《工程硕士数理统计.ppt

P.116- 2 (2) 解 计算公式 P.117- 6 P.117- 9 ② ③ 解 解一 解二 P.120- 27 表 * 数理统计 第三章习题 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2(2)解① 由 得 拒绝域 接受域为 当 时 ? Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解② 由 得 拒绝域 接受域为 当 时 ? Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解一 1061602004 陆文敏 由 得拒绝域 于是接受域为 当 时 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解二 1060502012 李 钧 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 右侧检验 左侧检验 双侧检验 其 中 U 检验法中 的计算公式 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 6 解 1060502004 沈 彬 本题属大样本.母体为二项分布 取统计量 假设 拒绝域 W : 接受 ,新工艺没有显著影响产品质量. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 9 解① 服用新药平均睡眠时间 所以新安眠药已达到新的疗效. 题目做成这样助教仍给打 上述求解者似乎没学过数理统计！ Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解② 标准差 假设 选统计量 拒绝域 W : 现 所以接受假设, 不达到新疗效. ? ? ? ? ? ? Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解③ 先检验两正态母体方差是否相等. 假设 取统计量 拒绝域 W ： 现 接受原假设,即两正态母体方差相等. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 假设 取统计量 拒绝域 W ： 统计量值 所以接受原假设, 新药未达到新疗效. ? Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 下面是大家作的各种假设,哪个对? 错 对 欠妥 错 Evaluation only. Created with Aspose.Slid