1. Difference Between Means Test (“t” statistic) Analysis of Variance (F statistic)

2. Difference Between Means (“t”) Test So far we’ve examined several statistics that can be used to test hypotheses: – Chi-Square (X 2 ), which requires all variables be categorical – Regression (R 2 ), which requires all variables be continuous – Logistic regression (b and Exp b), which requires a nominal dependent variable The difference between the means test (t) is used to test hypotheses with categorical independent and continuous dependent variables – Gender  height – Gender  cynicism (1-5 scale) We compare the means of two randomly drawn samples – The null hypothesis can be rejected if the difference, reflected in the magnitude of the t statistic, goes so far beyond what would be expected by sampling error, that there are less than five chances in 100 (p>.05) that the relationship between the variables is due to chance – This sampling error is called the “standard error of the difference between means” – the difference between all possible pairs of means, due to chance alone When using the t table we must know whether the hypothesis is 1-tailed (direction of effect predicted) or 2-tailed (direction not predicted) Major advantage: Remember that weak real-life effects can produce significant results? – When comparing means, we know their actual values. This lets us recognize situations where differences are, in the real world, trivial.

3. Calculating t 1. Obtain the “pooled sample variance” S p 2 (Simplified method – midpoint between the two sample variances) 2.Compute the S.E. of the Diff. Between Means 3.Compute the t statistic 4.Compute the “degrees of freedom” df = n 1 + n 2 - 2 (total number of cases in both samples minus 2) Actual (“obtained”) difference between means Predicted difference due to sampling error The t is a ratio: the greater the difference between means, the smaller the predicted error, the larger the t coefficient The larger the t, the more likely we are to reject the null hypothesis. According to the null, any relationship between variables, any difference between means, is due to chance. Our actual difference between means must be “significantly” larger than the difference we would obtain by chance. We use a table to determine whether the t is large enough to reject the null hypothesis (see next slide). We can reject the null if the probability that the difference between means is due to chance is less than five in one-hundred (p<.05). If the probability that the difference between the means is due to chance is five in one-hundred or larger (p>.05), the null hypothesis is true.

4. Classroom exercise - Jay’s police department H 1 : Male officers more cynical than females (1 - tailed) H 2 : Officer gender determines cynicism (2 - tailed) 1.Draw one sample of male officers, one of females 2.Compute each sample’s variance, then obtain the pooled sample variance 3.Compute the S.E. of the Difference Between Means 4.Calculate the t coefficient 5. C ompute the degrees of freedom df = n 1 + n 2 - 2 6.Check the t table (next slide) for significance. Confirm the (working) hypothesis if there are less than 5 chances in 100 that the null is true. Be sure to use the correct significance row (1-tailed or 2-tailed). Use the one-tailed test if the working hypothesis predicted the direction of the difference (H 1 ) - that is, that one group, male or female, would be significantly more cynical than the other. Use the two-tailed test if you predicted that cynicism would be significantly different, but not which group would be more cynical (H 2 ). One-tailed hypotheses require a smaller t to reach statistical significance. t-table

5. 1. Is hypothesis one-tailed (direction of change in the DV predicted) or two-tailed (direction not predicted)? H 1 : Males more cynical than females. This is one-tailed, so use the top row. H 2 : Males and females differ in cynicism. This is two-tailed, so use the second row. 2. df, “Degrees of Freedom” represents sample size – add the numbers of cases in both samples, then subtract two: df = n 1 + n 2 – 2 3. To call a t “significant” (thus reject the null hypothesis) the coefficient must be as large or larger than what is required at the.05 level; that is, we cannot take more than 5 chances in 100 that the difference between means is due to chance. For a one-tailed test, use the top row, then slide over to the.05 column. For a two-tailed test, use the second row, then slide to.05 column. If the t is smaller than the number at the intersection of the.05 column and the appropriate df row, it is non-significant. If the t is that size or larger, it is significant. Slide to the right to see if it is large enough to be significant at a more stringent level.

