Quantitative Research Methods In Political Science Lecture 8 PDF

Quantitative Research Methods in Political Science Lecture 8: Measures of Association for Nominal and Ordinal Variables Course Instructor: Michael E. Campbell Course Number: PSCI 2702 (A) Date: 11/07/2024 What are Measures of Association? Measures of association: “provide information about the strength and, where appropriate, the direction of relationships between variables in our data set – information that is more directly relevant for assessing the importance of relationships and testing the validity of our theories” (Healey, Donoghue, and Prus 2023, 250). Measures of association are the most powerful tool for documenting cause-and-effect relationships Can also help us make predictions – i.e., if variables are related, we can predict a score on one variable based on the score of another… Bivariate Association Two variables are associated if “the distribution of one of them changes under the various categories or scores of the other” (Healey, Donoghue, and Prus 2023, 252). For example, here we have a bivariate table of 173 factory workers… If we look at the columns, we can determine if association between the variables exists… The columns shows the pattern of scores on Y for each score on X Bivariate Association Cont’d This table displays the relationship between productivity and job satisfaction… Job Satisfaction = X Level of Productivity = Y Research Question: “Does job satisfaction effect the level of productivity?” Bivariate Association Cont’d If we look at the columns… 30 of 60 workers who reported “low” job satisfaction rank low on productivity 25 of the 61 who reported “moderate” job satisfaction rank moderate on productivity 27 of the 52 who reported “high” satisfaction rank high on productivity Bivariate Association Cont’d If we look at the columns… 30 of 60 workers who reported “low” job satisfaction rank low on productivity 25 of the 61 who reported “moderate” job satisfaction rank moderate on productivity 27 of the 52 who reported “high” satisfaction rank high on productivity Bivariate Association Cont’d If we look at the columns… 30 of 60 workers who reported “low” job satisfaction rank low on productivity 25 of the 61 who reported “moderate” job satisfaction rank moderate on productivity 27 of the 52 who reported “high” satisfaction rank high on productivity Bivariate Association Cont’d Looking across columns, we can see the effect of X on Y The within column frequencies are called “conditional distributions of Y” – because they display the distribution of scores on the dependent variable for each score on the independent variable So, from what we observe in the table, productivity and job satisfaction are associated The distributions of scores on Y (productivity) changes across the various conditions of X (satisfaction) To fully investigate the potential relationships between two variables, we need to ask: 1. Does association exist? Questions to Ask 2. If an association exists, how strong is it? 3. If association does exist, what are the pattern and/or the direction of the association? Does Association Exist? Association can be detected with chi square (any chi square score above 0.00 represents some level of association - but not necessarily statistical significance) We can also observe the “conditional distributions of Y” The best way to look at these distributions is to look at observed frequencies in percentages (not raw scores) Does Association Exist? Cont’d The best way to do this is to compute and compare column percentages (because it standardizes column totals on a scale of 100) This makes it easier to detect the conditional distributions of Y As we see here: 50% who ranked low on satisfaction, ranked low on productivity 40.98% who ranked moderate on satisfaction, ranked moderate on productivity 51.92% who ranked high on satisfaction, ranked high on productivity Does Association Exist? Cont’d If association did not exist, the conditional distributions of Y would not change across columns and the distribution of Y would be the same for each condition of X As you see here, the conditional distributions of Y are the same and we see that Looking for association between age group levels of productivity do not and productivity… change at all as the age group varies Perfect non-association is found How Strong is Association? Once we have established two variables are associated, we need to determine how strong the association is… Strength of association ranges from two extremes: 1. “Perfect non-association,” where conditional distributions of Y do not change at all (as seen on previous slide) 2. “Perfect association,” which occurs when each value of the dependent variable (Y) is All cases in each column are in a single associated with only one value of the independent variable (X) (see right) cell and there is no variation in Y for the given value of X How Strong Is Association? Cont’d When we find perfect association, it points towards a causal relationship – i.e., variation in one variable is responsible for variation in another We are unlikely to find perfect non-association or perfect association – we are more likely to find something between these two extremes This is why we use measures of association – they allow us to describe the strength of association between two variables with precision Strength of association can be described as: (1) strong, (2) moderate, (3) weak What is the Pattern and/or Direction of Association? The last thing we need to do when analyzing association is to determine which values or categories of one variable are associated with which values or categories of another… We see a pattern here 1. Low satisfaction associated with low productivity 2. Moderate satisfaction with moderate productivity 3. High satisfaction with high productivity This shows us directional association! What is the Pattern and/or Direction of Association? Associations between variables can be positive or negative… Positive association occurs when high scores on one variable are associated with high scores on the other (or when low scores are associated with low scores) In other words, when two variables move in the same direction it is a positive association – i.e., they will increase together or decrease together What is the Pattern and/or Direction of Association? Cont’d Conversely, a negative association occurs when the variables vary in opposite directions – i.e., high scores on one variable are associated with low scores on another (or vice-versa) If there is an increase in one variable and a decrease in another (or vice versa), then you have a negative relationship As we see here, as education increases, television viewership decreases… Measures of Association Measures of Association “characterize the strength (and for ordinal and interval- ratio variables, also the direction) of bivariate relationships in a single number” (Healey, Donoghue, and Prus 2023, 262). Measures of association will change depending on the variables level of measurement Today, we will look at: Phi and Cramer’s V for Nominal level variables Gamma, Somers’ d, Kendall’s tau-b, and Kendall’s tau-c for ordinal level data Note: if you have variables measured at different levels, convention states that you would use the MOA for the variable measured at the lowest level Chi Square When working with variables at nominal level, Based- researchers tend to use measures of association based on the value of chi square Measures If chi square is anything but zero, it shows that of there is some level of association between Association variables (but this does not mean it is statistically significant) The values of chi square can be transformed into other statistics that measure the strength of association between variables Last week, we looked at the likelihood of students in accredited programs finding employment as social workers after graduation We found a chi square value of 10.78 at alpha of 0.05 (indicating statistical significance) But knowing statistical significance tells us nothing about the strength of association… Chi Square-Based Measures of Association Cont’d We see here that the conditional distributions of Y (i.e., the conditional distributions of employment status) change as a function of X (i.e., accredited status) To assess the strength of association, we can compute Phi ()or Cramer’s V Chi Square-Based Measures of Association Cont’d Phi () Is a chi square based measure of association used for nominal variables It is appropriate for 2x2 tables The formula for Phi is: In this equation: = the chi square value n = the sample size Phi Cont’d Phi ranges from 0.00 (no association) to 1.00 (perfect association) The closer to 1.00, the stronger the relationship Therefore, With a chi square of 10.78 and a And with