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Answers Note: There are examples of single-sample z-tests and t-tests, related and independent t-tests, ANOVA, and correlation here. Focus on the ANOVA and correlation questions when practicing calculations, but make sure that you can correctly identify which test to do for each type of question, and that you can write out the hypotheses for each type. Also, if you think an answer may be incorrect, please let me know! 1. This is a single-sample test, and since you are given the population standard deviation you would do a z-test. Note that you won’t have to calculate z-tests for the final, but here is the answer in case you are interested. Standard error of the mean = 0.35 z = 7.14 Since the obtained value of z (7.14) is greater than or equal to the critical value (1.96), we reject the null hypothesis and conclude that limestone increases the life expectancy of fish. 2. You have to conduct a full ANOVA for this question. The summary table is as follows: Source Between Within df 3 16 SS 187.75 55.2 MS 62.58 3.45 F 18.14 Since the critical value of F = 3.24 and the obtained value of F = 18.14, you would conclude that there is a significant difference between the groups. You would then conduct post hoc analyses to determine which groups are different. Comparison Group 1-2 Group 1-3 Group 1-4 Group 2-3 Group 2-4 Group 3-4 X1 – X2 .8 6.6 6.4 5.8 5.6 .2 Q .8/.83 = .96 6.6/.83 = 7.95 6.4/.83 = 7.71 5.8/.83 = 6.99 5.6/.83 = 6.75 .2/.83 = .24 Result Not different Significantly different Significantly different Significantly different Significantly different Not different The critical value of Q = 4.05, therefore, we can conclude that Groups 1 and 3 are significantly different; Groups 1 and 4 are significantly different; Groups 2 and 3 are significantly different; and Groups 2 and 4 are significantly different. The other groups are not significantly different. 3. This question requires a correlational analysis, because it asks about the relationship between test scores. r = .80 The critical value of r = .811. Since the obtained value is not greater than or equal to the critical value, we do not reject the null hypothesis and conclude that the scores on exam 1 and exam 2 are not significantly related. Therefore there is a strong, positive correlation between exam 1 and exam 2. However, this correlation is not significant. r2 = .64 64% of the variance in the grades on Exam 2 is due to the grades on exam 1. 4. This is a single-sample test, and since you do not have the population standard deviation you need to conduct a t-test for single samples. You will not need to calculate single-sample ttests on the final exam, but here is the answer anyway. Estimated standard error = 0.49 t = 10.82 The critical value of t = 1.96 Since the obtained value of t (10.82) is greater than or equal to the critical value (1.96), we reject the null hypothesis and conclude that the sample differs from the population. 5. This is a correlational study. r = .90 The critical value of r = .63. Since the obtained value is greater than or equal to the critical value, we reject the null hypothesis and conclude that the scores from Judge 1 and Judge 2 are significantly related. Therefore there is a strong, positive correlation between exam 1 and exam 2. This correlation is significant. r2 = .81 81% of the variance in the scores from Judge 2 is due to the scores of Judge 1. 6. This is a related-samples t-test. Standard deviation = 1.43 Standard error = .45 Obtained value of t = 3.11 The critical value of t = 2.262 Therefore we reject the null hypothesis. 7. This is a single-sample test, and since you are given the population standard deviation you would do a z-test. Standard error of the mean = 1.4 z = 2.14 Since the obtained value of z (2.14) is greater than or equal to the critical value (1.96), we reject the null hypothesis and conclude that smoking cessation program is effective. 8. Because this question asks about differences between groups (rather than just the relationship between them), you would do a related-samples t-test. The obtained value of t is .21 (or –.21), and the critical value is 2.365 (df = 6). Therefore, you would fail to reject the null hypothesis, and conclude that there is no significant difference in attractiveness between people and their dogs. 9. This is a single-sample test, and since you do not have the population standard deviation you need to conduct a t-test for single samples. Estimated standard error = 408.25 t = 2.45 The critical value of t = 2.571 Since the obtained value of t (2.45) is not greater than or equal to the critical value (2.571), we do not reject the null hypothesis and conclude that the plant is being honest about its wastewater. 10. The ANOVA summary table is as follows: Source Between Within Total df 4 21 25 SS 648 2255.4 2903.4 MS 162 107.40 F 1.51 Since the critical value of F is 2.84, we fail to reject the null hypothesis, and conclude that there is no difference between the dosages. Therefore we don’t need to conduct post hoc tests.