D85习题答案与提示13p.ppt

第八章 第八章习题答案与提示(13) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1 . 只对_ 加以控制 , 2. 的检验, 拒绝原假设 而不考虑_第一类错误 称为显著性检验. 在假设检验中,当_时, 会犯第一类错误. 3. 要使犯两类错误的概率同时减小, 第二类错误 只有_.增大样本容量 4.从总体取一容量为100的样本,测得 若均未知,在 = 0.05下, 检验下列假设: Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 的拒绝域为 (双边检验 ) 落入 解: 用T检验法. 为未知, 由于 的拒绝域内, 故拒绝 由于 (1) . 检验统计量为 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 的拒绝域为 (双边检验 ) 落入 (2) . 为未知, 由于 的拒绝域内, 故拒绝 由于用 检验法. 检验统计量为 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 测得样本的平均值为97.7, 一种燃料中的辛烷等级服从正态分布, 问 5. 标准差为0.8.为78, 假定标准差与原来一样, 新燃料的辛烷平均等级是否比比原燃料辛烷等级偏低? ( = 0.05) 其平均等级 解 : 检验假设 今从一批新燃料中随机抽取25桶, (左边检验 ) 故采用用U 检验法. 为已知,由于 检验统计量为:的拒绝域为 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 检验统计量的样本值落入原假设 由于 的拒绝域内, 故拒绝即在显著性水平0.05下,接受 可认为新燃料油的辛烷等级偏低 . 6. 设蕃茄汁罐头中VC含量服从正态分布,按照规定VC 的平均含量不得少于21毫克.现从一批罐头中随机抽取 16罐,测得 问这批罐头VC含量是否合格 ? ( = 0.05) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 的拒绝域为 (右边检验 ) 落入 解: 用T 检验法. 为未知, 由于 的拒绝域内,接受 由于 即在显著性水平0.05下, 检验统计量为 依题意有 又 检验假设 故拒绝 可认为这批罐头VC的含量合格. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 测得苗高 某林厂采用两种方案作杨树育苗实验.7 . 标准差分别为下苗高均服从正态分布. 苗的高度有无影响? 已知两种方案 现各抽取60棵树苗作样本 , 厘米, 试以95%的可靠性估计两种方案对杨树厘米, 解: 检验假设 (双边检验 ) 故采用用U 检验法. 为已知,由于 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 的拒绝域为 落入 由于 的拒绝域内, 故拒绝 检验统计量为 依题意有 : 又 即有95%的把握认为两种方案对杨树苗高有影响. 接受 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 假定零件长度服从正态分布, 两台机床加工同一种零件, 解: 8 . 计算得 其长度(单位:毫米), ( = 0.05) 分别取6个和9个零件测量 是否可以认为两台机床加 的零件尺寸的方差无显著差异? 检验假设 (双边检验 ) 用F 检验法. 均未知, 由于 检验统计量为 的拒绝域为 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 未落入 由于 的拒绝域内, 故接受 依题意有 又 在显著性水平0.05下可认为两机床 拒绝 加工零件尺寸的方差无显著差异. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 假测定值总体服从正态分布, 某批矿砂的5个样品中的镍含量, 解: 9 . 3.25 , 3.27 , 3.24 , 3.26 , 3.24 . 下, 经测定为:(%) 但参数均为知,问在 = 0.01 能否接受假设: 这批矿砂的镍函量的均值为3.25. 用T 检验法. 未知,由于方差检验统计量为 的拒绝域为 (双边检验 ) 检验假设 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 未落入 由于 的拒绝域内,拒绝 即在显著性水平0.01下, 依题意有 又 经计算: 故接受 可认为这批矿砂的镍含量的均值 为3.25%. Evaluation on