the
y
intercept,
x
= 0.
5.
b
is the slope of the line of best fit (see
Figure 14-1
). The slope of the line indicates the amount of change in
y
for each one unit of change in
x
.
b
is also called the regression coefficient.
6.

Use the following formula to calculate your answer:
y
=
bx
+
a
y
= 0.502 (5) + 3.161 = 2.51 + 3.161 = 5.671 kilograms
Note: Flannigan et al. (2014) expressed the novel formula of
weight in kilograms = (0.502 × age in months) + 3.161 in the
title of
Figure 1
.
7.
Age in years is displayed on the
x
-axis and is used for the APLS UK formulas in
Figures 2
and
3
.
Figure 2
includes children 1 to 5 years of age, and
Figure 3
includes children 6 to 12 years of age. However, the novel
formulas developed by simple linear regression are calculated
with age in months. Therefore, the age in years must be
converted to age in months before calculating the
y
values with the novel formulas provided for
Figures 2
and
3
. For example, a child who is 2 years old would be converted to
24 months (2 × 12 mos./year = 24 mos.). Then the formulas in

Figures 2
and
3
could be used to predict
y
(weight in kilograms) for the different aged children. The
y
-axis on both
Figures 2
and
3
is weight in kilograms (kg).
8.
First calculate the child's age in months, which is 4 × 12
months/year = 48 months.
y
=
bx + a
= 0.176 (48) + 7.241 = 8.448 + 7.241 = 15.689 kilograms
Note the
x
value needs to be in age in months and Flannigan et al. (2014)
expressed the novel formula of weight in kilograms = (0.176 ×
age in months) + 7.241.
147
9.
Figure 1
has a positive slope since the line extends from the lower left
corner to the upper right corner and shows a positive
relationship. This line shows that the increase in
x
(independent variable) is associated with an increase in
y
(dependent variable). In the Flannigan et al. (2014) study, the

independent variable age in months is used to predict the
dependent variable of weight in kilograms. As the age in months
increases, the weight in kilograms also increases, which is the
positive relationship illustrated in
Figure 1
.
10.
According to Flannigan et al. (2014, p. 927), “The gold standard
for prescribing therapies to children admitted to Paediatric
Intensive Care Units (PICU) requires accurate measurement of
the patient's weight. . . . An accurate weight may not be
obtainable immediately because of instability and on-going
resuscitation. An accurate tool to aid the critical care team
estimate the weight of these children would be a valuable
clinical tool.” Accurate patient weights are an important factor
in preventing medication errors particularly in pediatric
populations. The American Academy of Pediatrics (AAP)'s
policy on Prevention of Medication Errors in the Pediatric
Inpatient Setting can be obtained from the following website:
https://www.aap.org/en-us/advocacy-and-policy/federal-
advocacy/Pages/Federal-
Advocacy.aspx#SafeandEffectiveDrugsandDevicesforChildren
. The Centers for Medicare & Medicaid Services, Partnership
for Patients provides multiple links to Adverse Drug Event
(ADE) information including some resources specific to
pediatrics at
http://partnershipforpatients.cms.gov/p4p_resources/tsp-
adversedrugevents/tooladversedrugeventsade.html
.
149
EXERCISE 14
Questions to Be Graded
Follow your instructor's directions to submit your answers to
the following questions for grading. Your instructor may ask

you to write your answers below and submit them as a hard
copy for grading. Alternatively, your instructor may ask you to
use the space below for notes and submit your answers online at
http://evolve.elsevier.com/Grove/Statistics/
under “Questions to Be Graded.”
Name:
_____________________________________________________
__ Class: _____________________
Date:
_____________________________________________________
______________________________
1.
According to the study narrative and
Figure 1
in the Flannigan et al. (2014) study, does the APLS UK formula
under- or overestimate the weight of children younger than 1
year of age? Provide a rationale for your answer.
2.
Using the values
a
= 3.161 and
b
Figure 1
, what is the predicted weight in kilograms (kg) for a child at 9
3.
Using the values
a
= 3.161 and
b
Figure 1

, what is the predicted weight in kilograms for a child at 2
4.
In
Figure 2
, the formula for calculating
y
(weight in kg) is Weight in kg = (0.176 × Age in months) +
7.241. Identify the
y
intercept and the slope in this formula.
150
5.
Using the values
a
= 7.241 and
b
Figure 2
, what is the predicted weight in kilograms for a child 3 years of
age? Show your calculations.
6.
Using the values
a
= 7.241 and
b
Figure 2
, what is the predicted weight in kilograms for a child 5 years of
age? Show your calculations.
7.
In

Figure 3
, some of the actual mean weights represented by blue line with
squares are above the dotted straight line for the novel formula,
but others are below the straight line. Is this an expected
finding? Provide a rationale for your answer.
8.
In
Figure 3
, the novel formula is (weight in kilograms = (0.331 × Age in
months) − 6.868. What is the predicted weight in kilograms for
a child 10 years old? Show your calculations.
9.
Was the sample size of this study adequate for conducting
simple linear regression? Provide a rationale for your answer.
10.
Describe one potential clinical advantage and one potential
clinical problem with using the three novel formulas presented
in
Figures 1
,
2
, and
3
in a PICU setting.
(Grove 139-150)
Grove, Susan K., Daisha Cipher.
Statistics for Nursing Research: A Workbook for Evidence-
Based Practice, 2nd Edition
. Saunders, 022016. VitalBook file.
The citation provided is a guideline. Please check each citation
for accuracy before use.
Exercise 19

Understanding Pearson Chi-Square
Statistical Technique in Review
The
Pearson Chi-square (
χ
2
)
is an inferential statistical test calculated to examine
differences among groups with variables measured at the
nominal level. There are different types of χ
2
tests and the Pearson chi-square is commonly reported in
nursing studies. The Pearson χ
2
test compares the frequencies that are observed with the
frequencies that were expected. The assumptions for the χ
2
test are as follows:
1.
The data are nominal-level or frequency data.
2.
The sample size is adequate.
3.
The measures are independent of each other or that a subject's
data only fit into one category (
Plichta & Kelvin, 2013
).
The χ
2
values calculated are compared with the critical values in the χ
2
table (see

