今日打卡 · 2021年03月08日
NO.PZ2015120604000117
问题如下：
A random sample is 100 CFA candidate's exam scoring. The mean of the scoring is 64. The standard deviation of the population scoring is 15 . The distribution is sopposed to be normal. The 95% confidence interval for the population mean should be:
选项：
A.61.06 to 66.94.
B.61.06 to 69.94.
C.65.06 to 66.94.
解释：
A is correct.
Because the variance of the population is know and n ≥ 30,so The confidence interval is:
64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.
NO.PZ2015120604000117 问题如下 A ranm sample is 100 CFA cante's exscoring. The meof the scoring is 64. The stanrviation of the population scoring is 15. The stribution is supposeto normal. The 95% confinintervfor the population meshoulbe: A.61.06 to 66.94. B.61.06 to 69.94. C.65.06 to 66.94. A is correct.Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.因为样本方差已知且n 30，我们直接使用置信区间计算64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94.
NO.PZ2015120604000117问题如下 A ranm sample is 100 CFA cante's exscoring. The meof the scoring is 64. The stanrviation of the population scoring is 15. The stribution is supposeto normal. The 95% confinintervfor the population meshoulbe: A.61.06 to 66.94.B.61.06 to 69.94.C.65.06 to 66.94. A is correct.Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.因为样本方差已知且n 30，我们直接使用置信区间计算64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94. 根据题目分数只可能是正数，拒绝域应该是在右边，95%的单尾应该对应的是1.65呀？
NO.PZ2015120604000117问题如下 A ranm sample is 100 CFA cante's exscoring. The meof the scoring is 64. The stanrviation of the population scoring is 15. The stribution is supposeto normal. The 95% confinintervfor the population meshoulbe: A.61.06 to 66.94. B.61.06 to 69.94. C.65.06 to 66.94. A is correct.Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.因为样本方差已知且n 30，我们直接使用置信区间计算64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94. 老师您好，PPT上的公司用的是sigma，所以我的理解是直接用population S好，为什么用SE呢？
NO.PZ2015120604000117 61.06 to 69.94. 65.06 to 66.94. A is correct. Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94. 因为样本方差已知且n>30，我们直接使用置信区间计算 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94. 如果只抽一次，一次数量大于等于30，那么，这一次数据的标准差等于总体标准差除以根号n么？如果抽100次，每一次数量都大于等于30，每一次都会有一个均值，这些均值又形成一个新的分布，这个新的分布的方差叫做标准误，也等于总体的标准差除以根号n吗？
NO.PZ2015120604000117 老师请问样本标准差和样本均值标准分别是怎么定义的，区别是什么？