6. Parking lot exercise Higher income  More expensive car 1.Transfer your panel’s data from the other coding sheet 2.Compute each sample’s variance, then calculate the pooled sample variance = = 1.49 3.Compute the S.E. of the Difference Between Means  1.49 ( 1 + 1 ) 10 10  1.49 (.2) = .3 =.55 …continued on next slide 1.4 1.4 21111121222111112122.6.4.4.4.4.4.6.4.6.6.36.16.16.16.16.16.36.16.36.36 2.4.27 3.3 45252144514525214451.7 1.7 1.3 1.7 1.3 2.3.7 1.7 2.3.49 2.9 1.7 2.9 1.7 5.3.49 2.9 5.3 24.17 2.7 2.97 2

7. 4.Calculate the t coefficient = = -3.5 Note: the sign, + or -, indicates the direction of the difference between groups. Keep that in mind! It turns out that, consistent with the hypothesis, the faculty lot has the more expensive cars. We had arbitrarily placed it second, so subtracting yields a negative t. 5.Compute the df (degrees of freedom) df = n 1 + n 2 - 2 = 10 + 10 - 2 = 18 6.Check the t table for significance. Note: Use the one-tailed test if the working hypothesis predicts the direction of the difference (which parking lot would have more expensive cars). Use the two-tailed test if the hypothesis predicts there will be a difference in car values between the lots, but not which lot would have the more expensive cars. 1.4 - 3.3.55 -1.9.55 df = 18 t = -3.5 Yes! There are less than five chances in one-thousand that the null hypothesis is true (one-tailed) or less than one in one-hundred that it is true (two-tailed)

8. More complex mean comparisons: Analysis of Variance

9. Dependent variable: continuous Independent variable(s): categorical Example: does officer professionalism vary between cities? (scale 1-10) Calculate the “F” statistic, look up the table. An “F” statistic that is sufficiently large can overcome the null hypothesis that the differences between the means are due to chance. When there are more than two groups: Analysis of Variance CityL.A.S.F.S.D. Mean853

10. Stratified independent variable(s) F statistic is a ratio of “between-group” to “within” group differences. To overcome the null hypothesis, the differences in scores between groups (between cities and, overall, between genders) should be much greater than the differences in scores within cities Between group variance (error + systematic effects of ind. variable) Within group variance (how scores disperse within each city) “Two-way” Analysis of Variance CityL.A.S.F.S.D. Mean – M1075 Mean - F632 Between Withi n

11. Homework

12. Two random samples of 10 patrol officers from the XYZ Police Department, each officer tested for cynicism (continuous variable, scale 1-5) Sample 1 scores: 3 3 3 3 3 3 3 1 2 5 -- Variance =.99 Sample 2 scores: 2 1 1 2 3 3 3 3 4 2 -- Variance =.93 Homework assignment

13. Pooled sample variance S p 2 Simplified method: midpoint between the two sample variances S p 2 = Standard error of the difference between means   x1 -  x2   x 1 -  x 2 =  S p 2 ( ) T-Test for significance of the difference between means  x 1 -  x 2 t = --------------   x 1 -  x 2 s 2 1 + s 2 2 2 1 n 1 n 2 +

14. CALCULATIONS Pooled sample variance:.96 Standard error of the difference between means:.44 t statistic: 1.14 df – degrees of freedom: (n1 + n2) – 2 = 18 Would you use a ONE-tailed t-test OR a TWO-tailed t- test? Depends on the hypothesis Two-tailed (does not predict direction of the change): Gender  cynicism One-tailed (predicts direction of the change): Males more cynical than females Can you reject the NULL hypothesis? (probability that the t coefficient could have been produced by chance must be less than five in a hundred) NO – For a ONE-tailed test need a t of 1.734 or higher NO – For a TWO-tailed test need a t of 2.101 or higher

15. Final exam practice

16. You will be given scores and variances for two samples and asked to decide whether their means are significantly different. You will be asked to state the null hypothesis. You will then compute the t statistic. You be given formulas, but should know the methods by heart. Please refer to week 15 slide show. To compute the t you will compute the pooled sample variance and the standard error of the difference between means. You will then compute the degrees of freedom (adjusted sample size) and use the t table to determine whether the coefficient is sufficiently large to reject the null hypothesis. – Print and bring to class: http://www.sagepub.com/fitzgerald/study/materials/appendices/app_f.pdf http://www.sagepub.com/fitzgerald/study/materials/appendices/app_f.pdf – Use the one-tailed test if the direction of the effect is specified, or two-tailed if not You will be asked to express using words what the t-table conveys about the significance (or non-significance) of the t coefficient Sample question: Are male CJ majors significantly more cynical than female CJ majors? We randomly sampled five males and five females. Males: 4, 5, 5, 3, 4 Females: 4, 3, 4, 4, 5 – Null hypothesis: No significant difference between cynicism of males and females – Variance for males (provided): 0.7 Variance for females (provided): 0.5 – Pooled sample variance =.6 SE of the difference between means =.49 t =.41 df = 8 – Check the “t” table. Can you reject the null hypothesis? NO – Describe conclusion using words: The t must be at least 1.86 (one-tailed test) to reject the null hypothesis of no significant difference in cynicism, with only five chances in 100 that it is true.