Appendix D
Critical Values of the χ
2
Distribution at the back of this text). If the result is greater
than or equal to the value in the table, significant differences
exist. If the values are statistically significant, the null
hypothesis is rejected (
Grove, Burns, & Gray, 2013
). These results indicate that the differences are probably an
actual reflection of reality and not just due to random sampling
error or chance.
In addition to the χ
2
value, researchers often report the degrees of freedom (
df
). This mathematically complex statistical concept is important
for calculating and determining levels of significance. The
standard formula for
df
is sample size (
N
) minus 1, or
df
=
N
− 1; however, this formula is adjusted based on the analysis
technique performed (
). The
df
formula for the χ
2
test varies based on the number of categories examined in the
analysis. The formula for
df
for the two-way χ

2
test is
df
= (
R
− 1) (
C
− 1), where
R
is number of rows and
C
is the number of columns in a χ
2
table. For example, in a 2 × 2 χ
2
table,
df
= (2 − 1) (2 − 1) = 1. Therefore, the
df
is equal to 1.
Table 19-1
includes a 2 × 2 chi-square contingency table based on the
findings of An et al. (2014) study. In
Table 19-1
, the rows represent the two nominal categories of alcohol
192
use and alcohol nonuse and the two columns represent the two
nominal categories of smokers and nonsmokers. The
df
= (2 − 1) (2 − 1) = (1) (1) = 1, and the study results were as
follows: χ
2
(1,
N
= 799) = 63.1;
p

< 0.0001. It is important to note that the
df
can also be reported without the sample size, as in χ
2
(1) = 63.1,
p
< 0.0001.
TABLE 19-1
CONTINGENCY TABLE BASED ON THE RESULTS OF AN
ET AL. (2014) STUDY
Nonsmokers
n
= 742
Smokers
n
= 57
*
No alcohol use
551
14
Alcohol use
†
191
43
*
Smokers defined as “smoking at least 1 cigarette daily during
the past month.”
†
Alcohol use “defined as at least 1 alcoholic beverage per month
during the past year.”
An, F. R., Xiang, Y. T., Yu., L., Ding, Y. M., Ungvari, G. S.,
Chan, S. W. C., et al. (2014). Prevalence of nurses' smoking
habits in psychiatric and general hospitals in China.
Archives of Psychiatric Nursing, 28
(2), 120.

If more than two groups are being examined, χ
2
does not determine where the differences lie; it only determines
that a statistically significant difference exists. A post hoc
analysis will determine the location of the difference. χ
2
is one of the weaker statistical tests used, and results are
usually only reported if statistically significant values are
found. The step-by-step process for calculating the Pearson chi-
square test is presented in Exercise 35.
Research Article
Source
Darling-Fisher, C. S., Salerno, J., Dahlem, C. H. Y., & Martyn,
K. K. (2014). The Rapid Assessment for Adolescent Preventive
Services (RAAPS): Providers' assessment of its usefulness in
their clinical practice settings.
Journal of Pediatric Health Care, 28
(3), 217–226.
Introduction
Darling-Fisher and colleagues (2014
, p. 219) conducted a mixed-methods descriptive study to
evaluate the clinical usefulness of the Rapid Assessment for
Adolescent Preventative Services (RAAPS) screening tool “by
surveying healthcare providers from a wide variety of clinical
settings and geographic locations.” The study participants were
recruited from the RAAPS website to complete an online
survey. The RAAPS risk-screening tool “was developed to
identify the risk behaviors contributing most to adolescent
morbidity, mortality, and social problems, and to provide a
more streamlined assessment to help providers address key
adolescent risk behaviors in a time-efficient and user-friendly
format” (
Darling-Fisher et al., 2014
, p. 218). The RAAPS is an established 21-item questionnaire
with evidence of reliability and validity that can be completed
by adolescents in 5–7 minutes.

“Quantitative and qualitative analyses indicated the RAAPS
facilitated identification of risk behaviors and risk discussions
and provided efficient and consistent assessments; 86% of
providers believed that the RAAPS positively influenced their
practice” (
, p. 217). The researchers concluded the use of RAAPS by
healthcare providers could improve the assessment and
identification of adolescents at risk and lead to the delivery of
more effective adolescent preventive services.
Relevant Study Results
In the
Darling-Fisher et al. (2014
, p. 220) mixed-methods study, the participants (
N
= 201) were “providers from 26 U.S. states and three foreign
countries (Canada, Korea, and Ireland).” More than half of the
participants (
n
= 111; 55%) reported they were using the RAAPS in their
clinical practices. “When asked if they would recommend the
RAAPS to other providers, 86 responded, and 98% (
n
= 84) stated they would recommend RAAPS. The two most
common reasons cited for their recommendation were for
screening (
n
= 76, 92%) and identification of risk behaviors (
n
= 75, 90%). Improved communication (
n
= 52, 63%) and improved documentation (
n
= 46, 55%) and increased patient understanding of their risk
behaviors (
n

= 48, 58%) were also cited by respondents as reasons to
recommend the RAAPS” (
, p. 222).
193
“Respondents who were not using the RAAPS (
n
= 90; 45%), had a variety of reasons for not using it. Most
reasons were related to constraints of their health system or
practice site; other reasons were satisfaction with their current
method of assessment . . . and that they were interested in the
RAAPS for academic or research purposes rather than clinical
use” (
, p. 220).
Chi-square analysis was calculated to determine if any
statistically significant differences existed between the
characteristics of the RAAPS users and nonusers.
Darling-Fisher et al. (2014)
did not provide a level of significance or α for their study, but
the standard for nursing studies is α = 0.05. “Statistically
significant differences were noted between RAAPS users and
nonusers with respect to provider types, practice setting,
percent of adolescent patients, years in practice, and practice
region. No statistically significant demographic differences
were found between RAAPS users and nonusers with respect to
race, age” (
, p. 221). The χ
2
results are presented in
Table 2
.
TABLE 2
DEMOGRAPHIC COMPARISONS BETWEEN RAPID
ASSESSMENT FOR ADOLESCENT PREVENTIVE SERVICE

