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Salkind, N. J. (2013). Statistics for people who (think they) hate statistics (5th ed.). SAGE

  • Chapter 5: Correlation
  • Chapter 15: Correlation
  • Chapter 16: Regression
  • Chapter 17: Frequency Association

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4/15/2015 Thinking about correlations and regression M. L. Zajicek-Farber, MSW, PhD CUA-NCSSS Shahan #112 Spring 2015 – Updated 4.15.2015 1   Working with a linear relationships between two variables (test scores with levels of income reward per week). What direction is implied by this relationship? Notice the positive direction between variables – positive correlation ! 100 90 80 Stats Test Score 70 60 50 40 30 20 10 0 $0 $20 $40 $60 $80 Reward in dollars per week of studying 2 1 4/15/2015 What direction of the correlation is implied by these results between these two variables? Stats Test Score 100 90 80 70 60 50 40 30 20 10 0 $0 $20 $40 $60 $80 Reward in dollars per week of studying Notice the curvilinear direction between variables – not a linear correlation ! 3 What direction of the correlation is implied by these results between these two variables? Stats Test Score 100 90 80 70 60 50 40 30 20 10 0 $0 $20 $40 $60 $80 Reward in dollars per week of studying Notice the negative – inverse direction between variables – negative correlation! 4 2 4/15/2015 What direction of the correlation is implied by these results between these two variables? Stats Test Score 100 90 80 70 60 50 40 30 20 10 0 $0 $20 $40 $60 $80 Reward in dollars per week of studying Notice the “NO” relationship between variables – no correlation ! 5 Pearson Correlation coefficient r A linear relationship between 2 variables:  Pearson correlation is denoted by statistic “r”  Both variables have to be “numbers” or continuous level of measurement (or interval or ratio level of measurement)  Data should be gathered from a parametric probability based (random) sample representative of a population.  Both variables should be normally distributed!  If these conditions cannot be met (for representative sample or normal distribution):  use Spearman Rank Order correlation coefficient test rho:   6 3 4/15/2015 Different types of correlation coefficients Level of Variable Measurement Type of Correlation Example Variable X Variable Y Nominal Nominal Phi Coefficient Voting Preference and Gender Nominal-Ordinal Ordinal Rank biserial coefficient Social class (low, medium, high) and Rank standing in high school Nominal Interval or Ratio Point biserial Type of support from family member (mother, father) and GPA in school Ordinal (rank) Ordinal (rank) Spearman rank coefficient Height in percentile rank and weight in percentile rank Interval/Ratio Interval/Ratio Pearson correlation coefficient Number of problems solved and age in 7 years Correlation in Bivariate Statistics  Univariate & Bivariate Statistics for Numeric Data ◦ Frequency distribution, mean, mode, range, standard deviation ◦ Numeric correlation between two variables  Correlation ….examines …whether  Warning: ◦ - the linear pattern of relationship between one variable (x) and another variable (y) – a linear correlation between two continuous variables ◦ - relative position of one variable correlates with relative distribution of another variable along a “straight” line on a graph ◦ A correlation can be shown with a graphical representation of the relationship between two variables by a straight line ◦ Correlation does not show proof of causality! ◦ Correlation cannot assume x causes y! 8 4 4/15/2015 Correlation: How much scores correlate linearly? IV = “resilience scores”  DV = self-esteem scores”  Selfesteem Resilience Correlation Or covariation of scores 9 Pearson’s Correlation Coefficient  “r” indicates… ◦ strength of relationship (strong, weak, or none) ◦ direction of relationship  positive (direct) – variables move in same direction  negative (inverse) – variables move in opposite directions  r ranges in value from –1.0 to +1.0 -1.0 Strong Negative 0.0 No Rel. +1.0 Strong Positive 10 5 4/15/2015 Direction of correlation  Positive (direct) The values of scores of both variables increase in the same direction (both are high or both are low). Or  Negative (indirect, or inverse) Scores of one variable increases while the scores of the other variable decreases. (when one is of high value the other is of low value). Or 11 What direction is the correlation?  A shorter length of hospitalization was related to a fewer number of symptoms in patients with acute heart disease. _positive _negative  As clients levels of anxiety increased, their level of problem solving decreased. _ positive _negative  As clients’ satisfaction decreased, the number of their complaints regarding dining services increased. _ positive _negative  The older the client’s age level, the greater the participation in socialization activities. _ positive _negative 12 6 4/15/2015 Pearson correlation r  Tests “linearity” of association between 2 variables  r ranges from -1 to 0 to +1 Note: the sign + or – indicates the direction Strength of correlation r = : Memorize the moderate value “0” …..means no correlation of r = .40 to .69 .01 to .39 = “weak” correlation .40 to .69 = “moderate” relationship .70 to .90 = “strong” correlation 1.00 or -1.00 = “identical /perfect relationship  13 What is the strength of the correlation r: r = .69  r = -.15  r = -.35  r = .55  r = -.78 1. Which is the weakest correlation? (-.15) 2. Which is the strongest correlation? (-.78) 3. Which correlations are positive and moderate in strength? (.69, and .55) 4. Which correlations are weak and negative? (-.15, and -.35)  14 7 4/15/2015 Hypothesis Testing: 1. 2. 3. 4. 5. 6. 7. 8. 15 State the Null hypothesis Ho: State the research tail-hypothesis, Find rcritical at p=.05 from the Table (when p-value is not provided for r-value!) Use N-2 = df Compare rcritical with robtained : If the rcritical in the table is < robtained : then Reject the Null Ho If the rcritical in the table is > robtained : Do not reject the Null Ho Draw conclusions about the correlation ** When the p-value for statistical significance is provided, base your interpretation on whether the p < .05 for evaluating whether the r-value is statistically significant! Considering r_critical: Using a Table Imagine that a study found a correlation of r = .69 between “test score GPA” and the “amount of dollar reward” per week given out by 10 randomly selected parents to improve youth GPA.  robtained = .69 and N = 10  rcritical : figure out df = N - 2 [df = 10 -2 = 8]  Go into Salkind Table B4 for critical values of r  rcritical (df=8) = .6319 at p = .05 (2-tailed)  rcritical (df=8) = .5494 at p = .05 (1-tailed)  Note: Some tables for rcritical give you N instead of df…..and this table has already been adjusted.  