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Review Question Answers Note: x’ is used as the symbol for the mean sqrt stands for square root 1. mean = 6.38 st dev = 1.61 2. Confidence interval (two tailed): = 0.05 df = n-1 = 4 t,df = 2.78 Interval = 6.38 +/- 2.78 * 1.61/sqrt(5) = 6.38 +/- 2.00 [4.38 to 8.38] 3. c (because the value 28.2 is an outlier which may influence the mean) 4. Overall mean = (5 * 53.8) + (4 * 47.4) + (5 * 52.2) + (4 * 43.8) / 18 = 49.7 mm 5. c (two independent sample t-test) 6. = 0.85, e = 2.71828, x = 1 P(X) = 0.851 * 2.71828-0.85 1! = 0.36 7. a (one sample KS test using Poisson’s) 8. d (errors terms must be homoscedastic – equal variance) 9. a (when many overcast days, we expect more cases of depression) 10. b 11. b (from table, b0 = 26.473, b1 = -0.492, equation is Y = 26.473 – 0.492X) 12. a (df are number of independents = 1 and n-2 = 33) 13. c (cannot calculate correlation with nominal data) 14. a (two tailed hypotheses test only for significant correlation, not positive or negative) Review Answers Page 1 15. c 16. Translate into Z score Z = ($30,000 - $38,500)/$12,643 = -0.67 Look up probability on Z table: P(0 to –0.67) = 0.2486 P(-0.67 to end) = 0.5 – 0.2486 = 0.2514 Probability of income less than $30,000 = 0.25 17. Translate into Z score Z = ($50,000 - $38,500)/$12,643 = 0.91 Z = ($60,000 - $38,500)/$12,643 = 1.70 Look up probability on Z table: P(0 to 0.91) = 0.3186 P(0 to 1.70) = 0.4554 P(0.91 to 1.70) = 0.4554 – 0.3186 = 0.1368 Probability of income between $50,000 and $60,000 = 0.14 18. Test: use one sample difference of means test because comparing sample value to known value Assumptions: data ($) are ratio, population distribution assumed to be normal, population mean ($36,750) is known Hypotheses: (non-directional) H0: sample mean equals known value (x’ = ) Ha: sample mean greater or less than known value (x’ ) Prob distribution: use t distribution Significance level: use = 0.05 (standard 95% confidence) Critical value: at = 0.05 and df (n-1) = 40-1 = 39, tc for two tailed test = 2.02 Review Answers Page 2 Test statistic: x’ (1999) = $38,500, (1995) = $36,750, s = $12,643, n = 40 t* = 38,500 – 36,750 12,643/sqrt(40) = 0.875 Decision: reject H0 if t* > tc t* (0.875) is less than tc (2.02) so cannot reject H0 Inference: we infer with 95% confidence that there is no significant difference between the 1999 mean income and the 1995 mean income in Victoria. 19. d (use formula for small sample because n=22) 20. e (use formula for sample containing individual observations) 21. c (paired t-test because each sample contains the same individuals, directional hypotheses to determine if internet use has increased) 22. Test: use one sample chi-square test with uniform expected frequencies to see if frequencies (number of students with access to computer) are different between the schools. Assumptions: schools are nominal categories, categories are mutually exclusive, frequencies (number of students) are absolute counts, expected frequencies for uniform distribution are: 290/5 = 58 (all expected frequencies greater than 5). Hypotheses: H0: there is no difference between the observed and expected frequencies Ha: there is a significant difference between observed and expected frequencies Probability distribution: use 2 distribution Significance level: use =0.05 (standard 95% confidence) Critical value: at = 0.05 and df = k-1 = 4, 2c = 9.49 Test statistic School Observed Eagle Bay 82 Elm Tree 43 Glen Valley 58 King George 36 Ridge Crest 71 Review Answers Expected 58 58 58 58 58 (O-E)2/E 9.93 3.88 0.00 8.34 2.91 Page 3 Total 290 25.07 2* = 25.07 Decision: Reject H0 if 2* > 2c 2* (25.07) is greater than 2c (9.49) so we reject H0 Inference: We infer with 95% confidence that there is a difference between the observed and expected (uniform) frequencies. 23. d (test will only tell you if there are differences, you cannot make any causal inferences) 24. p = allergy/total children = 210/608 = 0.35 25. Confidence interval (two tailed): = 0.05, /2=0.025 p=0.35 Z/2 = 1.96 Interval = 0.35 +/- 1.96 * sqrt(0.35 *(1-0.35)) 608 = 0.35 +/- 0.038 [0.31 to 0.39] 26. P(allergies given industrial city) = 89/209 = 0.43 27. P(rural area given allergies) = 43/210 = 0.20 28. a (chi-square because data are nominal, df = (k-1)*(l-1) = 1 * 2 = 2) 29. Answer d b f c g a e Inferential test One sample difference of means Regression Two or more sample Chi-square Spearman’s 1 sample chi square 1-sample K-S 2 sample difference of means 30. c Review Answers Page 4 31. Problem: this sample may not represent the true population of people who shop at the store. Because this is a non-probability sample, the retailer will not be able to use any inferential tests on the data. However, he could use descriptive statistics on his dataset if he wants. 32. c (directional because they are testing if the mean of sample A is greater than the mean of sample B) 33. a (at = 0.01 and df (n1+n2-2) = 15+15 –2 = 28, tc for one tailed test is 2.47). 34. Because the data are unreliable, the researcher should use a nonparametric test for multiple samples – the chi-square test which uses frequencies and categories instead of actual data values. 35. With unreliable data, the researcher should use a non-parametric test for differences between two samples – the two sample KS test (which also uses rankings). The researcher could also use the 2 or more sample Chi-square test but it would be better to downgrade to an ordinal test, rather than a nominal test (less information will be lost). Review Answers Page 5