USERS AND NONUSERS
Current user
Yes (%)
No (%)
χ
2
p
Provider type (
n
= 161)
12.7652,
df
= 2
< .00
Health care provider
64 (75.3)
55 (72.4)
Mental health provider
13 (15.3)
2 (2.6)
Other
8 (9.4)
19 (25.0)
Practice setting (
n
= 152)
12.7652,
df
= 1
< .00
Outpatient health clinic
20 (24.1)

36 (52.2)
School-based health clinic
63 (75.9)
33 (47.8)
% Adolescent patients (
n
= 154)
7.3780,
df
= 1
.01
≤50%
26 (30.6)
36 (52.2)
>50%
59 (69.4)
33 (47.8)
Years in practice (
n
= 157)
6.2597,
df
= 1
.01
≤5 years
44 (51.8)
23 (31.9)
>5 years
41 (48.2)
49 (68.1)
U.S. practice region (
n

= 151)
29.68,
df
= 3
< .00
Northeastern United States
13 (15.3)
15 (22.7)
Southern United States
11 (12.9)
22 (33.3)
Midwestern United States
57 (67.1)
16 (24.2)
Western United States
4 (4.7)
13 (19.7)
Race (
n
= 201)
1.2865,
df
= 2
.53
Black/African American
11 (9.9)
5 (5.6)
White/Caucasian
66 (59.5)
56 (62.2)
Other
34 (30.6)

29 (32.2)
Provider age in years (
n
= 145)
4.00,
df
= 2
.14
20–39 years
21 (25.6)
8 (12.7)
40–49 years
24 (29.3)
19 (30.2)
50+ years
37 (45.1)
36 (57.1)
χ
2
, Chi-square statistic.
df
, degrees of freedom.
Darling-Fisher, C. S., Salerno, J., Dahlem, C. H. Y., & Martyn,
K. K. (2014). The Rapid Assessment for Adolescent Preventive
Services (RAAPS): Providers' assessment of its usefulness in
their clinical practice settings.
Journal of Pediatric Health Care, 28
(3), p. 221.
194
Study Questions
1.
What is the sample size for the
study? How many study participants (percentage) are RAAPS

users and how many are RAAPS nonusers?
2.
What is the chi-square (χ
2
) value and degrees of freedom (
df
) for provider type?
3.
What is the
p
value for provider type? Is the χ
2
value for provider type statistically significant? Provide a
rationale for your answer.
4.
Does a statistically significant χ
2
value
provide evidence of causation between the variables? Provide a
rationale for your answer.
5.
What is the χ
2
value for race? Is the χ
2
value statistically significant? Provide a rationale for your
answer.
6.
Is there a statistically significant difference between RAAPS
users and RAAPS nonusers with regard to percentage adolescent
patients? In your own opinion is this an expected finding?
Document your answer.
195
7.

What is the
df
for U.S. practice region? Complete the
df
formula for U.S. practice region to visualize how
determined the appropriate
df
for that region.
8.
State the null hypothesis for the years in practice variable for
RAAPS users and RAAPS nonusers.
9.
Should the null hypothesis for years in practice developed for
Question 8 be accepted or rejected? Provide a rationale for your
answer.
10.
How many null hypotheses were accepted by
in
Table 2
? Provide a rationale for your answer.
196
1.
The sample size is
N
= 201 with
n
= 111 (55%) RAAPS users and
n

= 90 (45%) RAAPS nonusers as indicated in the narrative
results.
2.
The χ
2
= 12.7652 and
df
= 2 for provider type as presented in
Table 2
.
3.
The
p
= < .00 for the provider type. Yes, the χ
2
= 12.7652 for provider type is statistically significant as
indicated by the
p
value presented in
Table 2
. The specific χ
2
value obtained could be compared against the critical value in a
χ
2
table (see
Appendix D
Critical Values of the χ
2
Distribution at the back of this text) to determine the
significance for the specific degrees of freedom (
df
), but readers of research reports usually rely on the
p

value provided by the researcher(s) to determine significance.
Most nurse researchers set the level of significance or alpha (α)
= 0.05. Since the
p
value is less than alpha, the result is statistically significant.
You need to note that
p
values never equal zero as they appear in this study. The
p
values would not be zero if carried out more decimal places.
4.
No, a statistically significant χ
2
value does not provide evidence of causation. A statistically
significant χ
2
value indicates a significant difference between groups exists
but does not provide a causal link (
Grove et al., 2013
;
).
5.
The χ
2
= 1.2865 for race. Since
p
= .53 for race, the χ
2
value is not statistically significant. The level of significance is
set at α = 0.05 and the
p
value is larger than alpha, so the result is nonsignificant.
6.

Yes, there is a statistically significant difference between
RAAPS users and RAAPS nonusers with regard to percent of
adolescent patients. The chi-square value = 7.3780 with a
p
= .01.You might expect that nurses caring for more adolescents
might have higher RAAPS use as indicated in
Table 2
. However, nurses need to be knowledgeable of assessment and
care needs of populations and subpopulations in their practice
even if not frequently encountered. Two valuable sources for
adolescent care include the Centers for Disease Control and
Prevention (CDC) Adolescent and School Health at
http://www.cdc.gov/HealthyYouth/idex.htm
and the World Health Organization (WHO) adolescent health at
http://www.who.int/topics/adolescent_health/en/
.
7.
The
df =
3 for U.S. practice region is provided in
Table 2
. The
df
formula,
df
= (
R
− 1) (
C
− 1) is used. There are four “
R
” rows, Northeastern United States, Southern United States,
Midwestern United States, and Western United States. There are
two “
C
” columns, RAAPS users and RAAPS nonusers.

df =
(4 − 1)(2 − 1) = (3)(1) = 3.
8.
The null hypothesis: There is no difference between RAAPS
users and RAAPS nonusers for providers with ≤5 years of
practice and those with >5 years of practice.
197
9.
The null hypothesis for years in practice stated in Questions 8
should be rejected. The χ
2
= 6.2597 for years in practice is statistically significant,
p
= .01. A statistically significant χ
2
indicates a significant difference exists between the users and
nonusers of RAAPS for years in practice; therefore, the null
hypothesis should be rejected.
10.
Two null hypotheses were accepted since two χ
2
values (race and provider age) were not statistically significant
(
p
> 0.05), as indicated in
Table 2
. Nonsignificant results indicate that the null hypotheses are
supported or accepted as an accurate reflection of the results of
the study
.
199
EXERCISE 19

Name:
_____________________________________________________
__ Class: _____________________
Date:
_____________________________________________________
______________________________
1.
According to the relevant study results section of the
study, what categories are reported to be statistically
significant?
2.
What level of measurement is appropriate for calculating the χ
2
statistic? Give two examples from
Table 2
of demographic variables measured at the level appropriate for
χ
2
.
3.
What is the χ
2
for U.S. practice region? Is the χ
2
value statistically significant? Provide a rationale for your
answer.
4.