Decision for r = .69 ?  16 8 4/15/2015 This tables provide N with correct values 1-Tailed 17 Notice N 2-Tailed Notice df 1-tail 18 9 4/15/2015 Let’s figure out the result for r =.69. r (calculated) = 0.69  r (critical from table) at 2 tail = 0.6319 at df =8  r (critical from table) at 1 tail = 0.5494 at df =8       Compare r (critical) with r(calculated) Fro 2 Tail Ho: 0.6319 < 0.69 For 1 Tail Ho: 0.5494 < 0.69 Make conclusions: Since r(calculated) is “bigger” > than the table value r(critical) – then reject the Null Ho) hypothesis and conclude that the calculated r of 0.69 is a statistically significant result at both 2-tail and 1tail research hypotheses at p < 0.05. 19 Note: On statistical significance! If the analysis gives you r-value (correlation result) with a p-value such as  r = 0.35 at p = 0.012 for correlation between “resilience” and “self-esteem” - then –  Use the p-value to decide whether the correlation is statistically significant.  That is, here p = 0.012 - which is less than p < 0.05 and therefore, you can conclude that the r of 0.35 is statistically significant at p < 0.05!  20 10 4/15/2015 Note: On statistical significance! If the analysis gives you r-value (correlation result) without a p-value such as  r = 0.69 for correlation between “GPA scores” and “no. of rewards” - then –  Use the table for determining the critical rvalue at 2-tail or 1-tail at p = 0.05, and  Then – compare r(calculated) against r(critical) and if “calculated value r” > “table critical value r” – then Reject Ho, and conclude that the calculated r is statistically significant at p < 0.05.  21 Once you have a significant r-result (correlation coefficient), then that result must be interpreted covering several specific areas: • Direction of the correlation r coefficient • Interpretation of the direction in a normal English language meaning • Magnitude of r • Coefficient of determination or computation of rsquare • Explanation of shared variance (or r-square x 100%)22 11 4/15/2015 For a statistically significant (p < .05) correlation Indicate:  Direction of the relationship (positive or negative)  Interpret the direction in normal English language  Indicate the strength or magnitude of the correlation coefficient r  Calculate the Coefficient of Determination (r-square) and multiply it by 100%.  Interpret the coefficient of determination as the percent of shared variance that is explained in the significant correlation of the two variables. 23 Interpret r = 0.69 at p < 0.05 Direction: Positive  Meaning: Higher the students’ test scores are correlated with higher values of students’ rewards! Or, lower the students’ test scores are correlated with lower students’ rewards.  Magnitude of r: the value of 0.69 reflects a “moderate” magnitude of the strength of this correlation.  Coefficient of Determination ……continued on the next slide  24 12 4/15/2015 Note: r2 Linear = .69 x .69 = .479  r2 is a Coefficient of Determination!  Captures the shared amount variance in the covariation of the scores of X and Y variables.  Multiply r2 by 100% in order to get the percent of variance in a correlation coefficient explained by the coefficient of determination!  Coefficient of determination is 47.9% or the correlation between “GPA” and “Number of rewards” explains close to 48% of shared variance. 25 More Information on r-square r2  In a regular bivariate correlation, r2 represents just the percent of shared variance in the correlation of the two variables.  However, when we designate one of the variables as IV and the other as DV, then r2 actually is explains the percent of variance in the DV that can be specifically accounted for by the variance of IV.  This concept of IV variance explaining DV variance becomes important in a “prediction” or “statistical regression analysis.” 26 13 4/15/2015 Notice: Correlation and prediction and causation are separate concepts! Correlation refers to bi-variate correlation between 2 variables.  Prediction refers to “regression analysis” – which is an extension of correlation analysis with 2 or more variables for the specific purpose of explaining “which IV variables predict the DV variable.  Causation looks to a design of the study – and whether or not the study has (or has not) some form of experimental design!  27 Correlation and Causality !!! Correlation between two variables measures the relationship or correlation between two variables!  In NO way does correlation imply any causality between two variables!   However, the correlation that uses prediction analysis called “regression” analysis allows for stronger inference about the relationship between variables when one is designated as the X (independent variable) and the other as the Y (dependent variable). 28 14 4/15/2015 So, let’s look at another study:  A study examined a correlation between job satisfaction and burnout of 218 randomly selected employees in an organization. This study found a correlation:  r = - 0.409 at p < 0.001  The data are presented in a Table on the next slide!  29 A study examined a correlation between job satisfaction and30 burnout of 218 randomly selected employees in an organization. The data are presented in a Table below: Descriptive Statistics Mean Std. Deviation N Degree of Job Satisfaction 66.1606 8.82534 218 Degree of Burnout 21.5688 6.30466 218 Correlations Job Satisfaction Job Satisfaction Pearson Correlation 1 Sig. (2-tailed) **. Correlation is significant at the 0.01 level (2-tailed). 218 Pearson Correlation r= -.405** Sig. (2-tailed) p= .000 N -.405** .000 N Burnout Burnout 218 Note: Sig = .000 is p < .001 ! 218 1 218 15 4/15/2015 What is r for “satisfaction” with “burnout”? r = - .405 ** at p < .001  Note: ** indicates that this correlation is “statistically significant”  1) What is the direction of this correlation? negative 2) Translate the direction into “normal English” language to provide normal meaning! more satisfaction less burnout 3) What is the strength (magnitude) of this correlation? moderate strength 4) What is the coefficient of determination? r2 = .405 x .405 = 0.164025 x 100% = 16.40% 5) Explain the meaning of the coefficient of determination? This significant correlation explains 16% of variance that is shared between these two variable scores. 