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NO.PZ2015120604000142问题如下Here is a table shows a set of ta whiis subjeto the normstribution. Peter sets up a null hypothesis--- H0: μ=8.5. Construa confininterv 1% significanlevel.A.(8.3227, 9.5763)B.(7.6544, 8.5445)C.(8.3227, 9.4773) C is correct.X±2.58×σ=8.9±2.58×1.79/√64=(8.3227, 9.4773) 我以为只有需要中心极限定理的时候才需要除
NO.PZ2015120604000142问题如下Here is a table shows a set of ta whiis subjeto the normstribution. Peter sets up a null hypothesis--- H0: μ=8.5. Construa confininterv 1% significanlevel.A.(8.3227, 9.5763)B.(7.6544, 8.5445)C.(8.3227, 9.4773) C is correct.X±2.58×σ=8.9±2.58×1.79/√64=(8.3227, 9.4773) 是不是因为没有给样本的标准差，才用总体的标准差替代计算的？
NO.PZ2015120604000142 问题如下 Here is a table shows a set of ta whiis subjeto the normstribution. Peter sets up a null hypothesis--- H0: μ=8.5. Construa confininterv 1% significanlevel. A.(8.3227, 9.5763) B.(7.6544, 8.5445) C.(8.3227, 9.4773) C is correct.X±2.58×σ=8.9±2.58×1.79/√64=(8.3227, 9.4773) 题干没看懂 更没看懂 请全题
NO.PZ2015120604000142问题如下Here is a table shows a set of ta whiis subjeto the normstribution. Peter sets up a null hypothesis--- H0: μ=8.5. Construa confininterv 1% significanlevel.A.(8.3227, 9.5763)B.(7.6544, 8.5445)C.(8.3227, 9.4773) C is correct.X±2.58×σ=8.9±2.58×1.79/√64=(8.3227, 9.4773) 老师这个考的是Reang 5 Estimation里 confininterval的知识点吗？
NO.PZ2015120604000142 问题如下 Here is a table shows a set of ta whiis subjeto the normstribution. Peter sets up a null hypothesis--- H0: μ=8.5. Construa confininterv 1% significanlevel. A.(8.3227, 9.5763) B.(7.6544, 8.5445) C.(8.3227, 9.4773) C is correct.X±2.58×σ=8.9±2.58×1.79/√64=(8.3227, 9.4773) 从题目里面判断是two tail，所以我照着z 表找到99%的概率对应的z是1.645，为什么这样的思路不对呢