this, you now know the sample size of 100, our strength of association between the calculation would look like this… variables… Paired with observation of the column percentages, you can discern the direction of the relationship as well… Cramer’s V Cramer’s V is used for tables larger than 2x2 (i.e., tables with more than two columns and/or two rows) If we use Phi with tables larger than 2x2, it can exceed 1.00 (making interpretation difficult) The formula for Cramer’s V is: In this equation: = the chi square value n = the sample size = the minimum value of rows r – 1 (number of rows minus 1) and c – 1 (number of columns minus 1) Cramer’s V Cont’d In this example, we’re looking at the relationship between membership in student clubs and academic achievement (n = 75) There is an obtained Chi Square of 32.14, which is significant at 0.05 level Cramer’s V Cont’d We have a Cramer’s V score of 0.46 – indicating the strength of association Moreover, we can look at the column percentages to get a better idea of what’s going on… We see that members: Members of sports clubs have moderate academic achievement Members of non-sports clubs have high academic achievement Members of no clubs have low academic achievement Interpreting Strength If Phi () or Cramer’s V ranges from 0.00 to 0.10 – the association is weak If Phi () or Cramer’s V ranges from 0.11 to 0.30 – the association is moderate If Phi () or Cramer’s V are greater than 0.30 – the association is strong Measures Two types of ordinal variables: of 1. Continuous Ordinal Variables: ordinal variables with many scores that resemble interval-ratio level variables (e.g., a feeling thermometer) Association 2. Collapsed Ordinal Variables: ordinal level variables with just a few categories (no more than five or six) for Ordinal values or scores Level Today, we are interested in measures of association for Collapsed Ordinal Variables Variables For collapsed ordinal level variables, we use the following Measures of Association: 1. Gamma (G) 2. Somers’ d () 3. Kendall’s tau-b () 4. Kendall’s tau-c () Measures of Association Remember, we need to answer three questions for Ordinal when using measures of association: Level 1. Does association exist? Variables Cont’d 2. If an association does exist, how strong is it? 3. If association does exist, what are the pattern and/or the direction of the association? The Logic of Pairs Gamma, Somers’ d, Kendall’s tau-b, and Kendall’s tau-c all measure the strength and direction of association They do this by comparing each respondent to every other respondent These comparisons are called “pairs” – because they examine respondents rankings on both X and Y To find the total number of unique pairs, we use the following formula… Total number of unique pairs = The Logic of Pairs Cont’d Pairs can be divided into five sub-groups: 1. Similar Pair: a pair of respondents is similar if the respondent with the larger value on the independent variable also has a larger value on the dependent variable 2. Dissimilar Pair: a pair is dissimilar if the respondent with the larger value on the independent variable has a smaller value on the dependent variable 3. Tied on the IV (X): a pair is tied on the independent variable if both respondents have the same score on the independent variable but not the dependent variable 4. Tied on DV (Y): a pair is tied on the dependent variable if both respondents have the same score on the dependent variable but not the dependent variable 5. Tied on both variables: a pair is tied on both variables if they have the same independent and dependent variable scores The Logic of Pairs Cont’d Let’s say we are trying to determine a cause of burnout among elementary school teachers… Each variable has three response categories: We hypothesize that years of 1. 1.00 coded as low service is a potential cause of burnout 2. 2.00 coded as moderate 3. 3.00 coded as high We therefore have two variables: 1. Years of Service (X) 2. Level of Burnout (Y) The Logic of Pairs Cont’d Sample Size (n) = 5 The Logic of Pairs Cont’d To find the number of unique pairs, we use our formula… Total number of unique pairs = Total number of unique pairs = Total number of unique pairs = 10 We have 10 unique pairs The Logic of Pairs Cont’d If we examine the data closer, we can see the types of pairs we have… The Logic of Pairs Cont’d The Logic of Pairs Cont’d 1. Steven is ranked above Camil on Length of Service, and he is also ranked above Camil on burnout Therefore Steven-Camil is a Similar Pair 2. Joesph and Steven are ranked the same on the DV, but not the IV Therefore, Joesph-Steven are Tied on the Dependent Variable 3. Camil and Joseph have the same score on the IV, but not the DV Therefore, Camil-Joseph are Tied on the Independent Variable With knowledge of these pairs, we can compute our measures of 4. Karina and Steven have the same score on association… each variable Therefore, Karina-Steven are Tied on Analyzing Measures of Association for Ordinal Level Variables Gamma, Somers’ d, tau-b, and tau-c all “measure the strength of association between variables by considering the number of similar versus dissimilar pairs” (Healey, Donoghue, and Prus 2023, 268). Where these measures differ is how they treat tied pairs When the number of similar and dissimilar pairs is equal, the value of these statistics will be 0.00 (no association between X and Y) However, as the number of similar pairs increases relative dissimilar pairs (or vice-versa), the value of the statistic will approach 1.00 As the statistics approach 1.00, the strength of association increases When every pair is similar or dissimilar, the value of these statistics will be 1.00 (perfect association between X and Y) Interpreting Strength for Ordinal Level Measures Interpreting strength of association for Gamma, Somers’ d, tau-b, and tau-c is much like it was for Phi and Cramer’s V However… Direction of Relationship When we use measures of association for nominal level variables (phi or Cramer’s V), they only measure the strength of association They tell us nothing about directionality Measures of association for ordinal level variables are more sophisticated and can also tell us about the direction of the relationship (i.e., positive or negative) Direction of Relationship Cont’d Gamma, Somers’ d, tau-b, and tau-c have scores that range from -1.00 to +1.00 To determine directionality, look at the sign (+ or -) A plus (+) sign tells you it is a positive relationship (i.e., both variables are increasing or decreasing in value together) A minus sign (-) tells you that it is a negative relationship (i.e., as one variable increases, the other decreases, or vice-versa) Positive Versus Negative Relationships Visualized Above is a positive relationship Above is a negative relationship As values on X increase, values on Y To be a negative relationship as values on increase one variable increase, values on the other would decrease (or vice-versa) It would still be positive if values on X decreased and values on Y decreased A negative relationship occurs when variables move in different directions as A positive relationship means the variables they vary are moving in the same direction as they vary… Positive Versus Negative Relationships Visualized If tables are constructed with column variables increasing from left to right and the row variable increasing from top to bottom (as is convention) you can see the direction of the relationship by looking at the table A positive relationship will present itself with scores falling along a diagonal from upper left to lower right A negative relationship will present itself with a diagonal line moving from the upper right to the bottom left Gamma Gamma is the number of similar pairs as a proportion of all pars excluding ties… The formula for Gamma is: In this equation: = number of pairs of respondents ranked the same on both variables – i.e., similar pairs = number of pairs of respondents ranked the same on both variables – i.e., dissimilar pairs Ranges from -1.00 to +1.00 0.00 represents “No relationship” Gamma Cont’d Taking our previous example about years of service and burnout, we see there are 6 similar pairs and no dissimilar pairs Gamma Cont’d G = +1.00 This tells us it is a perfect, positive relationship… As the number of years worked increases, the level of burnout also increases But is it a perfect association? Remember, Gamma ignores tied pairs! The problem with Gamma is that it ignores tied pairs… This means Gamma can exaggerate the strength of association between two ordinal variables Somers’ d, Tau-b, and Tau-c were developed to Limitations correct for this of Gamma These are extensions of Gamma – so the numerator remain the same (i.e., ) However, the denominators are more complex Whereas tau-b and tau-c include in their