31 Limitations of Correlation • linearity: – can’t describe non-linear relationships – e.g., relation between anxiety & performance • truncation of range: – underestimate strength of relationship if you can’t see full range of x value • no proof of causation – third variable problem: • could be 3rd variable causing change in both variables • directionality: can’t be sure which way causality 32 “flows” 16 4/15/2015 Correlation and Prediction • Regression: Correlation with Prediction – predicting DV (y) based on IV (x) – e.g., predicting…. • “burnout” of employees (Y) variable • based on “job satisfaction” (X) variable • Is there a statistically significant correlation between the variables? and • Does the linear relationship between the variables predict the dependent variable Y from the variable X variable? • Regression is typically expressed as MRA (multiple regression analysis) • MRA usually has several IVs and 1 DV: 33 Consider: If age, job satisfaction and turnover are designated as three independent variables (IVs), and  job burnout is designated as the dependent (DV) - outcome variable, then  (a) is there a statistically significant correlation between these three variables and burnout, and  (b) Do variables of age, job satisfaction, and turnover scores predict the “burnout” of agency employees?  (c) What percent of burnout can be predicted from this model ?  (d) Which variable is the strongest predictor of “burnout?”  34 17 4/15/2015 Prediction Equation Y’ = a + b1X1 + b2X2 + b3X3 + ……bnXn Y’ is the predicted score of the Y (dependent variable – “burnout”) based on the known value of Xs (independent variables)  b is the raw (unstandardized) partial regression coefficient derived from the analysis, or the slope, or direction of the line (based on the variable relationship between Xs and Y)  X is the score of the variable being used as the predictor (independent variable)  a is the value of “burnout” when “job satisfaction” = 0 …also called the “intercept” or “constant”   35 MRA – Research Question and Hypotheses       RQ: Do variables of age, job satisfaction, and turnover in the agency predict the burnout of agency employees? Null Ho: Variables of age, job satisfaction, and turnover in the agency will not significantly predict the burnout of agency employees. 2-Tail Ha: Variables of age, job satisfaction, and turnover in the agency will significantly predict the burnout of agency employees. 1-Tail Ha1: Age will significantly positively predict burnout of employees; 1-Tail Ha2: Job satisfaction will significantly negatively predict burnout of employees; 1-TailHa3: Turnover will significantly positively predict burnout of employees. 36 18 4/15/2015 Note: When we do regression analysis, then we do not use correlation critical tables to interpret the results……rather……we will always be given the p-value associated with the regression results! So, let’s look at analytical results on the next slide 37 Results of SPSS MRA analysis Note: The Pearson Correlation r between “Burnout” and all other variables Note: r-values Note: p-values Note: Sample Size 38 19 4/15/2015 Identifying correlation from the previous slide  Age: r = -.077 at p = .130 (Not Significant!)  Satisfaction: r = -.405 at p < .001 (Significant)  Turnover: r = .577 at p < .001 ( Significant) Now, let’s look at the analytical output for MRA analysis on the next slide: 39 Additional Results of MRA analysis Note: Adjusted R2 for the entire model with all independent variables. Here: The three independent variables explained 35.8% of variance in “burnout.” Note: Each variable contributed a different percentage % to the explanation of the Adj. R-Square in this model. Note: The Sig. F Change tells you what happens to the model as you add variables to the equation in predicting “burnout.” Here: Age contributed less than 1%, while satisfaction explained 15.8% and turnover around 20%. Here: Age did not produce a sig. model, but job satisfaction and turnover variables did! 40 20 4/15/2015 Additional Results of MRA analysis 41 Read: Model 3 Note: The p-value for Note: B = b = raw Note: each independent unstandardized Beta is variable. partial correlation used to coefficient for each point to the Here: Age p=.197 independent variable “strongest Satisfaction p = .002 that is used for coefficient Turnover p = < .001 building the prediction equation Equation for Predicting Burnout: Y’ = 15.644 + (.049)(Age_score) + (-.138)(Satisfaction_score) + (.704)(Turnover_score) Note: Constant = a = 15.644 = Intercept value of predicted y’ or burnout when all other variables equal to zero. Another way to present MRA table with results: Table X. Prediction of Burnout based on Age, Job Satisfaction and Turnover of Employees (N=215) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Age -.077 .001 .006 .049 .073 .197 Job Satisfaction -.405** .156 .158 -.138 -.196 .002 Turnover .358 .203 .704 .509 .001 .577** Constant = 15.644 Adjusted R2 = 35.8% Notice the p-values of these predictor variables You should be able to realize that “age” was not a significant predictor of “burnout” whereas “job satisfaction” and “turnover” both statistically significantly predicted “burnout” because their p-values were p < .05! 42 21 4/15/2015 So what do these results tell us? Multiple Regression Analysis (MRA) found that employees’ “burnout” can be significantly predicted from three variable model (age, job satisfaction, and turnover) – look at slide 40and you should see we built the model step by step, including one variable at a time.  Notice on slide 40: the F-change showed you how each model p-value changed: age F change p = .260 (NS), and then when we added job satisfaction and turnover variables – the Fchange p-values for Sig. changed to .000 – telling us that p < 0.001!.  This three variable model explained 38.5% of variance in burnout (Adjusted R-Square = .385). – look at slide 40 and 42!  In this model, only job satisfaction and turnover were predictive of burnout (p < .05), whereas age did not significantly predict burnout (p= .197).  43 Continue with results:      Job satisfaction explained 15.8% and turnover explained 20.3% of the total variance in burnout. – Look at slide 40 under “adjusted R-square Change” column! However, turnover was the strongest predictor (Beta= .509) when compared with job satisfaction (Beta = -.196). – Look at slide 41 under the Beta column for “standardized coefficients” and just simply look for the highest value Beta regardless of the sign. Notice: the sign just tells you the direction of the Predictor and the Outcome variable! The relationship between burnout and turnover is such that a higher score of turnover predicted a higher score of burnout. The relationship between burnout and job satisfaction is negative such that a higher score in job satisfaction predicted a lower score in burnout. 