Statistical Analysis of Two Populations: Hypothesis Testing for Means and Variances, Assignments of Mathematics

Download Statistical Analysis of Two Populations: Hypothesis Testing for Means and Variances and more Assignments Mathematics in PDF only on Docsity! Chapter 7: Inferences about Population Variances 7.1 a. 0.01 b. 0.90 c. 1 - 0.99 = 0.01 d. 1 - 0.01 - 0.01 = 0.98 7.2 Let χ2a be the upper 100a−percentile from a Chi-square distribution. a. 21.92 b. 3.186 7.3 a. Let y be the quantity in a randomly selected jar: Proportion = P (y < 32) = P (z < 32−32.3.15 ) = 0.0228 ⇒ 2.28% b. The plot indicates that the distribution is approximately normal because the data values are reasonably close to the straight-line. c. 95% C.I. on σ : (√ (50−1)(.135)2 70.22 , √ (50−1)(.135)2 31.55 ) ⇒ (0.113, 0.168) d. Ho : σ ≤ 0.15 versus Ha : σ > 0.15 Reject Ho if (n−1)(s)2 (.15)2 ≥ 66.34 (50−1)(.135)2 (.15)2 = 39.69 < 66.34 ⇒ Fail to reject Ho and conclude the data does not support σ greater than 0.15. e. p-value = P ( (n−1)(s) 2 (.15)2 ≥ 39.69) Using the Chi-square tables with df = 49, 0.10 < p-value < 0.90 (Using a computer program, p-value = 0.8262). 7.4 a. The box plot is symmetric with whiskers of approximately the same length. The only possible deviation from a normal distribution would be that there is an outlier, however, with a large sample size, a single outlier is not completely inconsistent with samples from a normal population. b. 95% C.I. on σ : (√ (100−1)(11.35)2 128.42 , √ (100−1)(11.35)2 73.36 ) ⇒ (9.97, 13.19) c. Ho : σ ≤ 10 versus Ha : σ > 10 Reject Ho if (n−1)(s)2 (.15)2 ≥ 123.23 (100−1)(11.35)2 (10)2 = 127.53 > 123.23 ⇒ Reject Ho and conclude the data supports the contention that σ is greater than 10. 7.5 a. The box plot is symmetric but there are four outliers. Since the sample size is 150, a few outliers would be expected. However, four out of 150 may indicate the population distribution may have heavier tails than a normal distribution. This may cause the values of s to be inflated. 53 b. 99% C.I. on σ : (√ (150−1)(9.537)2 197.21 , √ (150−1)(9.537)2 108.29 ) ⇒ (8.290, 11.187) c. Ho : σ2 ≤ 90 versus Ha : σ2 > 90 With α = 0.05, reject Ho if (n−1)(s)2 90 ≥ 178.49 (150−1)(9.537)2 90 = 150.58 < 178.49 ⇒ Fail to reject Ho and conclude the data fails to support the statement that σ2 is greater than 90. 7.6 p-value = P (χ2 ≥ 150.58) = 0.4484 7.7 a. The box plot is symmetric with a single outlier. Since the sample size is 81, a few outliers would be expected. Thus, the normality of the population distribution appears to be satisfied. b. Ho : σ ≥ 2 versus Ha : σ < 2 With α = 0.05, reject Ho if (n−1)(s)2 (2)2 ≤ 60.39 (81−1)(1.771)2 (2)2 = 62.73 > 60.39 ⇒ Fail to reject Ho and conclude the data fails to support the contention that σ is less than 2. p-value = 0.0772 c. 95% C.I. on σ : (√ (81−1)(1.771)2 106.63 , √ (81−1)(1.771)2 57.15 ) ⇒ (1.534, 2.095) 7.8 We need to assume that the two samples were independently selected from normally dis- tributed populations. Ho : σ21 = σ 2 2 versus Ha : σ 2 1 6= σ22 With α = 0.10, reject Ho if s21 s22 ≤ 13.68 = 0.272 or s21 s22 ≥ 3.29 s21/s 2 2 = 0.583 ⇒ 0.272 < 0.583 < 3.29 ⇒ Fail to reject Ho and conclude the data does not support the contention that the population variances are different. 7.9 Ho : σ2A ≤ σ2B versus Ha : σ2A > σ2B With α = 0.05, reject Ho if s2A s2B ≥ 3.79 s2A/s 2 B = 3.15 < 3.79 ⇒ Fail to reject Ho and conclude the data does not support σ2A being greater than σ 2 B . 