calculations pairs tied on the IV and DV, Somers’ d includes only pairs tied on the DV Kendall’s Tau-b The formula for tau-b is: = In this equation: = the number of similar pairs = the number of dissimilar pairs = the number of ties pairs on the independent variable (X) = the number of pairs tied on the dependent variable (Y) Kendall’s Tau-b Cont’d In our example, there are 6 similar pairs and 0 dissimilar pairs There is 1 pair tied on the IV and 2 pairs tied on the DV Therefore… = = = +.80 +.80 indicates a strong and positive relationship between years worked and burnout Limitations of Tau-b Tau-b can only reach a value of ±1.00 when the IV and DV have the same number of categories This means your table should have the same number of rows and columns (see right)… But what if we had a 2x3 table? Tau-c Tau-c is designed so that it can reach a maximum of 1.00, even if the number of categories on the variables is unequal… The formula for tau-c is: In this equation: = the number of similar pairs = the number of dissimilar pairs = the number of respondents = the minimum value of the number of categories on the independent variable and the number of categories on the dependent variable Tau-c Cont’d Ranges from -1.00 to +1.00 0.00 represents “no relationship” -1.00 represents a perfect negative relationship +1.00 represents a perfect positive relationship Although the table in our example (i.e., burnout by years worked) has an equal number of categories on each variable, we can nonetheless compute tau-c Tau-c Cont’d To calculate this, we take our six similar pairs and a minimum value of 3 (the value is 3 because both variables have 3 categories) If the number of categories was lower on one variable, you would use that… The most important thing to remember with tau-c is that you generally use it This indicates a strong and when the number of response categories on your variables is unequal positive relationship between (you can spot this easily by looking if years worked and burnout your table has an unequal number of rows and columns) Somers’ d The formula for Somers’ d is: In this equation: = number of similar pairs = number of dissimilar pairs = number of pairs tied on the dependent variable (Y) Somers’ d Cont’d Ranges from -1.00 (perfect negative association) to +1.00 (perfect positive association) What makes Somers’ d special is that it is an asymmetric measure This means you can compute two different scores You can compute , where Y is treated as the dependent variable You can also compute , where X is treated as the dependent variable Therefore, Somers’ d will provide different predictive power between two variables depending on which variable you treat as X and which as Y Somers’ d Cont’d If we take our previous example, we see that there are 6 similar pairs, and 2 pairs tied on Y (i.e., level of burnout) Somers’ d Cont’d With , we see that there is a strong and positive relationship between our variables… As years worked increases, burnout also increases Like tau-b and tau-c the statistic is smaller than Gamma – this is because it accounts for tied pairs on the DV, whereas Gamma does not consider tied pairs at all Quantitative Research Methods in Political Science Lecture 10: Hypothesis Testing (t-Tests: One-Sample Case and Two-Sample Case) Course Instructor: Michael E. Campbell Course Number: PSCI 2702 (A) Date: 11/14/2024 Hypothesis Testing (one- and two-sample case) Hypothesis tests follow systematic five-step model: (1) Make assumptions and meet test requirements (2) State the null hypothesis (3) Select the sampling distribution and establish the critical region (4) Compute the test (obtained) statistic (5) Make a decision and interpret the results of the test Today, we will look at the one-sample case and the two-sample case techniques for testing with means (not proportions) If the population(s) standard deviation(s) are not known, we refer to these as one-sample t- tests and two-sample t-tests SPSS will always use the t distribution, because it converges with the Z distribution once it hits 120 degrees of freedom. Therefore, statistical software always uses the t distribution in its calculations. This is why we tend to refer to these as t-Tests, but technically if you use the Z-distribution, they are Z tests. For the sake of clarity, consider everything we’re doing here as a form of t-Test. The One-Sample Case Technique What can we do with One-Sample Case? With a one-sample case technique we can answer questions such as the following… Example 1: Are older adults in Ontario more or less likely to be victimized by crime than the population of Ontario in general? Example 2: Are the GPAs of university student athletes different from the GPAs of the student body as a whole? Example 3: Do students who participate in LSAT preparation courses score higher on the LSAT than the general population of LSAT writers? What can we do with One-Sample Case? Cont’d In each example, you work with a random sample and a population In example 1, it is older adults in Ontario (the population is all people in Ontario) In example 2, it is student athletes (the population is all students in the university you are studying) In example 3, it is students who participated in LSAT preparation courses (the population is all who took the LSAT) What can we do with One-Sample Case? Cont’d Our overarching goal, when using one-sample case technique, is to determine if the groups represented by the samples are different from the populations from which they are drawn on a specific trait… In example 1, the trait is victimization rates In example 2, the trait is GPA In example 3, the trait is LSAT scores Hypothesis Testing with One-Sample Case ( Example Research Question: “Do graduates of an LSAT preparation course have an average LSAT score that is different than others who write the LSAT? To answer this question, you need to compare the scores of all graduates of the LSAT preparation course with all LSAT test-takers… However, you have incomplete information on both of these groups Hypothesis Testing with One-Sample Case ( Cont’d The first thing you need to do is gather a random sample of graduates from records provided by companies conducting LSAT preparation courses You would also collect information on the entire population of LSAT test takers We see the sample mean is higher than the population mean… The information would be as follows…(see right) But is this simply happening by random chance? What We’re Looking Based on our sample, we want to know if the people who took the LSAT preparation course For score higher than those who did not take the LSAT preparation course We need to determine if the difference between our sample statistic and population value is statistically significant In other words, we want to know if the difference Explanations for Observed Differences Explanation 1: the difference between the population mean (152) and the sample mean (156) reflects a real difference in LSAT scores between those who took the preparation course and the entire population of test takers If this is the case, then the difference is statistically significant, and we would be unlikely to observe such a large difference by random chance alone Explanation 2: the difference between the population mean (152) and the sample mean (156) is the result of random chance This would mean that there is no important difference between those who took the preparation course and those who did not In other words, the preparation course has no impact on LSAT scores Explanations for Observed Differences Cont’d Explanation 2 represents our null hypothesis – because it is a statement of “no difference” Symbolically, the null hypothesis would look like this: in this case represents the average for all LSAT test takers who took the preparation course If we begin with the assumption that those who took the preparation course are not different than test takers in general, we can calculate the probability of getting the observed sample outcome () of 156 If the observed probability is less than 0.05 (meaning it is less than 5 out of 100), we can reject the null hypothesis How can We Determine This? We use our knowledge of the sampling distribution The mean of the sampling distribution will have the same mean as the population () Because the sampling distribution is normal in shape, it can also be interpreted as a distribution of probabilities: “the theoretical normal curve may also be thought of as a distribution of probabilities” (Healey, Donoghue, and Prus 2023, 140). Therefore, the mean of 156 is just one of thousands of possible sample means Using Z Scores (when is known) Using our knowledge of the Normal Curve, we can use Z scores to determine if a sample outcome with a probability less than 0.05 will allow us to reject the null When we