44 22 4/15/2015 Summarized results from the study: The researcher investigated whether 3 predictor variables (age, job satisfaction, and turnover) predicted the level of burnout in 215 agency employees, who were randomly selected from all agency employees for a survey study.  The findings from MRA analysis found that age did not significantly predict burnout, whereas both job satisfaction and turnover did significantly predict the burnout of the employees. The analytical model with three variables explained around 38% of burnout. The best significant predictor was the variable turnover, which explained around 20% of variance in burnout, while job satisfaction explained 16% of variance in burnout.  The finding of these results can be attributed to agency employees because the study used a probability sample. 45  Let’s look at this study: RQ:  Does gender, level of resilience, and income predict “life satisfaction”?  Ha (2-tail):  Gender, level of resilience, and income will predict “life satisfaction.”  Ho: Gender, level of resilience, and income will not predict “life satisfaction.”  46 23 4/15/2015 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 * p < 0.05, ** p < 0.001 Constant = 20.22 1. Which simple correlations are statistically significant ? 2. Which variable predictors in this model significantly predicted “life satisfaction”? 3. How much percent % variance did these predictors explain in “life satisfaction”? 4. Of the significant predictors how much % of variance did each predictor explain? 5. Which significant predictor is the “best” predictor of life satisfaction? 47 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender Male = 0 Female =1 .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 Predictor Variable Constant = 20.22 * p < 0.05, ** p < 0.001 1. Which simple correlations are statistically significant ? (a) Notice that Gender r = .351 with a p < 0.05 - which means “gender” is statistically significant, and weakly and positively correlated with Life Satisfaction (DV). And, female (with higher gender code) have higher life satisfaction! Males (with lower gender code) have lower life satisfaction! (b) Resilience (r = .421) has a significant (p < 0.001), positive, moderate correlation with life satisfaction. (c) Income has no significant simple correlation with life satisfaction (p > 0.05). 48 24 4/15/2015 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 * p < 0.05, ** p < 0.001 Constant = 20.22 2. Which variable predictors in this model significantly predicted “life satisfaction”? …look at the “Sig. column – and use the p-value in this column: (a) Gender (p < 0.05) and Resilience ( p < 0.001), but not Income (p > 0.05). 49 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 Constant = 20.22 * p < 0.05, ** p < 0.001 3. How much percent % variance did these predictors explain in “life satisfaction”? …….look at the Adjusted R-Square column: .32 x 100% = 32.00 % - which means this model with three predictors explained 32% of variance in life satisfaction! 50 25 4/15/2015 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 * p < 0.05, ** p < 0.001 Constant = 20.22 4. Of the significant predictors how much % of variance did each predictor explain? …….Look at the R-square change column: Gender contributed 15.1% and Resilience contributed 17.5%, while Income had almost no contribution with less than 1% 51 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 Constant = 20.22 * p < 0.05, ** p < 0.001 5. Which significant predictor is the “best” predictor of life satisfaction?......here look at the Beta column…and pick the best value(s) for the significant variables: Here, we can see that Resilience has a the best Beta (.544) compared to Gender with Beta = .323. 52 26 4/15/2015 Another way to present MRA table with results: Table X. Prediction of Life Satisfaction based on Gender, Resilience, and Income in Randomly Selected Adults (N= 500) Predictor Variable Simple r Adjusted R2 ∆ R2 Change Regression Coefficient b Beta Sig. Gender .351* .15 .151 .333 .323 .034 Resilience .421** .32 .175 .454 .544 .001 Income .011 .32 .001 .014 .050 .255 Constant = 20.22 * p < 0.05, ** p < 0.001 Overall, this study shows that gender and resilience were significant predictors of life satisfaction. Females had more life satisfaction than males. Income had no significant contribution to the prediction of life satisfaction. The model explained 32% of variance in life satisfaction. And, Resilience was a better stronger predictor of life satisfaction than gender. The results were generalizable from these participants to the targeted population of adults (originally in the sampling frame list). 53 27 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 1 Notes on Chi-Square Association • Chi-square [2 ] test is a test of association between or among frequencies of the variables. • Both the independent and dependent variables have to be of nominal or ordinal level of measurement. If one of the variables is of a higher level of measurement (i.e., interval or ratio), then the data have to be grouped in order to create categories of data (that are at ordinal level of measurement). • The null hypothesis addresses the association between variables by asserting that the differences were created by random sampling error. Note: the relationship that is being investigated is examining “association between frequencies” of the levels in variables and NOT a correlation! • Therefore, as in using other statistical tests, the null hypothesis is rejected when the probability of the chi-square [2 ] result is p < .05. • Use degrees of freedom (df) in order to find the critical value of chi-square from the table: df = (rows – 1) x (columns -1). • Remember that in order to conclude that the obtained calculated chi-square result is statistically significant, the calculated/obtained value of the chi-square result has to be bigger than the critical chi-square value in the table, at p = .05 (either at 2-tail or 1-tail hypothesis that is selected during the hypothesis testing process). • Note: When you have the p-value available with the chi-square result [or any other statistical test, for that matter], then the p-value is first judged on whether it is statistically significant [p Critical (2.71) = Reject the Null of No association (b) For 2-Tailed: Obtained (7.24) > Critical (3.84) = Reject