7.10 a. 95% C.I. on σOld : (√ (61−1)(0.231)2 83.30 , √ (61−1)(0.231)2 40.48 ) ⇒ (0.196, 0.281) 95% C.I. on σNew : (√ (61−1)(0.162)2 83.30 , √ (61−1)(0.162)2 40.48 ) ⇒ (0.137, 0.197) b. Ho : σ2Old ≤ σ2New versus Ha : σ2Old > σ2New With α = 0.05, reject Ho if s2Old s2New ≥ 1.53 s2Old/s 2 New = 2.033 > 1.53 ⇒ Reject Ho and conclude the data supports the statement that σ2New is less than σ 2 Old. 54 7.14 a. The box plots are given here: 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 Brand I Brand II Longevity of Two Brands of Tires Brand of Tire M ile s t o Ti re W ea ro ut (1 00 0 m ile s) The box plots and normal probability plots indicate that both samples are from nor- mally distributed populations. b. The C.I.’s are given here: Method n Mean 95% C.I. on µ St.Dev. 95% C.I. on σ I 10 38.79 (37.39, 40.19) 1.9542 (1.34, 3.57) II 10 40.67 (36.68, 44.66) 5.5791 (3.84, 10.19) c. A comparison of the population variances yields: Ho : σ2I = σ 2 II versus Ha : σ 2 I 6= σ2II With α = 0.01, reject Ho if s2I s2II ≤ 16.54 = 0.15 or s21 s22 ≥ 6.54 s21/s 2 2 = (5.5791) 2/(1.9542)2 = 8.15 > 6.54 ⇒ Reject Ho and conclude there is significant evidence that the population variances are different. A comparison of the population means using the separate variance t-test yields: Ho : µI = µII versus Ha : µI 6= µII t = 38.79−40.67q (1.9542)2 10 + (5.5791)2 10 = −1.01 with df=11 ⇒ p-value = 0.336 Fail to reject Ho and conclude that the data does not support a difference in the mean tread wear for the two brands of tires. However, Brand I has a more uniform tread wear as reflected by it significantly lower standard deviation. 7.15 a. 25x90% = 22.5 and 25x110% = 27.5 implies the limits are 22.5 to 27.5 b. The box plot and normal probability plot are given here: 57 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Normal Probability Plot Quantiles of Standard Normal Po te nc y o f T ab let -2 -1 0 1 2 23 24 25 26 27 23 .0 23 .5 24 .0 24 .5 25 .0 25 .5 26 .0 26 .5 27 .0 Potency of Antihistamine Tablets Tablets Po te nc y o f T ab let The box plot indicates a symmetric distribution with no outliers. The normal proba- bility plot shows the data values reasonably close to a straight line, although there is some deviation at both ends which indicates that the data may be a random sample from a distribution which has shorter tails than a normally distributed population. c. Range = 27.5-22.5 = 5 ⇒ σ̂ = 5/4 = 1.25 Ho : σ = 1.25 versus Ha : σ 6= 1.25 With α = 0.05, reject Ho if (n−1)s2 (1.25)2 ≤ 16.05 or (n−1)s 2 (1.25)2 ≥ 45.72 (30−1)(1.4691)2 (1.25)2 = 40.06 ⇒ 16.06 < 40.06 < 45.72 Fail to reject Ho and conclude there is insufficient evidence that the product standard deviation is greater than 1.25. Thus, it appears that the potencies are within the required bounds. 7.16 a. Ho : σ21 ≥ σ22 versus Ha : σ21 < σ22 With α = 0.05, reject Ho if s22 s21 ≥ 3.18 s22/s 2 1 = (5.9591) 2/(3.5963)2 = 2.75 < 3.18 ⇒ Fail to reject Ho and conclude there is not significant evidence that portfolio 2 has a larger variance than portfolio 1. 95% C.I. on σ 2 2 σ21 : ( (5.9591)2 (3.5963)2 (0.248), (5.9591)2 (3.5963)2 (4.03) ) ⇒ (0.68, 11.07) b. p-value = P (F(9,9) ≥ 2.75) ⇒ 0.05 < p-value < 0.10 c. The box plots are given here: 58 12 6 12 8 13 0 13 2 13 4 13 6 13 8 14 0 14 2 14 4 14 6 14 8 15 0 15 2 15 4 15 6 Portfolio 1 Portfolio 2 Comparison of Risk of Two Portfolios Type of Portfolio Ye ar ly Re tu rn s ( th ou sa nd s o f d oll ar s) From the box plots, the condition of normality appears to be satisfied for both port- folios. 7.17 Since the box plots indicate that data from both portfolios has a normal distribution. Also, the C.I. on the ratio of the variances contained 1 which indicates equal variances. Thus, a pooled variance t-test will be used as the test statistic. Ho : µ1 = µ2 versus Ha : µ1 6= µ2 t = 131.60−147.20 4.92 √ 1 10+ 1 10 = −7.09 with df=18 ⇒ p-value < 0.0005 Reject Ho and conclude that the data strongly supports a difference in the mean returns of the two portfolios. 59