convert raw scores into Z scores, it tells us “the percentage of the total area (or number of cases) above, below, or between scores in an empirical distribution” (Healey, Donoghue, and Prus 2023, 129). If we set alpha at 0.05, and we conduct a two- tailed test, we split the probability equally into both tails (i.e., 0.025 in each tail) This is associated with a Z score of Converting Raw Score to Z Score Formula from Lecture 4… This gives us… = +4.00 But now, we are working with a Therefore, a mean of 156 is associated sampling distribution, so the formula with a Z score of +4.00 must reflect the theoretical as opposed to empirical distribution This falls into the critical region under the normal curve Converting Raw Score to Z Score Cont’d Since +4.00 is higher than +1.96, it falls in the critical region and we know that it is statistically significant This means it would be highly unlikely to obtain a sample mean of 156 if the Null Hypothesis were true We can reject the null hypothesis… Therefore, “The sample of 100 graduates of an LSAT preparation course comes from a population that is significantly different for the population of LSAT writers as a The One-Sample Case in the Context of the Five-Step Model Step 1 – Make Assumptions and Step 2 – Make Assumptions and Meet Test Requirements Meet Test Requirements Remember, the null hypothesis a statement of “no difference” or “no Model: Random Sampling relationship” Level of Measurement is Interval-Ratio Sampling Distribution is ( Normal The null () suggests that the mean of the population of graduates who took the LSAT prep. course () is not different from the mean of the population of all test takers The One-Sample Case in the Context of the Five-Step Model Cont’d Step 3 – Select the Sampling Step 4 – Compute the Test Distribution and Establish the Statistic Critical Region Sampling Z Distribution Distribution: = +4.00 0.05 Z (critical) 1.96 The One-Sample Case in the Context of the Five-Step Model Step 5 – Make a Decision and Interpret the Results Z (critical) = 1.96 Z (obtained) = +4.00 Since the obtained score is greater than the critical score, we know it falls in the critical region Therefore, we can reject the Null… We can say, “the difference between the sample mean of 156 and the mean of 152 for the entire population of LSAT writers is statistically significant or unlikely to be caused by random chance alone” (Healey, Donoghue, and Prus 2023, 335). One-Tailed vs. Two-Tailed Tests When using one-sample case technique, we must decide if we want one- or two-tailed test This will determine if we divide the critical region into both tails (two- tailed), or only the upper tail (one-tailed), or only the lower tail (one- tailed) This decision depends on whether or not the research hypothesis () is directional One-Tailed vs. Two-Tailed Tests Cont’d If we are only looking for difference (either higher or lower) we use a two-tailed test… For example, “the population mean is not equal to the value stated in the null hypothesis () If you are interested in a specific direction, you use a one-tailed test… For example, if you predict that the graduates of an LSAT preparation course will have a higher LSAT score than the entire population or LSAT writers, your research hypothesis would be: ( indicates “greater than”) ( represents “less than or equal to”) One-Tailed vs. Two-Tailed Tests Cont’d Conversely, if we predicted that the graduated of the preparation course would have a lower score than the entire population of LSAT writers (or that the average LSAT score of those who took the preparation course would be less than 152), then the research and null hypotheses would appear as follows… ( represents “less than”) (symbolizes “greater than or equal to”) The decision to use a one- or two-tailed test will appear in STEP 3 of the five-step model, because this choice affects the size of the critical region… If we believe that the population characteristic is greater than the value stated in the null hypothesis, (i.e., if includes ), then we place all of the critical region in the upper tail If we believe that the population characteristic is less than what is stated in the null hypothesis (i.e., if includes ), then we place all of the critical region in the lower tail One-Tailed vs. Two-Tailed Tests Cont’d We use a one- tailed test when: 1. When the direction of the difference can be confidently predicted One-Tailed 2. When the vs. Two- researcher concerned is only Tailed with the difference in one-tail of the Tests sampling distribution Cont’d One-Tailed Test Example Using data from Community and Well-Being (CWB) Index (scale 0 to 100) Example Research Question: “has the socioeconomic development of First Nations communities improved since 2016?” We take random sample of 100 With this information, we can use the five- First Nation Communities from step model to test the null hypothesis that 2021 ( = 63.1) there is no difference between the level of socioeconomic well-being of First Nations CWB score for all first nations in between 2016 and 2021 2016 ( = 58.4 and = 10.3) One-Tailed Test Example Cont’d Step 1 – Make Assumptions and Step 2 – Make Assumptions and Meet Test Requirements Meet Test Requirements Model: Random Sampling Our research hypothesis is that the Level of CWB score for First Nations has Measurement is improved since 2016 (it is Interval-Ratio directional) Sampling Distribution is Normal One-Tailed Test Example Cont’d Step 3 – Select the Sampling Step 4 – Compute the Test Distribution and Establish the Statistic Critical Region Sampling Z Distribution Distribution: 0.05 Therefore… Z (critical) 1.65 One-Tailed Test Example Cont’d Step 5 – Make a Decision and Interpret the Results The value surpasses the critical score – therefore we can reject the null Notice Z (obtained) is positive (falls into upper tail) To find a difference of this size by random chance alone would be highly unlikely Therefore, according to this data (i.e., the CWB), we can say that the socioeconomic well-being of Test Type and Error Type I Error: incorrectly rejecting a null hypothesis that is true Type II error: incorrectly confirming a null hypothesis that is false Since a two-tailed test splits the critical region into each tail, each critical region is smaller and it becomes more difficult to reject the null Therefore, two-tailed tests are less likely to commit type I error The Student’s t Distribution and the One- Sample Case When we do not know the population standard deviation (), we must use the t distribution We must use the sample standard deviation in conjunction with the t distribution (see textbook Appendix B) to find the areas under the sampling distribution Must also consider degrees of freedom, as opposed to sample size When using the t distribution, the changes in the five-step model appear in Steps 3 and 4 The Student’s t Distribution and the One- Sample Case Cont’d Moreover, to find the critical value when using t distribution, you use a slightly different formula… Instead of: Now you would use: t (obtained) = As you can see, you are substituting the population standard deviation () for the sample standard deviation (s) You’re also using n-1 to correct for the bias of estimating the population standard deviation while using the sample standard deviation One-Sample t test example ( Unknown) Research Question: are the GPAs of Student Sociology sociology students different from those of the study body in general? Population Students Sample The population is “all students” and () = 2.78 the sample is “sociology students” = unknown s = 1.23 n = 30 One-Sample t test example ( Unknown) Cont’d Step 1 – Make Assumptions and Step 2 – Make Assumptions and Meet Test Requirements Meet Test Requirements Model: Random Sampling We only want to know if there is a Level of difference, so it is a two-tailed test Measurement is (non-directional) Interval-Ratio Sampling Distribution is Normal One-Sample t test example ( Unknown) Cont’d Step 3 – Select the Sampling Step 4 – Compute the Test Distribution and Establish the Statistic Critical Region t (obtained) = Sampling t Distribution t (obtained) = Distribution: 0.01 (99%) t (obtained) = df n – 1 (30-1=29) t (critical) 2.756 t (obtained) = +1.22 Notice we set alpha () at 0.01 here, and not the 0.05. This makes the critical region smaller… One-Sample t test example ( Unknown) Cont’d Step 5 – Make a Decision and Interpret the Results t t Since our obtained score does not fall in the critical region, we cannot reject the null hypothesis at a 0.01 level of significance Therefore, this tells us that the difference between the sample mean (2.78) and the population mean (2.50) is no greater than what would be expected if only random chance were operating Summary for Selecting a Sampling Distribution When testing