What is the
df
for provider type? Provide a rationale for why the
df
for provider type presented in
Table 2
is correct.
200
5.
Is there a statistically significant difference for practice setting
between the Rapid Assessment for Adolescent Preventive
Services (RAAPS) users and nonusers? Provide a rationale for
your answer.
6.
State the null hypothesis for provider age in years for RAAPS
users and RAAPS nonusers.
7.
Should the null hypothesis for provider age in years developed
for Question 6 be accepted or rejected? Provide a rationale for
your answer.
8.
Describe at least one clinical advantage and one clinical
challenge of using RAAPS as described by
.
9.
How many null hypotheses are rejected in the
study for the results presented in
Table 2

? Provide a rationale for your answer.
10.
A statistically significant difference is present between RAAPS
users and RAAPS nonusers for U.S. practice region, χ
2
= 29.68. Does the χ
2
result provide the location of the difference? Provide a
rationale for your answer
(Grove 191-200)
Exercise 29
Calculating Simple Linear Regression
Simple linear regression
is a procedure that provides an estimate of the value of a
dependent variable (outcome) based on the value of an
independent variable (predictor). Knowing that estimate with
some degree of accuracy, we can use regression analysis to
predict the value of one variable if we know the value of the
other variable (
Cohen & Cohen, 1983
). The regression equation is a mathematical expression of the
influence that a predictor has on a dependent variable, based on
some theoretical framework. For example, in
Exercise 14
,
Figure 14-1
illustrates the linear relationship between gestational age and

birth weight. As shown in the scatterplot, there is a strong
positive relationship between the two variables. Advanced
gestational ages predict higher birth weights.
A regression equation can be generated with a data set
containing subjects'
x
and
y
values. Once this equation is generated, it can be used to
predict future subjects'
y
values, given only their
x
values. In simple or bivariate regression, predictions are made
in cases with two variables. The score on variable
y
(dependent variable, or outcome) is predicted from the same
subject's known score on variable
x
(independent variable, or predictor).
Research Designs Appropriate for Simple Linear Regression
Research designs that may utilize simple linear regression
include any associational design (
Gliner et al., 2009
). The variables involved in the design are attributional,
meaning the variables are characteristics of the participant, such
as health status, blood pressure, gender, diagnosis, or ethnicity.
Regardless of the nature of variables, the dependent variable
submitted to simple linear regression must be measured as
continuous, at the interval or ratio level.
Statistical Formula and Assumptions
Use of simple linear regression involves the following
assumptions (
Zar, 2010
):
1.

Normal distribution of the dependent (
y
) variable
2.
Linear relationship between
x
and
y
3.
Independent observations
4.
No (or little) multicollinearity
5.
Homoscedasticity
320
Data that are
homoscedastic
are evenly dispersed both above and below the regression line,
which indicates a linear relationship on a scatterplot.
Homoscedasticity reflects equal variance of both variables. In
other words, for every value of
x
, the distribution of
y
values should have equal variability. If the data for the
predictor and dependent variable are not homoscedastic,
inferences made during significance testing could be invalid (
Cohen & Cohen, 1983
;
Zar, 2010
). Visual examples of homoscedasticity and heteroscedasticity

are presented in
Exercise 30
.
In simple linear regression, the dependent variable is
continuous, and the predictor can be any scale of measurement;
however, if the predictor is nominal, it must be correctly coded.
Once the data are ready, the parameters
a
and
b
are computed to obtain a regression equation. To understand
the mathematical process, recall the algebraic equation for a
straight line:
y
=
b
x
+
a
where
y
=
the dependent variable
(
outcome
)
x
=
the independent variable
(
predictor
)
b

=
the slope of the line
a
=
y
-intercept
(
the point where the regression line intersects the
y
-axis
)
No single regression line can be used to predict with complete
accuracy every
y
value from every
x
value. In fact, you could draw an infinite number of lines
through the scattered paired values (
Zar, 2010
). However, the purpose of the regression equa-tion is to
develop the line to allow the highest degree of prediction
possible—the line of best fit. The procedure for developing the
line of best fit is the
method of least squares
. The formulas for the beta (β) and slope (α) of the regression
equation are computed as follows. Note that once the β is
calculated, that value is inserted into the formula for α.
β
=
n
∑
x
y
−

∑
x
∑
y
n
∑
x
2
−
(
∑
x
)
2
α
=
∑
y
−
b
∑
x
n
Hand Calculations
This example uses data collected from a study of students
enrolled in a registered nurse to bachelor of science in nursing
(RN to BSN) program (
Mancini, Ashwill, & Cipher, 2014

). The predictor in this example is number of academic degrees
obtained by the student prior to enrollment, and the dependent
variable was number of months it took for the student to
complete the RN to BSN program. The null hypothesis is
“Number of degrees does not predict the number of months until
completion of an RN to BSN program.”
The data are presented in
Table 29-1
. A simulated subset of 20 students was selected for this
example so that the computations would be small and
manageable. In actuality, studies involving linear regression
need to be adequately powered (
Aberson, 2010
;
Cohen, 1988
). Observe that the data in
Table 29-1
are arranged in columns that correspond to
321
the elements of the formula. The summed values in the last row
of
Table 29-1
are inserted into the appropriate place in the formula for
b
.
TABLE 29-1
ENROLLMENT GPA AND MONTHS TO COMPLETION IN
AN RN TO BSN PROGRAM
Student ID
x
y
x
2
xy
(Number of Degrees)
(Months to Completion)

