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Exercise 29Calculating Simple Linear RegressionSimple 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 RegressionResearch 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 AssumptionsUse of simple linear regression involves the following assumptions (Zar, 2010):1. Normal distribution of the dependent (y) variable2. Linear relationship between x and y3. Independent observations4. No (or little) multicollinearity5. Homoscedasticity320Data 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=bx+awherey=the dependent variable(outcome)x=the independent variable(predictor)b=the slope of the linea=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 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. 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∑xy−∑x∑yn∑x 2 −(∑x) 2α=∑y−b∑xnHand CalculationsThis 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 321the 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 29-1ENROLLMENT GPA AND MONTHS TO COMPLETION IN AN RN TO BSN PROGRAMStudent IDxyx2xy(Number of Degrees)(Months to Completion)1117117229418301700419195016006111111701500801200911511510112112111141141211011013117117140200015294181621242417114114182104201911711720211422sum Σ2026730238The 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, 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 2b=−2.9Step 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)20α=16.25Step 3: Write the new regression equation_y=−2.9x+16.25322Step 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)+16.25y ̂ =13.35Thus, 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=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=3.52323The 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).324SPSS ComputationsThis 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.”325Interpretation of SPSS OutputThe following tables are generated from SPSS. The first table contains the multiple R and the R2 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.RegressionThe 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 0.002.326Final Interpretation in American Psychological Association (APA) FormatThe 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 R2 = 40.7%. Higher numbers of earned academic degrees significantly predicted shorter program completion time.327Study Questions1. 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.3287. 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.329Answers to Study Questions1. 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 25.330Data for Additional Computational Practice for the Questions to be GradedUsing 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 29-2AGE AT ENROLLMENT AND MONTHS TO COMPLETION IN AN RN TO BSN PROGRAMStudent IDxyx2xy(Student Age)(Months to Completion)1231752939122495762163241757640842696762345311696149663111961341732151,024480833121,089396933151,0894951034121,1564081134141,1564761235101,2253501335171,2255951439201,521780154091,6003601642121,7645041742141,7645881844101,9364401951172,601867202411576264sum Σ67726724,0059,089331EXERCISE 29 Questions to Be GradedName: _______________________________________________________ Class: _____________________Date: ___________________________________________________________________________________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.”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)?3325. 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 a rationale for your answer(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.Exercise 35Calculating Pearson Chi-SquareThe Pearson chi-square test (χ2) compares differences between groups on variables measured at the nominal level. The χ2 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 χ2 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 χ2 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 χ2 does not determine where the differences lie; it only determines that a significant difference exists. Further testing on pairs of groups with the χ2 would then be warranted to identify the significant differences.A two-way χ2 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 χ2 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 χ2 tests are reviewed in Exercise 19.Research Designs Appropriate for the Pearson χ2Research 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 χ2 calculations must be measured at the nominal level.410Statistical Formula and AssumptionsUse of the Pearson χ2 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, 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).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 χ2 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):ABCDWith any χ2 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_df=(R−1)(C−1)whereR=Number of rowsC=Number of columnsHand CalculationsA 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.411The 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 35-1ANTIBIOTIC RESISTANCE BY ANTIBIOTIC USEAntibiotic UseNo Recent UseResistant87Not resistant621The computations for the Pearson χ2 test are as follows:Step 1: Create a contingency table of the two nominal variables:Used AntibioticsNo Recent UseTotalsResistant8715Not resistant62127Totals142842←Total nStep 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 =666,792158,760χ 2 =4.20Step 3: Compute the degrees of freedom_df=(2−1)(2−1)=1Step 4: Locate the critical χ2 value in the χ2 distribution table (Appendix D) and compare it to the obtained χ2 value.The obtained χ2 value is compared with the tabled χ2 values in Appendix D. The table includes the critical values of χ2 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 χ2 table, the difference between the two variables is statistically significant. The critical χ2 for df = 1 is 3.84, and our obtained χ2 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%.412SPSS ComputationsThe 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.413Step 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.”414Interpretation of SPSS OutputThe following tables are generated from SPSS. The first table contains the contingency table, similar to Table 35-1 above. The second table contains the χ2 results.CrosstabsThe last table contains the χ2 value in addition to other statistics that test associations between nominal variables. The Pearson χ2 test is located in the first row of the table, which contains the χ2 value, df, and p value.Final Interpretation in American Psychological Association (APA) FormatThe following interpretation is written as it might appear in a research article, formatted according to APA guidelines (APA, 2010). A Pearson χ2 analysis indicated that antibiotic users had significantly higher rates of antibiotic resistance than those who did not use antibiotics, χ2(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.415Study Questions1. 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.4167. 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.417Answers to Study Questions1. 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 = 0.05.418Data for Additional Computational Practice for Questions to be GradedA 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.TABLE 35-2CANDIDURIA AND ANTIBIOTIC USE IN ADULTS WITH SPINAL CORD INJURIESCandiduriaNo CandiduriaTotalsAntibiotic use154358No antibiotic use03939Totals158297419EXERCISE 35 Questions to Be GradedName: _______________________________________________________ Class: _____________________Date: ___________________________________________________________________________________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.”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?4205. 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.Each exercise has 10 questions at the end which says Questions to be graded.I need those questions to be answer.