Understanding Simple Linear Regression: Statistical Technique in Review

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=dependentvariable(outcome)

“Weight in kg = (0.5 × age in months) + 4” and novel formula “Weight in kg = (0.502 × age in months) + 3.161” 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), p. 928.

145

147

150

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).

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

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.

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” (Darling-Fisher et al., 2014, 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” (Darling-Fisher et al., 2014, p. 222).

“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” (Darling-Fisher et al., 2014, 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” (Darling-Fisher et al., 2014, p. 221). The χ² results are presented in Table 2.

1. What is the sample size for the Darling-Fisher et al. (2014) study? How many study participants (percentage) are RAAPS users and how many are RAAPS nonusers?

2. What is the chi-square (χ²) value and degrees of freedom (df) for provider type?

3. What is the p value for provider type? Is the χ² value for provider type statistically significant? Provide a rationale for your answer.

4. Does a statistically significant χ² value provide evidence of causation between the variables? Provide a rationale for your answer.

5. What is the χ² value for race? Is the χ² 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.

7. What is the df for U.S. practice region? Complete the df formula for U.S. practice region to visualize how Darling-Fisher et al. (2014) 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 Darling-Fisher et al. (2014) in Table 2? Provide a rationale for your answer.

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 χ² = 12.7652 and df = 2 for provider type as presented in Table 2.

3. The p = < .00 for the provider type. Yes, the χ² = 12.7652 for provider type is statistically significant as indicated by the p value presented in Table 2. The specific χ² value obtained could be compared against the critical value in a χ² table (see Appendix D Critical Values of the χ² 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 χ² value does not provide evidence of causation. A statistically significant χ² value indicates a significant difference between groups exists but does not provide a causal link (Grove et al., 2013; Plichta & Kelvin, 2013).

5. The χ² = 1.2865 for race. Since p = .53 for race, the χ² 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.

9. The null hypothesis for years in practice stated in Questions 8 should be rejected. The χ² = 6.2597 for years in practice is statistically significant, p = .01. A statistically significant χ² 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 χ² 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.

Name: _______________________________________________________ Class: _____________________

Date: ___________________________________________________________________________________

1. According to the relevant study results section of the Darling-Fisher et al. (2014) study, what categories are reported to be statistically significant?

2. What level of measurement is appropriate for calculating the χ² statistic? Give two examples from Table 2 of demographic variables measured at the level appropriate for χ².

3. What is the χ² for U.S. practice region? Is the χ² 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.

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 Darling-Fisher et al. (2014).

9. How many null hypotheses are rejected in the Darling-Fisher et al. (2014) 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, χ² = 29.68. Does the χ² result provide the location of the difference? Provide a rationale for your answer

(Grove 191-200)

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.

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

TABLE 29-1

ENROLLMENT GPA AND MONTHS TO COMPLETION IN AN RN TO BSN PROGRAM

Step 1: Calculate b.
From the values in Table 29-1, we know that n = 20, Σx = 20, Σy = 267, Σx² = 30, and Σxy = 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=−2.9x+16.25

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).

Step 5: Determine whether the predictor significantly predicts y.

t=Rn−21−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=.638200−21−.407  ‾ ‾ ‾ ‾ ‾  √

t=3.52

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 R².
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 R² represents the percentage of variance explained in y by the predictor. Cohen describes R² 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 R² 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 R² 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).

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.”

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 = −2.9x + 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 = −2.9x + 16.25, compute the predicted months to program completion if a student’s number of earned degrees is 2. Show your calculations.

7. What was the correlation between the actual y values and the predicted 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 R² in the example? Provide a rationale for your answer.

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 values and the predicted 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 value at least as extreme as or as close to the one that was 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 R² 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.

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)?

5. Write the new regression equation.

6. How would you characterize the magnitude of the obtained R² 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

(Grove 319-332)

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.

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, χ² 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).

A	B
C	D

411

TABLE 35-1

ANTIBIOTIC RESISTANCE BY ANTIBIOTIC USE

	Antibiotic Use	No Recent Use
Resistant	8	7
Not resistant	6	21

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

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 =666,792158,760

χ 2 =4.20

Step 3: Compute the degrees of freedom:

df=(2−1)(2−1)=1

Step 4: Locate the critical χ² value in the χ² distribution table (Appendix D) and compare it to the obtained χ² value.

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.”

1. Do the example data meet the assumptions for the Pearson χ² 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 χ² value at least as extreme or as close to the one that was actually observed, assuming that the null hypothesis is true?

4. Using the numbers in the contingency table, calculate the percentage of antibiotic users who were resistant.

5. Using the numbers in the contingency table, calculate the percentage of non-antibiotic users who were resistant.

6. Using the numbers in the contingency table, calculate the percentage of resistant veterans who used antibiotics for more than 2 weeks.

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

1. Yes, the data meet the assumptions of the Pearson χ²:

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 χ² value at least as extreme as or as close to the one that was 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 χ² 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 χ² test (Goetz, Howard, Cipher, & Revankar, 2010). These data are presented in Table 35-2 as a contingency table.

Name: _______________________________________________________ Class: _____________________

Date: ___________________________________________________________________________________

1. Do the example data in Table 35-2 meet the assumptions for the Pearson χ² test? Provide a rationale for your answer.

2. Compute the χ² test. What is the χ² value?

3. Is the χ² significant at α = 0.05? Specify how you arrived at your answer.

4. If using SPSS, what is the exact likelihood of obtaining the χ² 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.

(Grove 409-420)

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.