1
1
17
1
17
2
2
9
4
18
3
0
17
0
0
4
1
9
1
9
5
0
16
0
0
6
1
11
1
11
7
0
15
0
0
8

0
12
0
0
9
1
15
1
15
10
1
12
1
12
11
1
14
1
14
12
1
10
1
10
13
1
17
1
17
14
0
20
0
0
15
2

9
4
18
16
2
12
4
24
17
1
14
1
14
18
2
10
4
20
19
1
17
1
17
20
2
11
4
22
sum Σ
20
267
30
238
The computations for the
b
and α are as follows:

Step 1: Calculate
b
.
From the values in
Table 29-1
, we know that
n =
20, Σ
x
= 20,
Σy
= 267, Σ
x
2
= 30, and Σ
xy
= 238. These values are inserted into the formula for
b
, as follows:
b
=
20
(
238
)
−
(
20
)
(
267
)
20
(
30
)

−
20
2
b
=
−
2.9
Step 2: Calculate α.
From Step 1, we now know that
b
= −2.9, and we plug this value into the formula for α.
α
=
267
−
(
−
2.9
)
(
20
)
20
α
=
16.25
Step 3: Write the new regression equation:
y

=
−
2
.9
x
+
16
.25
322
Step 4: Calculate
R
.
The multiple
R
is defined as the correlation between the actual
y
values and the predicted
y
values using the new regression equation. The predicted
y
value using the new equation is represented by the symbol
ŷ
to differentiate from
y,
which represents the actual
y
values in the data set. We can use our new regression equation
from Step 3 to compute predicted program completion time in
months for each student, using their number of academic
degrees prior to enrollment in the RN to BSN Program. For
example, Student #1 had earned 1 academic degree prior to
enrollment, and the predicted months to completion for Student
1 is calculated as:
y

̂
=
−
2.9
(
1
)
+
16.25
y
̂
=
13.35
Thus, the predicted
ŷ
is 13.35 months. This procedure would be continued for the
rest of the students, and the Pearson correlation between the
actual months to completion (
y
) and the predicted months to completion (
ŷ
) would yield the multiple
R
value. In this example, the
R
= 0.638. The higher the
R
, the more likely that the new regression equation accurately
predicts
y
, because the higher the correlation, the closer the actual

y
values are to the predicted
ŷ
values.
Figure 29-1
displays the regression line where the
x
axis represents possible numbers of degrees, and the
y
axis represents the predicted months to program completion (
ŷ
values).
FIGURE 29-1
REGRESSION LINE REPRESENTED BY NEW REGRESSION
EQUATION.
Step 5: Determine whether the predictor significantly predicts
y
.
t
=
R
n
−
2
1
−
R
2
‾
‾

‾
‾
√
To know whether the predictor significantly predicts
y
, the beta must be tested against zero. In simple regression, this
is most easily accomplished by using the
R
value from Step 4:
t
=
.638
200
−
2
1
−
.407
‾
‾
‾
‾
‾
√
t
=
3
.52

323
The
t
value is then compared to the
t
probability distribution table (see
Appendix A
). The
df
for this
t
statistic is
n
− 2. The critical
t
value at alpha (α) = 0.05,
df
= 18 is 2.10 for a two-tailed test. Our obtained
t
was 3.52, which exceeds the critical value in the table, thereby
indicating a significant association between the predictor (
x
) and outcome (
y
).
Step 6: Calculate
R
2
.
After establishing the statistical significance of the
R
value
,
it must subsequently be examined for clinical importance. This
is accomplished by obtaining the

coefficient of determination
for regression—which simply involves squaring the
R
value. The
R
2
represents the percentage of variance explained in
y
by the predictor. Cohen describes
R
2
values of 0.02 as small, 0.15 as moderate, and 0.26 or higher as
large effect sizes (
Cohen, 1988
). In our example, the
R
was 0.638, and, therefore, the
R
2
was 0.407. Multiplying 0.407 × 100% indicates that 40.7% of
the variance in months to program completion can be explained
by knowing the student's number of earned academic degrees at
admission (
Cohen & Cohen, 1983
).
The
R
2
can be very helpful in testing more than one predictor in a
regression model. Unlike
R
, the
R
2
for one regression model can be compared with another
regression model that contains additional predictors (

Cohen & Cohen, 1983
). The
R
2
is discussed further in
Exercise 30
.
The standardized beta (β) is another statistic that represents the
magnitude of the association between
x
and
y
. β has limits just like a Pearson
r
, meaning that the standardized β cannot be lower than −1.00 or
higher than 1.00. This value can be calculated by hand but is
best computed with statistical software. The standardized beta
(β) is calculated by converting the
x
and
y
values to
z
scores and then correlating the
x
and
y
value using the Pearson
r
formula. The standardized beta (β) is often reported in
literature instead of the unstandardized
b
, because
b
does not have lower or upper limits and therefore the
magnitude of

b
cannot be judged. β, on the other hand, is interpreted as a
Pearson
r
and the descriptions of the magnitude of β can be applied, as
recommended by
Cohen (1988)
. In this example, the standardized beta (β) is −0.638. Thus, the
magnitude of the association between
x
and
y
in this example is considered a large predictive association (
Cohen, 1988
).
324
SPSS Computations
This is how our data set looks in SPSS.
Step 1: From the “Analyze” menu, choose “Regression” and
“Linear.”
Step 2: Move the predictor, Number of Degrees, to the space
labeled “Independent(s).” Move the dependent variable, Number
of Months to Completion, to the space labeled “Dependent.”
Click “OK.”
325
Interpretation of SPSS Output
The following tables are generated from SPSS. The first table
contains the multiple
R
and the
R
2
values. The multiple
R

is 0.638, indicating that the correlation between the actual
y
y
values using the new regression equation is 0.638. The
R
2
is 0.407, indicating that 40.7% of the variance in months to
program completion can be explained by knowing the student's
number of earned academic degrees at enrollment.
Regression
The second table contains the ANOVA table. As presented in
Exercises 18
and
33
, the ANOVA is usually performed to test for differences
between group means. However, ANOVA can also be performed
for regression, where the null hypothesis is that “knowing the
value of
x
explains no information about
y
”. This table indicates that knowing the value of
x
explains a significant amount of variance in
y
. The contents of the ANOVA table are rarely reported in
published manuscripts, because the significance of each
predictor is presented in the last SPSS table titled
“Coefficients” (see below).
The third table contains the
b
and
a
values, standardized beta (β),
t