the Null of No association Notice that our obtained result for 7.24 is still significant at 1-tailed critical chisquare at p = .005! • Draw a conclusion: The results of this study show that there is a statistically significant association between student gender and level of comfort in learning research (2 (1, N=75) = 7.24, p < .05). Based on observed proportions, males are more likely to report “high” level of comfort (71%) when compared to females, who are more likely to report a “low” level of comfort (62%) in learning research. Example 2: A study examined behaviors of 120 college students. The study was interested in testing an association [relationship] between students’ self-reported behaviors and class participation. The students’ self-reported behaviors were measured by their [“YES/NO”] responses that occurred in the past 30 days. The class participation was measured by instructors’ verification [“YES/NO”] of whether the student cut class more than two times in the past 30 days. Examine the findings in Table 1 and answer the questions below: Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 Table 1 6 Number and Percentage of Students Answering YES to Classroom Behaviors Based on Class Participation (N=120) Variable Cutting Class Not Cutting Class [YES] Behavior (YES) Got Drunk Sped Broke the law Told a significant lie Was feeling depressed Read a book for pleasure Visited family * p< .05, ** p < .002 [NO] 68 n 57% 52 n 43% Chi-square 59 63 35 14 7 25 40 87% 93% 51% 21% 10% 37% 59% 24 39 10 8 5 15 48 46% 75% 19% 15% 10% 29% 92% 22.79** 7.19* 13.07** 0.53 0.02 0.83 ????? Another way to represent rows and columns is to create a cross-tabulation table: Behavior Getting Drunk YES NO Totals Class Participation Cutting Class YES (n=68) Cutting Class NO (n=52) Totals 59 (87%) 24 (46%) 83 (69%) (71%) (29%) 100% 9 (13%) 28 (54%) 37 (31%) (24%) (76%) 100% 68 (100%) 52(100%) 120 (100%) 57% 43% 100% Q1. What percent of the sample tends to “cut class”? 68/ 120 = 56.67 rounded to 57% Q2. What percent of the sample “gets drunk” ? 83 / 120 = 69.16% Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 Q3. What percent of those who “Cut Class” in the past month did not get drunk? 9/68 = 13.23% Q4. What percent of those who “Did Not Cut Class” in the past month did not get drunk? 28 / 52 = 53.85% …or rounded to 54% Q5. Who is more likely to “get drunk” in the past month, those who “cut class” or those “who do not cut class”? Indicate the percentages that you are comparing: Start out with the “getting drunk” row….71% cut class and 29% did not! Q6. Based on the chi-square result in Table 1, is there a statistically significant association between these two variables (getting drunk and class cutting)? Yes: chi-square = 22.79 at p = < .022, which means the p-level < .05 for determining statistical significance, and in turn that means the chi-square result is statistically significant. And, that also means that there is a significant association between getting drunk and cutting class categories! And, then, we would need to look at the proportions or percentages in the table in order to see how the association (or dependency between categories) plays out! Q7. State the null hypothesis between “Getting drunk” and “Class Cutting”: For example: There will be no statistically significant association between “getting drunk” and “cutting class” levels. Q8. Utilizing the table on the previous page, complete the cross-tabulation table for “Visiting Family” and “Class Participation” Behavior Class Participation Totals Visiting Family YES NO Farber NCSSS/CUA 2015 Class Cutting Not Class Cutting YES (n=68) NO (n=52) 40 48 Totals 7 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 68 (100%) Q9. 52(100%) 8 120 State a possible one-tailed research hypothesis between “Visiting family” and “Class Participation” : Q10. Examining results in Table 1, calculate whether there is a statistically significant association between “Visiting Family” and “Class Cutting”? Explain: For this example – you will have to compute the chi-square value from the table here! Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 Chi-Square Critical Table: 9 Alpha set for 1-Tail Directional Hypothesis. Alpha set for 2-Tail Non-directional hypothesis Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 Using SPSS Output Farber NCSSS/CUA 2015 10 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 11 Example #1: A study is investigating whether there is a statistically significant association between the type of residence of the care recipient [RESIDE] and whether or not the caregiver uses respite care services [RESPITE]. RESPITE * RESIDE Crosstabulation RESIDE RESPITE NO caregiver elsewhere 91 232 72.7% 85.8% 77.3% 53 15 68 27.3% 14.2% 22.7% 194 106 300 100.0% 100.0% 100.0% Count % within RESIDE Total 141 Count % within RESIDE Total Resides Count % within RESIDE YES Resides with Chi-Square Tests Value Pearson Chi-Square Continuity Correctionb Likelihood Ratio df Asymp. Sig. Exact Sig. Exact Sig. (2-sided) (2-sided) (1-sided) 6.781a 1 .009 6.051 1 .014 7.171 1 .007 Fisher's Exact Test Linear-by-Linear Association N of Valid Cases .009 6.759 1 .006 .009 300 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 24.03. b. Computed only for a 2x2 table • • • Here we have the results of the Chi-Square statistic. Notice that here we have several different Chi-Square test statistical results: Pearson Chi-Square is the most frequently reported result! • However, technically - because this table represents a 2 x 2 table, the “Continuity Correction” (or Yates Correction Factor noted in your textbook) version of Chi-Square is provided. Its value is 6.051 and the p level equals .014. • Do we have a statistically significant association here? ……….p < .05! • Another thing to notice in this table is the statement that “O cells have expected count less than 5.” If there were cells with an expected count of less than 5, we would have to use the “Fisher’s Exact Test” statistic instead of Chi-Square. Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 12 Now, let’s look at the magnitude of the obtained chi-square result: Symmetric Measures Value Nominal by Nominal Phi Cramer's V -.150 .009 .150 .009 N of Valid Cases • • • • • • • Approx. Sig. 300 This final table tells us the strength of the association using either Phi (for 2 x 2 tables), or Cramer’s V (for larger tables). Therefore, we would report the Phi coefficient value for this computer run (you ignore the minus sign). If the X2 is not significant, you don’t report these findings or the percentages in the cells! < .39 = weak association .40 - .69 = moderate association .70+ strong association INTERPRETATION: “There is a statistically significant association between the residence of the care recipient and whether or not the caregiver has used respite services [2(1, N = 300) = 6.05, p = .014]. Specifically, a higher percentage of caregivers who had the care recipient living with them used respite services compared to caregivers of care recipients who lived elsewhere (27.3% vs. 14.2%, respectively). This is a weak relationship based on a Phi coefficient of .15.” NOTE: For a 2 x 2 cross-tabs, you report the percentages in the cells that are most pertinent to the focus of your study (focusing on the DV or “use of respite services” in this case), reading across categories of the IV. • Here, we would want to report the column percentages (not counts) of caregivers who had used respite services (“YES” responses). • Also, you generally mention the IV first, followed by the DV, especially if you have stated a specific hypothesis concerning the association between the 2 variables. This is why we phrased the first two sentences to mention the care recipient’s living situation first, followed by the use of respite services second rather than the other way around. Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 Example #2: Chi-Square This study investigates whether there is an association between the caregiver’s gender [SEX] and the level of hours of caregiving provided to the care recipient per week [CAREHRS]. 13 CAREHRS * SEX Crosstabulation SEX MALE CAREHRS Low Count % within SEX RECODED Moderate Count % within SEX RECODED High Count % within SEX RECODED Total Count % within SEX RECODED FEMALE Total 12 84 96 35.3% 31.6% 32.0% 13 88 101 38.2% 33.1% 33.7% 9 94 103 26.5% 35.3% 34.3% 34 266 300 100.0% 100.0% 100.0% Chi-Square test Asymp. Sig. (2Value df sided) 1.058a 2 .589 1.098 2 .578 Linear-by-Linear Association .718 1 .397 N of Valid Cases 300 Pearson Chi-Square Likelihood Ratio a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 10.88. What can we conclude about this association and why? Yes – you should be seeing that there is no statistical significant association between gender and care hours! Farber NCSSS/CUA 2015 SSS 590 – Online – Week 7 – Learning About Chi-Square Association – Dr. Farber 2015 14 Example #3: Chi-Square This study investigates whether there is an association between the level of caregiving burden [Stress] (recoded as “low,” “moderate,” or “high”) and levels of depression [DEPRESSION] (recoded as “low” or “high”). DEPRESSION * BURDEN Crosstabulation Stress Low DEPRESSION LOW Count % within Burden Levels HIGH Count % within Burden Levels Total Count % within Burden Levels Moderate High Total 79 56 35 170 78.2% 57.7% 34.3% 56.7% 22 41 67 130 21.8% 42.3% 65.7% 43.3% 101 97 102 300 100.0% 100.0% 100.0% 100.0% Chi-Square Tests Asymp. Sig. (2Value df sided) 39.903a 2 .000 Likelihood Ratio 41.330 2 .000 Linear-by-Linear Association 39.713 1 .000 Pearson Chi-Square N of Valid Cases 300 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 42.03. Symmetric Measures Value Approx. Sig. Phi .365 .000 Cramer's V .365 .000 Nominal by Nominal N of Valid Cases 300 What do these results tell us and why? Yes, these results point to statistical significance, and the effect size is modest (use Cramer’s V), and then look at the proportions – and you should see that “low depression” is more associated with “low stress” and “high depression” is dependent on “high stress” ….and so on. Farber NCSSS/CUA 2015 SSS 590: WEEK 7, ASSIGNMENT 4 1 Names: THIS ASSIGNMENT IS WORTH 100 POINTS. IT IS COMPRISED OF EIGHT ITEMS, ALL OF WHICH HAVE MULTIPLE PARTS. PLEASE TYPE YOUR RESPONSES DIRECTLY INTO THIS DOCUMENT AND USE A BLUE–COLORED FONT. 1. Below are the results of a chi–square test which was conducted to determine if there was an association between meeting the diagnostic criteria for Depression (DEP) and meeting the diagnostic criteria for General Anxiety Disorder (GAD) among adults receiving services from Agency XYZ. (2 points for each sub–item; 10 points total) GAD*DEP CROSSTABULATION Meets Diagnostic Criteria for DEP Meets Diagnostic Criteria for GAD No Yes Total No Count % within GAD 22 55.0% 18 45.0% 40 100.0% Yes Count % within GAD 8 12.3% 57 87.7% 65 100.0% Count % within GAD 30 28.6% 75 71.4% 105 100% Total CHI–SQUARE TESTS Pearson Chi–Square Likelihood Ratio Linear–by–Linear Association N of Valid Cases Value df Asymp. Sig. (2–sided) 22.115a 22.094 21.904 105 1 1 1 .000 .000 .000 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 11.43. SYMMETRIC MEASURES Nominal by Nominal N of Valid Cases Phi Cramer's V Value Approx. Sig. .459 .459 105 .000 .000 SSS 590: WEEK 7, ASSIGNMENT 4 2 a. Present the Ha (2–tailed) for this study. b. Present the Ho for this study. c. Interpret the results of the chi–square test and specify if Ho should be rejected or retained. d. If applicable, interpret the crosstabulations (re: differences among the groups). e. If applicable, interpret the magnitude of the association. 2. Agency RST tracked the number of hours employees exercised last week (mean=35, SD=4.8). Show all your calculations for each item below. No credit will be given without this information. (2 points for each sub–item; 10 points total) a. How many hours did an employee with a z–score of –2.1 exercise? Round your answer to the nearest whole number (re: no decimals). b. What percentage of employees exercised between 27 and 44 hours? Round your answer to the nearest whole number (re: no decimals). c. How many hours did an employee who scored 2.5 SD below the mean exercise? Round your answer to the nearest whole number (re: no decimals). d. What percentage of employees exercised between 33 and 37 hours? Round your answer to the nearest whole number (re: no decimals). SSS 590: WEEK 7, ASSIGNMENT 4 3 e. What is the associated percentile for an employee who exercised for 40 hours? Round your answer to the nearest whole number (re: no decimals). 