sample means, you need to make a choice between the theoretical distribution to establish the critical region If the population standard deviation () is known, and the sample size is sufficiently large (n 100), or if the sample is taken from a population that is normally distributed, use Z distribution If the population standard deviation () is unknown, then use the t distribution The Two-Sample Case Technique The Two-Sample Case Unlike a one-sample case, which is considers the significance of the difference between a sample value and population value… The two-Sample t-Test considers the significance of the difference between two separate populations Also referred to as “Independent Samples t-Test” Example Research Question: “Is there a difference in pay between university and high school graduates who work full-time?” You need two random samples (to compare both sample means) The Two- Sample You can use the information to infer population patterns Case Example Central question when using two-sample case: “Is the difference between samples large enough to allow us to conclude (with a known probability of error) that the populations represented by the samples are different?” (Healey, Donoghue, and Prus 2023, 363) If we find a large enough difference between the sample means, we can say that the difference is not occurring by random chance alone… The Two- Sample Case Example Cont’d Two-Sample Case Example ( Known) Cont’d We will see three major differences in the five-step model (relative the one-sample case) 1. Step 1 - You need two random samples (must be selected independently from one another) 2. Step 3 – hypothesis is still a statement of “no difference,” “greater than or equal to,” or “less than or equal to,” but it is between the two population means (not simply between sample and population) 3. Step 3 – The sample outcome is now the difference between the sample statistics (as opposed to between sample and the population) If the null hypothesis is true, and there is no difference between these two populations, then “the difference between the population means is zero, the mean of the sampling distribution is zero, and the huge majority of differences between the sample means are zero (or, at any rate, very small in value)” (Healey, Donoghue, and Prus 2023, 364). Two-Sample Case Example ( Known) Cont’d When is known, we still use the Z distribution, but to compute the test statistics, we must slightly alter the formula to obtain Z (to account for the two populations and two samples)… To solve for we use… Z (obtained) = = In this equation: = the difference in the sample means = the standard deviation of the sampling distribution of the differences in sample means Two-Sample Case Example ( Known) Cont’d Example Research Question: “Is there a difference in pay between university and high school graduates who work full-time?” We collect two random samples (both n = 150)… Sample 1 (university graduates) Sample 2 (high school graduates) = 50 000 = 40 000 = 12 000 = 9000 n = 150 n = 150 Two-Sample Case Example ( Known) Cont’d Step 1 – Make Assumptions and Step 2 – Make Assumptions and Meet Test Requirements Meet Test Requirements Model: Independent Null hypothesis states that the Random Samples populations represented by the Level of samples have no difference (not Measurement is directional) Interval-Ratio Sampling : Distribution is Normal Two-Sample Case Example ( Known) Cont’d Step 3 – Select the Sampling Step 4 – Compute the Test Distribution and Establish the Statistic Critical Region = = Sampling Z Distribution Distribution: 0.05 Therefore… Z (critical) 1.96 Z (obtained) = Z (obtained) = Z (obtained) = Two-Sample Case Example ( Known) Cont’d Step 5 – Make a Decision and Interpret the Results Z (obtained) = Z (critical) = Therefore, Z falls in the critical region, and we know that the observed differences in means as large as the one found between the two samples is highly unlikely if the null hypothesis is true… This means we can reject the null, because we know that there is a statistically significant difference between the wage gap of high school graduates and university graduates Two Sample Case ( Unknown) If the population standard deviations are unknown (as they usually will be), we must use the t distribution to find the areas under the sampling distribution The formula is: We must also perform an additional calculation and make an additional = assumption to use the t distribution with two sample means In this equation: = the pooled estimate of the standard The calculation concerns the degrees of deviation of the sampling distribution of freedom (which when using two samples is the differences in sample means equal to ) and = the size of the first and second samples We must also assume that the variances of and = variance (i.e., standard deviation the populations are equal (in order to form a squared) of the first and second samples pooled estimate of the standard deviation of the sampling distribution of the differences in sample means). Two Sample Case( Unknown) Cont’d After you compute the pooled estimate of the sampling distribution of the sample mean differences (i.e., ), substitute it directly into the denominator for the formula for t (obtained)… t (obtained) = Two Sample Case Example ( Unknown) Example Research Question: “Is there a difference in the GPA between graduate students enrolled in psychology and those enrolled in sociology?” We collect two random samples (Sample 1: n = 39 / Sample 2: n = 42)… Sample 1 (sociology) Sample 2 (psychology) = 83.74 = 82.16 = 6.59 = 9.35 = 39 = 42 Two Sample Case Example ( Unknown) Cont’d Step 1 – Make Assumptions and Step 2 – Make Assumptions and Meet Test Requirements Meet Test Requirements Model: Independent Null hypothesis states that the Random Samples populations represented by the Level of samples have no difference (not Measurement is directional) Interval-Ratio Population : variances are equal: Sampling Distribution is Normal Two Sample Case Example ( Unknown) Cont’d Step 3 – Select the Sampling Step 4 – Compute the Test Distribution and Establish the Statistic = Critical Region = Sampling t Distribution = = (8.241)(0.222) = 1.83 Distribution: 0.05 Therefore… T (obtained) = Degrees of = (39+42-2 = Freedom 79) T (obtained) = = +0.86 Z (critical) 2.000 Two Sample Case Example ( Unknown) Cont’d Step 5 – Make a Decision and Interpret the Results t (obtained) = +0.86 t (critical) = 2.00 Therefore, the test statistic does not fall in the critical region, and we fail to reject the null hypothesis This indicates that the observed difference between the sample means is trivial and is likely to result of random chance (i.e., sociology and psychology master’s students are not significantly different in GPAs) Cohen’s d Is a measure of effect size that tells us the magnitude of the difference between two groups on a given variable (i.e., the magnitude between the difference of two means) Is symbolized by d The greater the magnitude, the larger the effect and the stronger the relationship between the two variables Commonly used to determine effect size when using two-sample case technique Cohen’s d Cont’d The formula is for Cohen’s d is: Cohen’s d = In this equation: = the mean for group 1 = the mean for group 2 = the pooled standard deviation Cohen’s d Cont’d There are two formulas to determine the pooled standard deviation (one for when the sample sizes are the same, and one for when the sample sizes are different) When sample sizes are different: When sample sizes are the same: In this equation: = the standard deviation for group 1 In this equation: = the standard deviation for group 2 = the standard deviation for group 1 = the standard deviation for group 2 = sample size group 1 = sample size group 2 Cohen’s d Cont’d In our previous example of wage group difference between university and high school graduates… First, solve for pooled standard Group 1 (university graduates) had a mean deviation: wage of $50 000 and a standard deviation of $12 000 Group 2 (high school graduates) had a mean wage of $40 000 and a standard deviation of $9000 = 10 606.60 Both had sample sizes of 150 ( = 150 and = 150) Cohen’s d Cont’d Then solve for Cohen’s d: Cohen’s d = Cohen’s d = As Cohen’s d increases in value, it indicates greater Cohen’s d = 0.94 effect size – i.e., it increases in size as the difference between two means increases Therefore, a d of 0.94 indicates that the difference between the mean wages of university and high school graduates is large and important t-Tests in SPSS One-Sample t-Test in SPSS Using data from the Canadian Community Health Survey (CCHS), we want to know if people who have poor health work less hours per week than that of the general Canadian population. We assume that people in the general population work 40 