, and exact
p
value. The
a
is listed in the first row, next to the label “Constant.” The β is
listed in the second row, next to the name of the predictor. The
remaining information that is important to extract when
interpreting regression results can be found in the second row.
The standardized beta (β) is −0.638. This value has limits just
like a Pearson
r
, meaning that the standardized β cannot be lower than −1.00 or
higher than 1.00. The
t
value is −3.516, and the exact
p
value is 0.002.
326
Final Interpretation in American Psychological Association
(APA) Format
The following interpretation is written as it might appear in a
research article, formatted according to APA guidelines (
APA, 2010
). Simple linear regression was performed with number of
earned academic degrees as the predictor and months to
program completion as the dependent variable. The student's
number of degrees significantly predicted months to completion
among students in an RN to BSN program, β = −0.638,
p
= 0.002, and
R
2
= 40.7%. Higher numbers of earned academic degrees
significantly predicted shorter program completion time.
327
Study Questions

1.
If you have access to SPSS, compute the Shapiro-Wilk test of
normality for months to completion (as demonstrated in
Exercise 26
). If you do not have access to SPSS, plot the frequency
distributions by hand. What do the results indicate?
2.
State the null hypothesis for the example where number of
degrees was used to predict time to BSN program completion.
3.
In the formula
y = bx + a,
what does “
b
” represent?
4.
In the formula
y = bx + a,
what does “
a
” represent?
5.
Using the new regression equation,
ŷ
= −2.9
x
+ 16.25, compute the predicted months to program completion
if a student's number of earned degrees is 0. Show your
calculations.
6.

Using the new regression equation,
ŷ
= −2.9
x
+ 16.25, compute the predicted months to program completion
if a student's number of earned degrees is 2. Show your
calculations.
328
7.
What was the correlation between the actual
y
y
values using the new regression equation in the example?
8.
What was the exact likelihood of obtaining a
t
value at least as extreme as or as close to the one that was
actually observed, assuming that the null hypothesis is true?
9.
How much variance in months to completion is explained by
knowing the student's number of earned degrees?
10.
How would you characterize the magnitude of the
R
2
in the example? Provide a rationale for your answer.
329
1.
The Shapiro-Wilk

p
value for months to RN to BSN program completion was 0.16,
indicating that the frequency distribution did not significantly
deviate from normality. Moreover, visual inspection of the
frequency distribution indicates that months to completion is
approximately normally distributed. See SPSS output below for
the histograms of the distribution:
2.
The null hypothesis is: “The number of earned academic
degrees does not predict the number of months until completion
of an RN to BSN program.”
3.
In the formula
y = bx + a,
“
b
” represents the slope of the regression line.
4.
In the formula
y = bx + a,
“
a
” represents the
y
-intercept, or the point at which the regression line intersects
the
y
-axis.
5.
The predicted months to program completion if a student's
number of academic degrees is 0 is calculated as:

ŷ
= −2.9(0) + 16.25 = 16.25 months.
6.
The predicted months to program completion if a student's
number of academic degrees is 2 is calculated as:
ŷ
= −2.9(2) + 16.25 = 10.45 months.
7.
The correlation between the actual
y
y
values using the new regression equation in the example, also
known as the multiple
R,
is 0.638.
8.
The exact likelihood of obtaining a
t
actually observed, assuming that the null hypothesis is true, was
0.2%. This value was obtained by looking at the SPSS output
table titled “Coefficients” in the last value of the column
labeled “Sig.”
9.
40.7% of the variance in months to completion is explained by
knowing the student's number of earned academic degrees at
enrollment.
10.
The magnitude of the
R

2
in this example, 0.407, would be considered a large effect
according to the effect size tables in
Exercises 24
and
25
.
330
Data for Additional Computational Practice for the Questions to
be Graded
Using the example from
Mancini and colleagues (2014)
, students enrolled in an RN to BSN program were assessed for
demographics at enrollment. The predictor in this example is
age at program enrollment, and the dependent variable was
number of months it took for the student to complete the RN to
BSN program. The null hypothesis is: “Student age at
enrollment does not predict the number of months until
completion of an RN to BSN program.” The data are presented
in
Table 29-2
. A simulated subset of 20 students was randomly selected for
this example so that the computations would be small and
manageable.
TABLE 29-2
AGE AT ENROLLMENT AND MONTHS TO COMPLETION
IN AN RN TO BSN PROGRAM
Student ID
x
y
x
2
xy
(Student Age)
(Months to Completion)
1

23
17
529
391
2
24
9
576
216
3
24
17
576
408
4
26
9
676
234
5
31
16
961
496
6
31
11
961
341
7
32
15
1,024
480
8
33

12
1,089
396
9
33
15
1,089
495
10
34
12
1,156
408
11
34
14
1,156
476
12
35
10
1,225
350
13
35
17
1,225
595
14
39
20
1,521
780
15
40
9

1,600
360
16
42
12
1,764
504
17
42
14
1,764
588
18
44
10
1,936
440
19
51
17
2,601
867
20
24
11
576
264
sum Σ
677
267
24,005
9,089
331
EXERCISE 29
Name:

_____________________________________________________
__ Class: _____________________
Date:
_____________________________________________________
______________________________
1.
If you have access to SPSS, compute the Shapiro-Wilk test of
normality for the variable age (as demonstrated in
Exercise 26
). If you do not have access to SPSS, plot the frequency
distributions by hand. What do the results indicate?
2.
State the null hypothesis where age at enrollment is used to
predict the time for completion of an RN to BSN program.
3.
What is
b
as computed by hand (or using SPSS)?
4.
What is
a
as computed by hand (or using SPSS)?
332
5.