3. Below are the results of a chi–square test which was conducted to determine if there was an association between educational attainment (EDU) and meeting the diagnostic criteria for Alcohol Use Disorder (AUD) among adults receiving services from Agency XYZ. (2 points for each sub–item; 10 points total) AUD*EDU CROSSTABULATION Highest Level of EDU Completed Total High Some College School/GED College Graduate Meets Diagnostic No Criteria for AUD Count % within AUD 8 22.9% 9 25.9% 18 51.4% 40 100.0% Yes Count % within AUD 27 38.6% 26 37.1% 17 24.3% 65 100.0% Count % within AUD 35 33.3% 35 33.3% 35 33.3% 105 100% Total CHI–SQUARE TESTS Pearson Chi–Square Likelihood Ratio Linear–by–Linear Association N of Valid Cases Value df Asymp. Sig. (2–sided) 7.800a 7.645 6.367 105 2 2 1 .020 .022 .012 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 11.67. SYMMETRIC MEASURES Nominal by Nominal N of Valid Cases Phi Cramer's V Value Approx. Sig. .273 .273 .020 .020 105 SSS 590: WEEK 7, ASSIGNMENT 4 4 a. Present the Ha (2–tailed) for this study. b. Present the Ho for this study. c. Interpret the results of the chi–square test and specify if Ho should be rejected or retained. d. If applicable, interpret the crosstabulations (re: differences among the groups). Note, more than one interpretation may be required. e. If applicable, interpret the magnitude of the association. 4. The mean level of new flu cases per week in Star County is 22 (SD=1.5). However, among a sample of 144 residents who live in Mira City, which is located in Star County, the mean level of new flu cases per week is 26. (Points for each sub-item are provided below; 10 points total) a. Present the Ha (2–tailed) for this study. (2 points) b. Present the Ho for this study. (2 points) c. Calculate the z–test. Show all your calculations. No credit will be given without this information. (3 points) SSS 590: WEEK 7, ASSIGNMENT 4 5 d. Draw conclusions based upon comparing your calculated z with the critical z of +/–1.96. Indicate if the null hypothesis should be retained or rejected. Explain whether the mean number of flu cases per week in Star County and Deneb City are significantly different. (3 points) 5. Below are the results of a Multiple Regression Analysis (MRA) that was conducted to determine which of the following variables could be used to predict the number of cancelled therapy sessions among youth in a residential treatment center: age, sex, having earned off–campus privileges, treatment group, the number of serious behavioral incidents (SBI), and the quality of the therapist–client relationship (Quality). Overall, the model explained 30.3% of the variance in the number of cancelled therapy sessions. (Points for each sub-item are provided below; 20 points total) COEFFICIENTS Unstandardized Standardized Coefficients Coefficients Model B Std. Error (Constant) 4.379 .882 Age –.001 .071 Sexa –1.328 Privilegesb t Sig. 4.967 .000 –.001 –.013 .990 .292 –.369 –4.547 .000 .432 .413 .120 1.045 .298 Groupc –.549 .359 –.153 –1.527 .130 SBI .070 .034 .238 2.075 .040 Quality –.202 .071 –.268 –2.838 .005 a Beta Sex (0=Male, 1=Female) b Earned Off–Campus Privileges (0=No, 1=Yes) c Treatment Group (0=Routine Treatment, 1=New Treatment) a. Present the Ha (2–tailed) for this study. (2 points) b. Present the Ho for this study. (2 points) SSS 590: WEEK 7, ASSIGNMENT 4 6 c. Identify which variables did not significantly predict the number of cancelled therapy sessions among youth in a residential treatment center. (2 points) d. Identify which variables significantly predict the number of cancelled therapy sessions among youth in a residential treatment center. Which variable was the strongest predictor? (2 points) e. For each of the variables that significantly predict the number of cancelled therapy sessions among youth in a residential treatment center, specify and interpret the direction of its relationship with the outcome variable. (12 points) 6. For each item below, address the following: (a) specify and interpret the direction of the relationship between the variables; (b) indicate the magnitude/strength of the relationship between the variables; and (c) calculate the shared variance between the variables. In regards to (c), show all your calculations and take your answer out to one decimal point. No credit will be given without this information. (4 points for each sub–item; 20 points total) a. There is a significant correlation between infants’ lengths of hospitalization in the NICU and their parents’ levels of anxiety (r = 0.74, p < 0.05). b. There is a significant correlation between parents’ attendance at PTA meetings and their children’s year in school (r = –0.41, p < 0.05). c. There is a significant correlation between family composition (0=Non–Single Parents, 1=Single Parents) and levels of family–related stress (r = 0.53, p < 0.05). d. There is a significant correlation between levels of leadership and organizational performance rates (r = 0.37, p < 0.05). SSS 590: WEEK 7, ASSIGNMENT 4 7 e. There is a significant correlation between parent (0=Fathers, 1=Mothers) and levels of perceived control (r = –0.20, p < 0.05). 7. A study was conducted among adolescents with recent cancer diagnoses who participated in a six week intervention that was designed to enhance their coping skills. Prior to their enrollment, participants’ levels of resilience were measured via a 30 item, 5–point Likert–scale. Higher scores are associated with higher levels of resilience. At pre–test, resilience scores were normally distributed (mean=85, SD=2.5). (2 points for each sub–item; 10 points total) a. What is the median? b. What is the mode? c. Within what range of two scores around the mean did 68% of participants score? Show all your calculations. No credit will be given without this information. d. Within what range of two scores around the mean did 95% of participants score? Show all your calculations. No credit will be given without this information. e. What is the associated score for a participant whose score was 3SD below the mean? Show all your calculations. No credit will be given without this information. 8. Upon conclusion of the above study, participants’ levels of resilience were measured once again (mean=90, SD=5.5). a. Within what range of two scores below the mean did 34% of participants score? Show all your calculations. No credit will be given without this information. SSS 590: WEEK 7, ASSIGNMENT 4 8 b. What is the associated score for a participant whose score was 3SD below the mean? Show all your calculations. No credit will be given without this information. c. Within what range of scores fell 2SD below and above the mean? Show all your calculations. No credit will be given without this information. d. Within what range of scores fell between 1 SD and 3 SD above the mean? Show all your calculations. No credit will be given without this information. e. In which distribution of resilience scores (pre–test or post–test) are scores more heterogeneous? Provide a brief rationale for your response.
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Explanation & Answer