hours a week… If we find that the number of hours worked for the sample of those with poor health is significantly less than that of the general population, we can conclude that individuals in poor health in Canada tend to work fewer hours per week than the population as a whole Since we are comparing a sample to the population, we conduct a One-Sample t-Test One-Sample t-Test in SPSS Cont’d In this box, we find descriptive statistics We see that 73 people are in “poor health” The mean number of hours worked among these people is 38.20, which is different to our assumed mean of 40 However, is the difference in means significant? One-Sample t-Test in SPSS Cont’d We see that we have a t (obtained) of -.878 We also have 72 degrees of freedom To test for significance, we can use the Appendix B to compare this to the critical value with an alpha of 0.05 – the critical value is -1.671 With that information, we know the test statistic does not fall in the critical region and is therefore not statistically significant… One-Sample t-Test in SPSS Cont’d However, we know it is a directional hypothesis, because we want to know if people in poor health work fewer hours… (i.e., do they work less than the general population?) Since it is directional, we can look at the significance level under “One-Sided p” We see that the p value is 0.192 This is larger than 0.05 Therefore, the results are not statistically significant, and we cannot reject the null The difference in means we observed was likely the result of random chance Two-Sample Case (Independent t-Test) in SPSS In this example, we will conduct a hypothesis test using the Two-Sample Case technique (i.e., Independent Sample t-Test) in SPSS Here, we will use data from the Canadian General Social Survey (GSS) We want to know if there is a statistically significant difference between the sample means for males and females in the number of hours volunteered If there is, we can conclude that the populations (all Canadian adult males and female) are different on this variable/trait Two-Sample Case (Independent Sample t- Test) in SPSS Cont’d We see that the 270 males in the sample who volunteered have an average 103.67 volunteer hours, with a standard deviation of 206.14 hours We see that the 348 females had an average of 164.15 volunteer hours, with a standard deviation of 308.47 hours Therefore, the mean number of volunteer hours worked are different and on average, males volunteered fewer hours than females But is the difference between these population means statistically significant? Two-Sample Case (Independent Sample t-Test) in SPSS Cont’d We need to assume equal variances in the population… SPSS provides a formal test for this: “Levene’s Test for the Equality of Variances” If the value is above 0.05, we can assume equal variances; if below, we cannot assume equal variance Our value is below 0.05, so we cannot use the row labelled “Equal Variances Assumed” Instead, we use the row “Equal Variances Not Assumed” Two-Sample Case (Independent Sample t-Test) in SPSS Cont’d To determine if the difference in means is statistically significant, we can compare t (obtained) to t (critical) Or we can simply look at statistical significance… Since we only wanted to know if there was a difference between populations, it is a two-tailed test Therefore, we look at the significance under “Two-Sided p” We see that it is 0.04 (below 0.05) – this indicates statistical significance We can reject the null hypothesis… Therefore, there is a statistically significant difference between the number of volunteer hours worked for males and females in the Canadian population Two-Sample Case (Independent Sample t- Test) in SPSS Cont’d Moreover, we see under “Point Estimate” that there is a Cohen’s d score of -.224 This tells us the observed effect size is medium This means that the difference between males and females on the number of hours worked in medium in size The value is negative, because males have a lower mean than females and males were treated as “Group 1” as opposed to “Group 2” Quantitative Research Methods in Political Science Lecture 10: Hypothesis Testing (Interval-Ratio) with Pearson’s Correlation Course Instructor: Michael E. Campbell Course Number: PSCI 2702 (A) Date: 11/21/2024 Introduction Unlike previous hypothesis tests, which treated the independent variable (X) as nominal or ordinal, here we look at tests for two variables measured at interval-ratio level Specifically, we will look at scatterplots and Pearson’s correlation (i.e., the computation of Pearson’s r) We are still concerned with three questions with hypothesis testing: a. Is there a relationship between variables? b. If a relationship exists, how strong is the relationship c. What direction is the relationship We also need to consider statistical significance (to determine if the relationship we see between variables also exists in the population from which our sample is drawn) Please note that this entire lecture is based on the same example throughout Scatterplots Is similar to a bivariate table in terms of the information it provides (it visually represents the relationship between two variables) Is usually the first thing analyzed when examining the relationship between two interval- ratio level variables Allows for a visual inspection of the relationship between two interval-ratio variables The independent variable (X) is arrayed along horizontal axis (much like X is in columns in bivariate table) The dependent variable (Y) variable is arrayed along the vertical axis (much like the Y is in rows in bivariate table) Scatterplots Cont’d Example Research Question: “Does the number of children a family has affect the number of hours the male partner from that family spends on housework each week?” Two variables: 1. Number of children family has (X) 2. Number of hours male partner spends on housework (Y) We have a sample of 12 families (See right)… First column is family identifier Second column is number of children Third column is number of hours spent on housework by male partner Both variables are measured at interval-ratio level X Y Scatterplo ts Cont’d We see that Family A has a score of 1 on the independent variable (X), which corresponds to a score of 1 on the dependent variable (Y) We see that Family B has a score of 1 on the independent variable (X) which corresponds to a score of 2 on the dependent variable (Y) Scatterplots Cont’d If we take each of these data points and place them on a scatterplot, it looks like this… We see the X arrayed along horizontal axis and Y arrayed along vertical axis… Every dot represents an observation that corresponds to a pair of values from the two variables… For Family A, with 1 child and 1 hour spend on housework, we see how the dot/observation is placed on the scatterplot Scatterplots Cont’d We had 12 families in our sample, and each had a score on the different variables… Therefore, we have 12 dots on the scatterplot corresponding with the values on those variables The pattern of the dots/observations tells us about the relationship between the two variables Notice the straight line that cuts through the dots… Scatterplots Cont’d This is the Regression Line… It will either touch every dot, or come as close as possible to every single dot If it were placed anywhere differently, it would not come as close to all dots as possible Think back to the least-squares principle (see PowerPoint 3, Chapter 3) This line is very important for interpreting the relationship between variables Scatterplots serve several purposes: 1. They provide impressionistic information about the existence, strength, and direction of the relationship between variables Scatterplot 2. They can be used to check the relationship for linearity (i.e., how well the patterns of s Cont’d dots/observations can be approximated with a straight line) 3. They can be used to predict the score of a case on one variable from the score of a case on another variable A Scatterplo t in SPSS If we were to use the same data and generate a scatterplot in SPSS , it would look like this… Identifying the Existence of a Relationship Two variables are related if the distributions of Y change according to the conditions of X The dots above each score on X are the conditional distributions of Y – i.e., the dots represent the scores on the dependent variable for each value of X We know that there is some sort of relationship here, because these conditional values change as X changes… Identifying the Existence of a Relationship Cont’d You can also look at the regression line If it lies at an angle (as it does here), we know that there is a relationship… Had there been no relationship, the conditional distribution of Y would not have changed as a function of X and the regression line would not be at an