Write the new regression equation.
6.
How would you characterize the magnitude of the obtained
R
2
value? Provide a rationale for your answer.
7.
How much variance in months to RN to BSN program
completion is explained by knowing the student's enrollment
age?
8.
What was the correlation between the actual
y
y
values using the new regression equation in the example?
9.
Write your interpretation of the results as you would in an APA-
formatted journal.
10.
Given the results of your analyses, would you use the calculated
regression equation to predict future students' program
completion time by using enrollment age as
x
? Provide
(Grove 319-332)

Exercise 35
Calculating Pearson Chi-Square
The
Pearson chi-square test (χ
2
)
compares differences between groups on variables measured at
the nominal level. The χ
2
compares the frequencies that are observed with the
frequencies that are expected. When a study requires that
researchers compare proportions (percentages) in one category
versus another category, the χ
2
is a statistic that will reveal if the difference in proportion is
statistically improbable.
A
one-way χ
2
is a statistic that compares different levels of one variable only.
For example, a researcher may collect information on gender
and compare the proportions of males to females. If the one-way
χ
2
is statistically significant, it would indicate that proportions of
one gender are significantly higher than proportions of the other
gender than what would be expected by chance (
Daniel, 2000
). If more than two groups are being examined, the χ
2
does not determine where the differences lie; it only determines
that a significant difference exists. Further testing on pairs of
groups with the χ

2
would then be warranted to identify the significant differences.
A
two-way χ
2
is a statistic that tests whether proportions in levels of one
nominal variable are significantly different from proportions of
the second nominal variable. For example, the presence of
advanced colon polyps was studied in three groups of patients:
those having a normal body mass index (BMI), those who were
overweight, and those who were obese (
Siddiqui, Mahgoub, Pandove, Cipher, & Spechler, 2009
). The research question tested was: “Is there a difference
between the three groups (normal weight, overweight, and
obese) on the presence of advanced colon polyps?” The results
of the χ
2
test indicated that a larger proportion of obese patients fell into
the category of having advanced colon polyps compared to
normal weight and overweight patients, suggesting that obesity
may be a risk factor for developing advanced colon polyps.
Further examples of two-way χ
2
tests are reviewed in
Exercise 19
.
Research Designs Appropriate for the Pearson χ
2
Research designs that may utilize the Pearson χ
2
include the randomized experimental, quasi-experimental, and
comparative designs (
Gliner, Morgan, & Leech, 2009
). The variables may be active, attributional, or a combination
of both. An active variable refers to an intervention, treatment,

or program. An attributional variable refers to a characteristic
of the participant, such as gender, diagnosis, or ethnicity.
Regardless of the whether the variables are active or
attributional, all variables submitted to χ
2
calculations must be measured at the nominal level.
410
Statistical Formula and Assumptions
Use of the Pearson χ
2
involves the following assumptions (
Daniel, 2000
):
1.
Only one datum entry is made for each subject in the sample.
Therefore, if repeated measures from the same subject are being
used for analysis, such as pretests and posttests, χ
2
is not an appropriate test.
2.
The variables must be categorical (nominal), either inherently
or transformed to categorical from quantitative values.
3.
For each variable, the categories are mutually exclusive and
exhaustive. No cells may have an
expected
frequency of zero. In the actual data, the
observed
cell frequency may be zero. However, the Pearson χ
2
test is sensitive to small sample sizes, and other tests, such as
the Fisher's exact test, are more appropriate when testing very
small samples (

Daniel, 2000
;
Yates, 1934
).
The test is distribution-free, or nonparametric, which means that
no assumption has been made for a normal distribution of values
in the population from which the sample was taken (
Daniel, 2000
).
The formula for a two-way χ
2
is:
χ
2
=
n
[
(
A
)
(
D
)
−
(
B
)
(
C
)
]
2

(
A
+
B
)
(
C
+
D
)
(
A
+
C
)
(
B
+
D
)
The contingency table is labeled as follows. A contingency
table is a table that displays the relationship between two or
more categorical variables (
Daniel, 2000
):
A
B
C
D
With any χ
2
analysis, the degrees of freedom (
df
) must be calculated to determine the significance of the value

of the statistic. The following formula is used for this
calculation:
d
f
=
(
R
−
1
)
(
C
−
1
)
where
R
=
Number of rows
C
=
Number of columns
Hand Calculations
A retrospective comparative study examined whether longer
antibiotic treatment courses were associated with increased
antimicrobial resistance in patients with spinal cord injury (
Lee et al., 2014
). Using urine cultures from a sample of spinal cord–injured
veterans, two groups were created: those with evidence of
antibiotic resistance and those with no evidence of antibiotic
resistance. Each veteran was also divided into two groups based
on having had a history of recent (in the past 6 months)
antibiotic use for more than 2 weeks or no history of recent

antibiotic use.
411
The data are presented in
Table 35-1
. The null hypothesis is: “There is no difference between
antibiotic users and non-users on the presence of antibiotic
resistance.”
TABLE 35-1
ANTIBIOTIC RESISTANCE BY ANTIBIOTIC USE
Antibiotic Use
No Recent Use
Resistant
8
7
Not resistant
6
21
The computations for the Pearson χ2 test are as follows:
Step 1: Create a contingency table of the two nominal variables:
Used Antibiotics
No Recent Use
Totals
Resistant
8
7
15
Not resistant
6
21
27
Totals

14
28
42
←Total
n
Step 2: Fit the cells into the formula:
χ
2
=
n
[
(
A
)
(
D
)
−
(
B
)
(
C
)
]
2
(
A
+
B
)
(

C
+
D
)
(
A
+
C
)
(
B
+
D
)
χ
2
=
42
[
(
8
)
(
21
)
−
(
7
)
(
6
)

]
2
(
8
+
7
)
(
6
+
21
)
(
8
+
6
)
(
7
+
21
)
χ
2
=
666
,
792
158
,

760
χ
2
=
4.20
Step 3: Compute the degrees of freedom:
d
f
=
(
2
−
1
)
(
2
−
1
)
=
1
Step 4: Locate the critical χ
2
value in the χ
2
distribution table (
Appendix D
) and compare it to the obtained χ
2
value.

