Hypothesis Testing and Chi-Square Test in Inferential Statistics, Exams of Psychology

Download Hypothesis Testing and Chi-Square Test in Inferential Statistics and more Exams Psychology in PDF only on Docsity! 1 Hypothesis Testing and Chi Square STAT Chap 13 PSYC 201 – Psychological Research I Inferential Statistics are used to make three kinds of inferences Point estimates A statistic (e.g., the mean) computed from a single sample of subjects is used to estimate the mean of the population of subjects Interval estimates or Confidence Intervals An interval, with limits at either end, with a specified probability (usually 95%) of including the parameter being estimated (e.g., the population mean) Hypothesis testing or Statistical Significance A process whereby you decide whether the differences or relationships observed in the data are due to chance fluctuations or whether they are sufficiently large enough to consider them “significant” or not due to chance We’re ready for hypothesis testing! We’re interested, right now, in going over the general strategy of inferential statistics with regards to hypothesis testing And, we’re going to apply it to the chi square test used to analyze categorical data 2 Inferential statistics are based on probability statements Probability statements are statistical statements that Describe the data Predict what we don’t know, on the basis of current data So: inferential statistics are all about making inferences about the population based on sample data Therefore: it is very important to acquire a sample of subjects or observations that truly reflects the population E.g: Population = all fourth grade children in the U.S. Sample = 100 fourth grade children from all elementary schools in Montgomery County Sampling How you select the participants in the sample is extremely important: In order to generalize from sample to population, the sample must be representative of the population If the sample has not been generated randomly (each possible sample of that size has an equal chance of being selected), then the laws of probability won’t apply as well to the sample Therefore: your statistical conclusions will be faulty 5 Step 3 (continued) Determine the rejection region using α = .05 You will usually need the degrees of freedom (df) to obtain the critical value for the test statistic Then check in the appropriate Appendix in your STAT book (page 296) If you’re doing a Goodness of Fit test: df = k – 1 k = number of categories df = 4 – 1 = 3 χ 2 critical = 7.81 If you’re doing a Test of Independence: df = (r-1)(c-1) r = number of rows & c = number of columns The Hypothesis Test: When requested to set up the Hypothesis Test, the following example is what is expected: H0: χ 2 = 0 α = .05 H1: χ 2 ≠ 0 df = 4 – 1 = 3 χ 2 critical = 7.81 The critical value for your chi square problem is the value of χ 2 that you have to obtain, or bigger, when you do your calculations in order to reject the null hypothesis Step 4 (Data analysis) Once the hypothesis test has been set up, then you check the data, enter it into the computer (as needed), summarize it, and compute inferential statistics to determine the obtained value of the test statistic If done on the computer using SPSS, the program will give you the exact probability of obtaining that value of the test statistic for that number of subjects, assuming that the null hypothesis is true It is called Sig., short for significance level 6 The chi square test Chi square is used predominantly for categorical or nominal data Specifically, it is used when one or all of the variables are nominal in nature Rates of occurrence, frequency data E.g., number of people using a cell phone vs. not using it E.g., number of blonds, redheads, or brunettes in class E.g., percentage of people receiving social security Chi Square The chi square test is a “distribution-free” statistic, meaning that it is not based on the normal curve (nonparametric). Chi square is an interesting nonparametric test that allows you to determine if what you observe in a distribution of frequencies would be what you would expect to occur by chance alone. There is a two-sample test that includes two dimensions, such as whether political party affiliation is independent of gender Example problem #1 A director of a small psychotherapy clinic is trying to plan hiring of temporary staff to assist with intake and is wondering if there is any difference in the use of the clinic at different seasons of the year. Based on last years data showing how many clients were admitted during each season, should the director conclude that season makes a difference? 51163328 FallSummerSpringWinter 7 One-sample chi square test: Goodness of Fit One category: season, with 4 levels The rationale is that in any set of occurrences, you can easily compute what you would expect by chance: Simply divide the total number of occurrences by the number of classes or categories or levels Therefore, 28 + 33 + 16 + 51 = 128 128 / 4 = 32 Chance? If season makes no difference, or if the only thing operating is chance, then you would expect 32 clients each season (the null hypothesis) How different are the observed numbers from the expected numbers? Compute the chi square value and compare it to the critical value of chi square determined earlier to make the decision Χ2 = Σ (O – E)2 E Χ2 is the chi-square value Computing chi square Σ is the summation sign O is the observed frequency E is the expected frequency 10 Outcomes of Decision Making Type II Error p = β Correct Decision p = 1 - α Do not reject Null Correct Decision p = 1 - β = power Type I Error p = α Reject Null Null is FalseNull is TrueTrue state of affairs Your decision ↓ Psychologists generally consider the Type I error more serious (you assume that there is an effect in the population when in fact there is not) How do you decrease the risk of a Type I error? We’ll get into this a bit later when we talk about Statistical Power Step 7 (Write the results in APA-style) When you write up the results in an APA-style paper you must include the following information for each analysis conducted: Describe the data analyzed Describe the statistical test used State whether or not the results were significant Report the STAT statement (Χ2(1, N = 387) = 1.37, p = .242) Report the descriptive statistics Provide a conclusion 11 Chi square Test of Independence When there are 2 variables, you set up a contingency table Example: In a study on the phenomenon of “blaming the victim” in prosecutions for rape, two variables were examined: 1) whether the defense alleged that the victim was partially at fault (Low vs. High fault), and 2) whether or not the defendant was found guilty or not. (Based somewhat on Pugh, 1983). The Hypothesis Test for our Test of Independence example: H0: χ 2 = 0 α = .05 H1: χ 2 ≠ 0 df = (r-1)(c-1) = (2-1)(2-1) = 1 x 1 = 1 df = 1 χ 2 critical = 3.84 The number of observations in each cell are presented below Column 1 Column 2 Guilty Not Guilty Row E O X 2 E O X 2 Totals Low 153 24 177 Row 1 Fault 105 76 181 Row 2 High Fault Column 258 100 Totals 12 Expected frequencies In order to obtain the expected frequency for each cell in the contingency table you do the following: Eij or fe = (row total) * (column total) N After you fill in the expected value for each cell, then you just apply the formula again to obtain each cell’s chi square value (Σ ((O – E)2/E)): Column 1 Column 2 Guilty Not Guilty Row E O X 2 E O X 2 Totals Low 153 24 177 Row 1 Fault 128 49.4 105 76 181 Row 2 High Fault 130 50.6 Column 258 100 358 Totals Column 1 Column 2 Guilty Not Guilty Row E O X 2 E O X 2 Totals Low 153 24 177 Row 1 Fault 128 5.0742 49.4 13.0915 105 76 181 Row 2 High Fault 130 4.9621 50.6 12.8022 Column 258 100 358 Totals chi square = X2 = 5.07 + 13.1 + 4.96 + 12.8 = 35.9300