Can anyone help with 2 assignments, ten questions each.

Chapter 29 Questions to be graded:

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)?

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 values and the predicted 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 a rationale for

your answer.

Data Set for Chapter 29:

Chapter 29 Content:

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 infl uence 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

EXERCISE

29

320 EXERCISE 29 • Calculating Simple Linear Regression

Copyright © 2017, Elsevier Inc. All rights reserved.

Data that are homoscedastic are evenly dispersed both above and below the regression

line, which indicates a linear relationship on a scatterplot. Homoscedasticity refl ects 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 signifi cance 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 bx 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 infi nite number of lines through

the scattered paired values ( Zar, 2010 ). However, the purpose of the regression equation

is to develop the line to allow the highest degree of prediction possible—the

line of best fi t. The procedure for developing the line of best fi t 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 .



−

−





n xy x y

n x2 x 2



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

Calculating Simple Linear Regression • EXERCISE 29 321

Copyright © 2017, Elsevier Inc. All rights reserved.

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 .

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, Σ y = 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 30 202

( ) ( )( )

( )

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 .

−−267 2 9 20

20

( . )( )

16.25

Step 3: Write the new regression equation:

y −2.9x 16.25

TABLE 29-1 ENROLLMENT GPA AND MONTHS TO COMPLETION IN AN RN TO BSN PROGRAM

Student ID

x y

(Number of Degrees) (Months to Completion) x 2 xy

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

322 EXERCISE 29 • Calculating Simple Linear Regression

Copyright © 2017, Elsevier Inc. All rights reserved.

Step 4: Calculate R .

The multiple R is defi ned 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:

ˆy −2.9(1) 16.25

ˆy 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.

y (Months to completion)

19

18

17

16

15

14

13

12

11

10

0.0 0.5 1.0 1.5 2.0

x (Number of academic degrees earned)

ŷ = –2.9x + 16.25

Step 5: Determine whether the predictor signifi cantly predicts y .

t R

n

R

−

−

1 2

To know whether the predictor signifi cantly 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

Calculating Simple Linear Regression • EXERCISE 29 323

Copyright © 2017, Elsevier Inc. All rights reserved.

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 signifi cant association between the predictor ( x ) and outcome ( y ).

Step 6: Calculate R 2 .

After establishing the statistical signifi cance of the R value , it must subsequently be

examined for clinical importance. This is accomplished by obtaining the coeffi cient 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 R2 can be very helpful in testing more than one predictor in a regression model.

Unlike R , the R2 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 EXERCISE 29 • Calculating Simple Linear Regression

Copyright © 2017, Elsevier Inc. All rights reserved.

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.”

Calculating Simple Linear Regression • EXERCISE 29 325

Copyright © 2017, Elsevier Inc. All rights reserved.

INTERPRETATION OF SPSS OUTPUT

The following tables are generated from SPSS. The fi rst table contains the multiple R and

the R 2 values. The multiple R is 0.638, indicating that the correlation between the actual

y values and the predicted y values using the new regression equation is 0.638. The R2 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

Model Summary

Model R R Square Adjusted R

Square

Std. Error of the

Estimate

1 .638a .407 .374 2.608

Enrollment

Correlation between actual

y values and predicted y

values

Correlation between actual y

values and predicted y

values, squared

a. Predictors: (Constant), Number of Earned Academic Degrees at

ANOVAa

Model Sum of Squares df Mean Square F Sig.

Regression 84.100 1 84.100 12.363 .002b

Residual 122.450 18 6.803

Total 206.550 19

a. Dependent Variable: Number of Months to Complete Program

b. Predictors: (Constant), Number of Earned Academic Degrees at Enrollment

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 signifi cant amount of variance in y . The contents of the ANOVA

table are rarely reported in published manuscripts, because the signifi cance of each predictor

is presented in the last SPSS table titled “Coeffi cients” (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 fi rst 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 EXERCISE 29 • Calculating Simple Linear Regression

Copyright © 2017, Elsevier Inc. All rights reserved.

Coefficientsa

Model Unstandardized Coefficients Standardized

Coefficients

t Sig.