The obtained χ
2
value is compared with the tabled χ
2
values in
Appendix D
. The table includes the critical values of χ
2
for specific degrees of freedom at selected levels of
significance. If the value of the statistic is equal to or greater
than the value identified in the χ
2
table, the difference between the two variables is statistically
significant. The critical χ
2
for
df
= 1 is 3.84, and our obtained χ
2
is 4.20, thereby exceeding the critical value and indicating a
significant difference between antibiotic users and non-users on
the presence of antibiotic resistance.
Furthermore, we can compute the rates of antibiotic resistance
among antibiotic users and non-users by using the numbers in
the contingency table from Step 1. The antibiotic resistance rate
among the antibiotic users can be calculated as 8 ÷ 14 = 0.571 ×
100% = 57.1%. The antibiotic resistance rate among the non-
antibiotic users can be calculated as 7 ÷ 28 = 0.25 × 100% =
25%.
412
SPSS Computations
The following screenshot is a replica of what your SPSS
window will look like. The data for subjects 24 through 42 are
viewable by scrolling down in the SPSS screen.
413

Step 1: From the “Analyze” menu, choose “Descriptive
Statistics” and “Crosstabs.” Move the two variables to the right,
where either variable can be in the “Row” or “Column” space.
Step 2: Click “Statistics” and check the box next to “Chi-
square.” Click “Continue” and “OK.”
414
Interpretation of SPSS Output
The following tables are generated from SPSS. The first table
contains the contingency table, similar to
Table 35-1
above. The second table contains the χ
2
results.
Crosstabs
The last table contains the χ
2
value in addition to other statistics that test associations
between nominal variables. The Pearson χ
2
test is located in the first row of the table, which contains the χ
2
value,
df
, and
p
value.
Final Interpretation in American Psychological Association
(APA) Format
The following interpretation is written as it might appear in a
research article, formatted according to APA guidelines (
APA, 2010
). A Pearson χ

2
analysis indicated that antibiotic users had significantly higher
rates of antibiotic resistance than those who did not use
antibiotics, χ
2
(1) = 4.20,
p
= 0.04 (57.1% versus 25%, respectively). This finding suggests
that extended antibiotic use may be a risk factor for developing
resistance, and further research is needed to investigate
resistance as a direct effect of antibiotics.
415
Study Questions
1.
Do the example data meet the assumptions for the Pearson χ
2
test? Provide a rationale for your answer.
2.
What is the null hypothesis in the example?
3.
What was the exact likelihood of obtaining a χ
2
value at least as extreme or as close to the one that was
4.
Using the numbers in the contingency table, calculate the
percentage of antibiotic users who were resistant.
5.
percentage of non-antibiotic users who were resistant.
6.

percentage of resistant veterans who used antibiotics for more
than 2 weeks.
416
7.
percentage of resistant veterans who had no history of antibiotic
use.
8.
What kind of design was used in the example?
9.
What result would have been obtained if the variables in the
SPSS Crosstabs window had been switched, with Antibiotic Use
being placed in the “Row” and Resistance being placed in the
“Column”?
10.
Was the sample size adequate to detect differences between the
two groups in this example? Provide a rationale for your
answer.
417
1.
Yes, the data meet the assumptions of the Pearson χ
2
:
a.
Only one datum per participant was entered into the
contingency table, and no participant was counted twice.
b.

Both antibiotic use and resistance are categorical (nominal-level
data).
c.
For each variable, the categories are mutually exclusive and
exhaustive. It was not possible for a participant to belong to
both groups, and the two categories (recent antibiotic user and
non-user) included all study participants.
2.
The null hypothesis is: “There is no difference between
antibiotic users and non-users on the presence of antibiotic
resistance.”
3.
The exact likelihood of obtaining a χ
2
actually observed, assuming that the null hypothesis is true, was
4.0%.
4.
The percentage of antibiotic users who were resistant is
calculated as 8 ÷ 14 = 0.5714 × 100% = 57.14% = 57.1%.
5.
The percentage of non-antibiotic users who were resistant is
calculated as 7 ÷ 28 = 0.25 × 100% = 25%.
6.
The percentage of antibiotic-resistant veterans who used
antibiotics for more than 2 weeks is calculated as 8 ÷ 15 =
0.533 × 100% = 53.3%.
7.

The percentage of resistant veterans who had no history of
antibiotic use is calculated as 6 ÷ 27 = 0.222 × 100% = 22.2%.
8.
The study design in the example was a retrospective
comparative design (
Gliner et al., 2009
).
9.
Switching the variables in the SPSS Crosstabs window would
have resulted in the exact same χ
2
result.
10.
The sample size was adequate to detect differences between the
two groups, because a significant difference was found,
p
= 0.04, which is smaller than alpha = 0.05.
418
Data for Additional Computational Practice for Questions to be
Graded
A retrospective comparative study examining the presence of
candiduria (presence of
Candida
species in the urine) among 97 adults with a spinal cord injury
is presented as an additional example. The differences in the use
of antibiotics were investigated with the Pearson χ
2
test (
Goetz, Howard, Cipher, & Revankar, 2010
). These data are presented in
Table 35-2
as a contingency table.
TABLE 35-2

CANDIDURIA AND ANTIBIOTIC USE IN ADULTS WITH
SPINAL CORD INJURIES
Candiduria
No Candiduria
Totals
Antibiotic use
15
43
58
No antibiotic use
0
39
39
Totals
15
82
97
419
EXERCISE 35
Name:
_____________________________________________________
__ Class: _____________________
Date:
_____________________________________________________
______________________________
http://evolve.elsevier.com/Grove/statistics/
1.

Do the example data in
Table 35-2
meet the assumptions for the Pearson χ
2
test? Provide a rationale for your answer.
2.
Compute the χ
2
test. What is the χ
2
value?
3.
Is the χ
2
significant at α = 0.05? Specify how you arrived at your
answer.
4.
If using SPSS, what is the exact likelihood of obtaining the χ
2
420
5.
percentage of antibiotic users who tested positive for
candiduria.
6.
percentage of non-antibiotic users who tested positive for
candiduria.
7.

percentage of veterans with candiduria who had a history of
antibiotic use.
8.
percentage of veterans with candiduria who had no history of
antibiotic use.
9.
Write your interpretation of the results as you would in an APA-
formatted journal.
10.
Was the sample size adequate to detect differences between the
two groups in this example? Provide a rationale for your
answer.
(Grove 409-420)
Need answers of questions to be graded at the end of each
exercise.

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