Attached.

SSS 590: WEEK 7, ASSIGNMENT 4

1

Names:
THIS ASSIGNMENT IS WORTH 100 POINTS. IT IS COMPRISED OF EIGHT ITEMS,
ALL OF WHICH HAVE MULTIPLE PARTS. PLEASE TYPE YOUR RESPONSES
DIRECTLY INTO THIS DOCUMENT AND USE A BLUE–COLORED FONT.
1. Below are the results of a chi–square test which was conducted to determine if there
was an association between meeting the diagnostic criteria for Depression (DEP) and
meeting the diagnostic criteria for General Anxiety Disorder (GAD) among adults
receiving services from Agency XYZ.
(2 points for each sub–item; 10 points total)
GAD*DEP CROSSTABULATION
Meets Diagnostic
Criteria for DEP
Meets Diagnostic
Criteria for GAD

No

Yes

Total

No

Count
% within GAD

22
55.0%

18
45.0%

40
100.0%

Yes

Count
% within GAD

8
12.3%

57
87.7%

65
100.0%

Count
% within GAD

30
28.6%

75
71.4%

105
100%

Total

CHI–SQUARE TESTS

Pearson Chi–Square
Likelihood Ratio
Linear–by–Linear Association
N of Valid Cases

Value

df

Asymp. Sig.
(2–sided)

22.115a
22.094
21.904
105

1
1
1

.000
.000
.000

a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 11.43.

SYMMETRIC MEASURES
Nominal by Nominal
N of Valid Cases

Phi
Cramer's V

Value

Approx. Sig.

.459
.459
105

.000
.000

SSS 590: WEEK 7, ASSIGNMENT 4

2

a. Present the Ha (2–tailed) for this study.
Alternative hypothesis (Ha): There is an association between meeting the diagnostic criteria
for Depression (DEP) and meeting the diagnostic criteria for General Anxiety Disorder
(GAD) among adults receiving services from Agency XYZ.
b. Present the Ho for this study.
Null hypothesis (H0): There is no association between meeting the diagnostic criteria for
Depression (DEP) and meeting the diagnostic criteria for General Anxiety Disorder (GAD)
among adults receiving services from Agency XYZ.
c. Interpret the results of the chi–square test and specify if Ho should be rejected or retained.
The Chi-Square test value obtained from our analysis above is 2 = 22.115a, degrees of
freedom (DF) is 1 and the corresponding probability value (p-value) is 0.000. Since the
obtained p-value is less than alpha (α = 0.05), we “reject” the null hypothesis and conclude
that the test was statistically significant at 5% level of significance. This implies that there
is an association between meeting the diagnostic criteria for Depression (DEP) and meeting
the diagnostic criteria for General Anxiety Disorder (GAD) among adults receiving
services from Agency XYZ.
d. If applicable, interpret the cross-tabulations (re: differences among the groups).
The contingency table shows that among the 40 adults who did not meet the diagnostic
criteria for General Anxiety Disorder (GAD, only 22 (55%) and 18 (45%) didn’t and meet
the diagnostic criteria for Depression (DEP) respectively. Moreover, in the remaining 65
adults who meet with the diagnostic criteria for General Anxiety Disorder (GAD, only 8
(12.3%) and 57 (87.7%) didn’t and meet the diagnostic criteria for Depression (DEP)
respectively.
e. If applicable, interpret the magnitude of the association.
The above test of association has a contingency table matrix of 2*2, therefore to interpret
the magnitude of the association we will apply the Phi value. The Phi value is generated as
0.459, this value signifies that the association was moderate between the diagnostic criteria
for Depression (DEP) and diagnostic criteria for General Anxiety Disorder (GAD) among
adults receiving services from Agency XYZ.

SSS 590: WEEK 7, ASSIGNMENT 4

3

2. Agency RST tracked the number of hours employees exercised last week (mean=35,
SD=4.8). Show all your calculations for each item below. No credit will be given
without this information.
(2 points for each sub–item; 10 points total)
a. How many hours did an employee with a z–score of –2.1 exercise? Round your answer
to the nearest whole number (re: no decimals).
(x− ẋ)
z-score = σ where z-score is -2.1, x is the number of hours the employees exercised, ẋ
(mean) is equal to 35 and standard deviation (σ) is equal to 4.8 respectively.
Therefore, -2.1 =

(x− 35)
4.8

 x= {(-2.1 * 4.8) + 35}
 x= 24.92
 The number of hours employees exercised is 25 hours (re: no decimals).
b. What percentage of employees exercised between 27 and 44 hours? Round your answer
to the nearest whole number (re: no decimals).
(x− ẋ)
(44− 35)
9
Z-score = σ ; z-score (44) = 4.8 = 4.8 = 1.875
Z-score (27) =

(27− 35)
4.8

=

(−8)
4.8

= -1.6667

 P( -1.6667 < z < 1.875) = p(z = 1.875) – p(z= -1.6667)
 Where, p(z = -1.6667) = 1 – (-1.6667) = 2.6667
 Therefore, P( -1.6667 < z < 1.875) = 1.875 – 2.6667 = -0.7917
We check the z-score (-0.79) in the Normal Distribution Table to see the corresponding
probability value related to the score. The probability is 1 – 0.21476 = 0.78524, therefore
the percentage of employees exercised between 27 and 44 hours is 79% (re: no decimals).
c. How many hours did an employee who scored 2.5 SD below the mean exercise? Round
your answer to the nearest whole number (re: no decimals).
(x− ẋ)
z-score = σ where z-score is -2.5, x is the number of hours the employees exercised, ẋ
(mean) is equal to 35 and standard deviation (σ) is equal to 4.8 respectively.
...


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