B Std. Error Beta

(Constant) 16.250 1.010 16.087 .000

Number of Earned

Academic Degrees at

Enrollment

-2.900 .825 -.638 -3.516 .002

Observe that the first row contains the a The exact p value is 0.002

and the second row contains the b

Standardized beta (b); note that

b is equal to R when there is

only one predictor in model

a Dependent Variable: Number of Months to Complete Program

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 signifi cantly 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 signifi cantly predicted shorter

program completion time.

Chapter 35 Questions to be graded:

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 signifi cant 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 value at least as extreme as or as

close to the one that was actually observed, assuming that the null hypothesis is true?

5. Using the numbers in the contingency table, calculate the percentage of antibiotic users who

tested positive for candiduria.

6. Using the numbers in the contingency table, calculate the percentage of non-antibiotic users

who tested positive for candiduria.

7. Using the numbers in the contingency table, calculate the percentage of veterans with candiduria

who had a history of antibiotic use.

8. Using the numbers in the contingency table, calculate the 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.

Chapter 35 Data set:

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 χ2 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 signifi cant, it would indicate that

proportions of one gender are signifi cantly 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

signifi cant difference exists. Further testing on pairs of groups with the χ 2 would then be

warranted to identify the signifi cant differences.

A two-way χ 2 is a statistic that tests whether proportions in levels of one nominal variable

are signifi cantly 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 χ2 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.

EXERCISE

35

Chapter 35 Content:

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 χ2 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:

2

−



n A D B C

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 χ2 analysis, the degrees of freedom (d f ) must be calculated to determine the

signifi cance of the value of the statistic. The following formula is used for this

calculation:

df 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.

Calculating Pearson Chi-Square • EXERCISE 35 411

Copyright © 2017, Elsevier Inc. All rights reserved.

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.”

The computations for the Pearson χ 2 test are as follows:

Step 1: Create a contingency table of the two nominal variables:

TABLE 35-1 ANTIBIOTIC RESISTANCE BY ANTIBIOTIC USE

Antibiotic Use No Recent Use

Resistant 8 7

Not resistant 6 21

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:

2

−



n A D B C

A B C D A C B D

2

2 42 8 21 7 6

8 7 6 21 8 6 7 21

−



2 666 792

158 760

,

2 4.20

Step 3: Compute the degrees of freedom:

df 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 specifi c degrees of freedom at selected levels of signifi

cance. If the value of the statistic is equal to or greater than the value identifi ed in the

χ 2 table, the difference between the two variables is statistically signifi cant. 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 signifi cant 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 EXERCISE 35 • Calculating Pearson Chi-Square

Copyright © 2017, Elsevier Inc. All rights reserved.

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.

Calculating Pearson Chi-Square • EXERCISE 35 413

Copyright © 2017, Elsevier Inc. All rights reserved.

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 EXERCISE 35 • Calculating Pearson Chi-Square

Copyright © 2017, Elsevier Inc. All rights reserved.

INTERPRETATION OF SPSS OUTPUT

The following tables are generated from SPSS. The fi rst table contains the contingency

table, similar to Table 35-1 above. The second table contains the χ 2 results.

Crosstabs

Resistant to Antibiotics * Antibiotic Use for 2 Weeks or More Crosstabulation

Count

Antibiotic Use for 2 Weeks or More Total

No Antibiotic

Use

History of >2

Weeks of

Antibiotic Use

Resistant to Antibiotics

Not Resistant 21 6 27

Resistant 7 8 15

Total 28 14 42

Chi-Square Tests

Value df Asymp. Sig. (2-

sided)

Exact Sig. (2-

sided)

Exact Sig. (1-

sided)

Pearson Chi-Square 4.200a 1 .040

Continuity Correctionb 2.917 1 .088

Likelihood Ratio 4.135 1 .042

Fisher's Exact Test .085 .045

Linear-by-Linear Association 4.100 1 .043

N of Valid Cases 42

Observe that the c² value is 4.20 The exact p value is .04

a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 5.00.

b. Computed only for a 2x2 table

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 fi rst 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 signifi cantly 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 fi nding 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.