angle… Instead, the regression line would be parallel to the horizontal axis Identifying the Strength of a Relationship The determine the strength of the relationship, you need to look at the way the dots are clustered around the regression line… The more the dots cluster around the regression line, the stronger the association If all the dots line on the regression line, it is a perfect relationship Identifying the Direction of a Relationship The direction of the relationship can be ascertained by observing the direction of the regression line In our case, there is a positive relationship: as the number of children (X) increases, so too does the number of hours the male partner spends on housework (Y) Had the relationship have been negative, the regression line would have gone in the other direction If negative, this would have indicated that high scores on one variable were associated with low scores on the other Identifying the Strength of a Relationship Cont’d Three types of relationships here: 1. Figure A shows a perfectly positive relationship between two variables 2. Figure B shows a perfectly negative relationship between two variables 3. Figure C shows a non-relationship (i.e., a “zero relationship”) between two variables Identifying the Strength of a Relationship Cont’d However, perfect relationships are rare, so we are more likely to scatterplots that look like this… Scatterplots Without Regression Lines Sometimes, a scatterplot will not have a regression line, so you will have to estimate the direction and strength of the relationship based on the pattern of dots alone… To do this, you need to identify the general direction of the dots/observations… You can then draw a rough estimation of the regression line through the data Scatterplots Without Regression Lines For example, we can see that there is a slight positive relationship between these data, because the conditional distributions of Y increase as X increases Using Scatterplots to Test for Linearity If a straight line can be drawn through the datapoints, then it is a “linear relationship” If a straight line cannot be drawn, it is not a linear relationship This would be bad, because linear models (like linear regression) would lead to inaccurate predictions because the model would not be able to capture the complexity of the data This is why we tend to generate scatterplots first – so that we can determine if the relationship is linear Scatterplots can also be used to predict the scores of cases on one variable based on the scores on the other variable Using Scatterplot For example, let’s say we wanted to predict the number of hours of household work the male s for partner would do if a family had 6 children Prediction Remember, our sample only had information on families with up to 5 children… To do this, we need to extend the axes of the scatterplot, as well as the length of the regression line Using Scatterplots for Prediction Cont’d Symbolically, the predicted score of Y is To find (by hand and not using computations) we need to find the relevant score on X that would be associated with 6 children We then need to draw a straight line parallel to the Y axis from that point on the regression line… From there, we draw another straight line, parallel to the X axis, across the Y axis The predicted score would then be found at the point where the regression line cross the Y axis Using Scatterplots for Prediction Cont’d Based on this, we can predict that, in a family with 6 children, the male partner would devote about five hours a week to housework But we did this by hand, so it is only roughly accurate… Using Scatterplots for Prediction Cont’d To ensure accuracy of predictions, it is crucial that the regression comes as close as possible to all datapoints in the scatterplot This brings us back to the least-squares principle (see PowerPoint 3 + Chapter The distance between the line and all data 3) points must be minimized to make the more accurate linear approximation of the data… Remember, the dots above each value on X are considered conditional distributions of Y Within each conditional distribution of Y, the mean is the point around which the variation of scores is at a minimum Using Scatterplots for Prediction Cont’d “If the regression line is drawn so that it touches each conditional mean of Y, it is the straight line that comes as close as possible to all possible scores” (Healey, Donoghue, and Prus 2023, 436). To find the conditional means, you need to sum all Y values for each value of X and divide by the number of cases For example, four families had 1 child (X=1) The male partners in these families devoted 1, 2, 3, and 5 hours of housework Using Scatterplots for Prediction Cont’d “Therefore, X=1 and Y = 1, 2, 3, 5 1+2+3+5 = 11 The number of cases where X=1 is 4 So, 11/4 = 2.75 Therefore, the conditional mean of Y for X = 1 is 2.75 Using Scatterplots for Prediction Cont’d Here, we see all of the conditional means for Y for the various values on X With this information, we can begin computing the best-fitting regression line for summarizing the relationship between X and Y This is accomplished by drawing a lie through the conditional means of Y Doing so will minimize the spread of the observation points and come closest to all the scores as possible It is, therefore, a “line of best fit” Using Scatterplots for Prediction Cont’d We see the conditional means of Y mapped out here… But these dots do not make a straight line… So, how do we draw a straight line through these so that it will come closest to all data points? To do this, we must use the formula for the “least-squares” regression line (or the line of best fit) The “Least-Squares” Regression Line The formula for the “least squares” regression line is: In this equation: Y = the score on the dependent variable = the Y intercept (or the point where the regression line crosses the Y axis) = the slope of the regression line or the amount of change produced in Y by a unit change in X X = the score on the independent variable The “Least- Two new concepts are forwarded in this formula: Squares” 1. The Y intercept Regression This is the point where the regression line crosses the vertical axis Line Cont’d The intercept is equal to the value of Y when X is at 0.00 2. The Slope (b) of the “least-squares” regression line This is the amount of change produced in Y by a unit change in X You can think of the slope as “a measure of the effect of the X variable on the Y variable” (Healey, Donoghue, and Prus 2023, 437) The Intercept (a) and Slope (b) If there is strong association between variables, then changes in the value of X will result in significant changes in the value of Y Therefore, if there is strong association between the variables, the slope (b) will have a high value (conversely, the weaker the association the lower the value) If the variables are completely unrelated, the value of the slope will be 0.00 (thus a “Zero Relationship”) – this would indicate the line has no slope Both the slope (b) and the intercept (a) are expressed using the units of the independent variable, allowing for direct and intuitive interpretation Using the “least squares” formula to produce the “least squares” regression line, we are in the best position to make accurate predictions of the values of Y Computing a and b To solve the terms in our equation (), we follow a series of steps… Remember that a represents the Y intercept (or the point that the regression line crosses the Y axis) b represents the slope of the regression line (or the amount of change produced by Y by a one-unit change in X) You must first compute b to solve for a Computing b The formula for b (i.e., the slope) is: In this equation: = the number of cases = the summation of the cross-products of scores = the summation of X scores = the summation of Y scores = the summation of the squared scores of X Computing b Cont’d We have all of the information in this table to solve the equation for b… The first two columns detail the original scores for X and Y The third and fourth columns contain the squared scores of X and Y The fifth column contains all the cross-products of the scores on the table Computing b Cont’d What does a slope value of 0.069 tell us? It indicates for every one-unit change in X, there will be a 0.69-unit increase in Y In our case, this means that for every additional child a family has, the male partner will spend 0.69 hours more doing housework But remember, we need to complete the “least squares” regression line – so we’re not done yet! Computing a Once you calculate the slope (b), you need to calculate the intercept To do this, we use the following formula: To find the mean values of Y and X, take the total values in the first two columns and divide by the sample size (separately) 32/12 = = 2.67 40/12 = = 3.33 Computing a Cont’d Once you have the means, take them, and the slope value (b = 0.69) and input them into the equation to solve for a… Computing a Cont’d The intercept value of 1.49 tells us that the “least-squares” regression line will cross the Y intercept at the point where Y equals 1.49 This is the estimated value of Y when X is at 0.00 We see this when we look at our a = 1.49 scatterplot (see right) Completing the “Least-Squares” Regression Line Equation With all of this information, we can now complete our “least-squares” regression line equation… This will allow us to make accurate predictions to estimate or predict scores on Y for any value of X = When we did this freehand, we predicted that a score on Y (male partner’s hours of housework) for a family with six children (X = = 6) was about five hours per week But with our formula, we get a more accurate = 5.63 prediction… Based on our “least-squares” regression line, we predict that in a family of 6, the male partner will spend 5.63 hours on household work each week Results of If we wanted to predict the number of hours the male partner would spend on housework each week, if they had seven children, we would simply change the X value in the equation to represent this… the “Least- Squares” = Equation This would give us a value of 6.32 – meaning that in a family with 7 children, we predict the male partner will spend 6.32 hours on housework per week Note that these are not perfect predictions, because they are not perfect relationships… The stronger the relationship between two variables, the more accurate the predictions will be… The Correlation Coefficient (Pearson’s r) The Correlation Coefficient (Pearson’s r) The slope (b) of the “least-squares” regression line is a measure of the effect of X on Y… It is the amount of change produced in Y by a one-unit change in X Therefore, as b increases in value, the strength of the relationship changes… However, b does not range from 0.00 to 1.00, so it is an awkward measure of association To resolve this issue, we use “Pearson’s r” The Correlation Coefficient (Pearson’s r) Cont’d Pearson’s r is a measure of association for two-interval level variables Pearson’s r varies from -1.00 to +1.00 Therefore… -1.00 indicates a perfect negative association +1.00 represents a perfect positive 0.00 to 0.10 = weak association association 0.00 represents no association 0.11 to 0.30 = moderate association However, we rarely find perfect 0.31 to 1.00 = strong association relationships… Calculating Pearson’s r To calculate r, we use the following formula… Calculating Pearson’s r Cont’d We see here that we have a Pearson’s r of 0.50 – indicating a strong and positive relationship between our variables… But is it statistically significant? Does the relationship exist in Testing Pearson’s r for Significance When we work with random samples, we want to know if our results are statistically significant… This will tell us if the relationship exists in the population from which the sample was drawn… Therefore, in this case, we want to test our Pearson’s r for statistical significance (to see if relationship we found between our variables also exists in the population) Thus far, we know r = 0.50… But is it statistically significant? Testing Pearson’s r for Significance Cont’d We can use our five-step hypothesis testing model to do this… Our null hypothesis is that there is no linear-relationship between the two variables in the population – i.e., the number of children a family has (Y) and the number of hours the male partner spends on housework (Y) The population parameter is represented by (pronounced Rho) Population standard deviation is unknown, and sample size is small, so we use t- distribution Testing Pearson’s r for Significance Cont’d Because we’re using Pearson’s r, there are some changes to the assumptions made in Step 1 of the five-step model… 1. We’re using two interval variables, so we need to assume that both are normal in distribution (referred to as bivariate normal distribution) 2. We need to assume that the relationship between variables is roughly linear in form 3. We need to assume that the relationship between variables is Homoscedastic Homoscedasticity Homoscedasticity is good thing when testing for r A homoscedastic relationship occurs when “the variance of Y scores is uniform for all values of X” (Healey, Donoghue, and Prus 2023, 452). In other words, this means that the scores of Y are evenly spread above and below the regression line for the entire length of the line It assumes that the variance of errors (i.e., the distance of the observations from the line of best fit – called “residuals”) is constant across all levels of the independent variable This means we can assume the residuals are drawn from a population with constant variance Homoscedasticity Cont’d To assume homoscedasticity means that the model fits the data well and the statistical tests for regression coefficients are valid… Generally, we can look at the scatterplot and determine if the relationship conforms to the assumptions of linearity and homoscedasticity The datapoints should fall in a “roughly symmetrical, cigar-shaped pattern whose shape can be approximated with a straight line” (Healey, Donoghue, and Prus 2023, 453). Homoscedasticity Cont’d The alternative to homoscedasticity is heteroscedasticity… But you do not need to worry about that for this course… Testing r for Significance Using Five-Step Model Step 1 – Make Assumptions and Step 2 – State Null Hypothesis Meet Test Requirements Model: Random Sampling () Level of Measurement is Interval-Ratio Here, the null assumes that there Bivariate Normal is no relationship in the Distributions population… Linear Relationship The research hypothesis is the Homoscedasticity opposite, and states that there is a relationship in the population Sampling Distribution is Normal Testing r for Significance Using Five-Step Model Cont’d Step 3 – Select the Sampling Distribution and Establish the Step 4 – Compute the Test Critical Region Statistic Sampling t Distribution t (obtained) = Distribution: 0.05 t (obtained) = Degrees of n – 2 = 10 Freedom t (obtained) = (0.50)(3.65) t (critical) t (obtained) = 1.83 Remember, t distribution relies on df We have two variables so df = n-2 Testing r for Significance Using Five-Step Model Cont’d Step 5 – Make a Decision and Interpret the Results t (critical) = t (obtained) = 1.83 The test statistic is not larger than the critical value, therefore we cannot reject the null hypothesis This means the r value of 0.50 could have occurred by random chance alone if the null hypothesis is true and the variables are unrelated in the population Testing Hypothesis using SPSS Output We begin by analyzing the scatterplot… First, we see our IV (X) on the horizontal axis and our DV (Y) on the vertical axis We see that the conditional values of Y do vary as a function of X (i.e., as X changes, Y also changes) Based on the pattern of the dots, as well as the slope of the regression line, we also see that it is a positive relationship (as X increases, Y increases) We also see that the observations generally cluster around the regression line, indicating that the strength of association between our variables is likely strong (i.e., above 0.31) Testing Hypothesis using SPSS Output The box in the middle of the regression line provides us with our a and b values We see that the intercept begins at 1.5 (consistent with our earlier findings) We also see that b = 0.69 – indicating that for every one-unit increase on X, there will be a 0.69-unit increase on Y (again, consistent with our earlier findings) Testing Hypothesis using SPSS Output Cont’d Here, we see that the Pearson’s r value is 0.499 (which can be rounded off to 0.50, demonstrating consistency with our previous calculations)… The r value is positive, supporting our observation that the relationship between variables was positive The r value of 0.499 indicates that the strength of association between variables is strong So, the relationship between variables is strong and positive Testing Hypothesis using SPSS Output Cont’d But we also see that the significance level (beside Sig. (2-tailed) is 0.099 Since this is above the conventional 0.05 alpha level, we cannot reject the null We found a strong, positive association hypothesis… between our variables, but we cannot reject the null because we did not find Therefore, while we found strong statistical significance… association between the two variables in Our r value of 0.499 is likely found by our sample, we cannot reject the null hypothesis… random chance if the null hypothesis is true

Quantitative Research Methods In Political Science Lecture 8 PDF

Document Details

Tags

Related

Summary

Full Transcript

Upgrade to continue