答案没有问题，我想问的是答案解释中其他的问题。
问题是“答案中的原假设H0，为什么不是u0>=6%,Alternative假设是：u<6%呢？”
因为题目中提到要验证的是u<=6%,我们把最想reject的放在原假设上，那不应该是u0>=6%吗？答案解释中为啥相反？谢谢啦
问题如下图：
选项：
A. 
B. 
C. 
解释：
NO.PZ2017092702000122问题如下 analyst is examining a large sample with unknown population variance. Whiof the following is the most appropriate test to test the hypothesis ththe historicaverage return on inx is less thor equto 6%? A.One-tailez-test B.Two-tailez-test C.One-taileF-test A is correct.If the population samplehunknown variananthe sample is large, a z-test muse Hypotheses involving \"greater than\" or \"less than\" postulations are one-si(one-taile. In this situation, the null analternative hypotheses are stateH0: μ ≤ 6% anHμ 6%, respectively. A one-tailet-test is also acceptable in this case, anthe rejection region is on the right si of the stribution.检验历史的平均收益是否小于等于6%，这样的假设就是单尾的一个概念且对于检验总体均值而言，（总体）方差已知用z，（总体）方差未知用t，非正态（总体）小样本不可估计。所以选择单尾 z-test 根据口诀总体方差未知用t，这题目不就说总体方差未知吗？怎么又用z？
NO.PZ2017092702000122 Two-tailez-test One-taileF-test A is correct. If the population samplehunknown variananthe sample is large, a z-test muse Hypotheses involving \"greater than\" or \"less than\" postulations are one-si(one-taile. In this situation, the null analternative hypotheses are stateH0: μ ≤ 6% anHμ > 6%, respectively. A one-tailet-test is also acceptable in this case, anthe rejection region is on the right si of the stribution. 检验历史的平均收益是否小于等于6%，这样的假设就是单尾的一个概念 且对于检验总体均值而言，（总体）方差已知用z，（总体）方差未知用t，非正态（总体）小样本不可估计。 所以选择单尾&z-test 已知不才用z吗？未知不是用t吗
NO.PZ2017092702000122 大样本下，如果方差未知，检验均值既可以用T检验，也可以用Z检验。 请问课件哪里哪页
原假设不是我们想要拒绝的吗？题干中说想要检验AR是不是小于等于6%，那Ho应该是AR大于6%，考虑到等号要放在Ho里，那Ho应该是大于等于6%，Ha是小于6%？求解答，感谢！
您好 请问A是比C好嘛？还是说C是有问题的？
