Fall 2005 Stats Exam Solutions: Normal Dist., Confidence Intervals, ANOVA, Regression - Pr, Exams of Data Analysis & Statistical Methods

Download Fall 2005 Stats Exam Solutions: Normal Dist., Confidence Intervals, ANOVA, Regression - Pr and more Exams Data Analysis & Statistical Methods in PDF only on Docsity! Stat./For./Hort. 571 Final Exam, Fall 2005 Brief Solutions 1. Let W = weight of a sweet corn ear ∼ N(11.1, 3.12). (a) The 8th percentile is 1.41 standard deviation below the mean, or 11.1 − 1.41 × 3.1 = 6.74 oz. (b) P (W > 12.0) = P (Z > 12.0−11.1 3.1 ) = P (Z > 0.29) = 0.3859. (c) With n = 12, W̄ ∼ N(11.1, 3.12/12) and thus P (W̄ > 12.0) = P (Z > 12.0−11.1 3.1/ √ 12 ) = P (Z > 1.00) = 0.1587. (d) The number of ears weighing more than 12.0 oz is the random variable Y ∼ B(12, 0.3859). Thus P (Y ≤ 1) = 0.614112 + 12 × 0.3858 × 0.614111 = 0.0247. (e) For Y ∼ B(288, 0.3859), use YNA ∼ N(µ, σ2) where µ = 288 × 0.3859 = 111.1392, σ2 = 288 × 0.3859 × 0.6141 = 68.2506 = 8.26142 . Thus P (Y ≥ 125) ≈ P (YNA ≥ 125) = P (Z > 125−111.1392 8.2614 ) = P (Z > 1.68) = 0.0465. Normal approximation is OK because 288 × 0.3859 > 5 and 288 × 0.6141 > 5. 2. (a) Let µ denote the population mean LVEF. Since ȳ = 6.05/27 = 0.2241, s = p 0.1661/(27 − 1) = 0.07993, the 99% confidence interval for µ is ȳ ± tα/2,n−1 s√n , which is 0.2241± 2.779× 0.07993/ √ 27, or 0.22± 0.04 = [0.18, 0.26]. (b) Let p denote the population proportion of subjects with LVEF greater than 0.25. Since p̂ = 7/27, the 95% confidence interval for p is p̂ ± zα/2 p p̂(1 − p̂)/n, which is 0.2593 ± 1.96 p 0.2593 × 0.7407/27, or 0.26 ± 0.17 = [0.09, 0.43]. (c) Circle #2 and comment on the robustness of t statistics against non-normality. 3. (a) Compute MSError as the average of the k = 6 sample variances, which is 79.09 on 18 df. Thus SSError is 18 × 79.09 = 1423.71 and SSTrt = SSTotal − SSError = 1370.29 on 5 df. The ANOVA table is Source df SS MS Trt 5 1370.29 274.06 Error 18 1423.71 79.09 Total 23 2794 – For H0 : µi’s are all equal versus HA : not all µi’s are equal, the observed f = 274.06/79.09 = 3.46. Compare with F on df = (5, 18), the p-value P (F5,18 ≥ 3.46) is between 0.01 and 0.05. Reject H0 at 5% level and there is moderate evidence against H0. (b) Consider one of the four assumptions: correct model, in- dependence within and among treatment groups, equal variance, and normality. 4. (a) Compute ȳ − (ȳA + ȳB)/2, which are −2.01, 2.06, 10.39, 19.37 for wells C, D, E, and F, respectively. The difference that would lead to a rejection of H0 is t0.04/4,18 q 79.09 4 (1 + (− 1 2 )2 + (− 1 2 )2) = 2.552 × 5.45 = 13.9. Thus reject H0 for well F but do not reject H0 for wells C, D, and E. (b) At this level of significant, there is no evidence that sul- fide concentration is higher at wells C, D, and E, but there is evidence at well F, all compared to the mean sulfide concentration of wells A and B. (c) These contrasts can help examining various issues such as whether the industrial plant is increasing the sulfide concentration in the ground water. 5. (a) The fitted slope is b̂1 = P i (xi−x̄)(yi−ȳ) P i (xi−x̄) 2 = −41.9/5540 = −0.007563 and the fitted intercept is b̂0 = ȳ − b̂1x̄ = 0.759 − (−0.007569) × 48.9 = 1.1288. (b) For H0 : b1 = 0 versus HA : b1 < 0, use an f-test. Since SSTotal = 0.756 and SSRegression = b̂1 P i(xi − x̄)(yi − ȳ) = (−0.007569) × (−41.9) = 0.3168, we have SSError = 0.756 − 0.3168 = 0.4392. The observed f = MSReg/MSErr = 0.3168 0.4292/39 = 28.13. Compare with F on df = (1, 39), the p-value P (F1,39 ≥ 28.13) is less than 0.001 and thus the final p-value is less than 0.0005 (for a one-sided alternative). Reject H0 at 5% level and there is very strong evidence against H0. (c) Use ŷ = b̂0+b̂1x ∗ = 1.1288−0.007569x∗ . When x∗ = 16, ŷ = 1.0078 and when x∗ = 50, ŷ = 0.7507. (d) Caution against extrapolation at x∗ = 16. Grade Distribution 100: 1 90-99:27 80-89:47 70-79:25 mean = 80.8, median = 84 60-69:13 below: 8