问题如下图：
选项：
A. 
B. 
C. 
解释：
这道题可以这么理解吗?总体方差未知用t，n>30，所以可以z,所以虽然公式用的是均值加减ta/2(s/s根号n),所以就使用公式z分布吗？
星星_品职助教 · 2019年09月01日
同学你好，
这道题不建议这么理解。因为题干中已经明确的说了用Z-statistics来计算，所以直接用Z对应的critical value就行。
我看这道题和你上一次问的PZ2015120604000137都在想应该用什么分布，我感觉这可能是因为题库中和课后题里有题目没有说清楚到底用哪种分布所导致的。尤其是经常解题时用了熟悉的1.65,1.96那些数字后，发现答案解析实则用的是t-分布。
但具体用哪种分布，或者哪个critical value，在真实考试中会非常明确的告诉的。真实考试的出题方式就类似这道题，直接就说用Z-stats去算。这样也就不用纠结到底是什么分布了。
加油
NO.PZ2015120604000130问题如下We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis :A.- 10.88% to 19.13%.B.10.88% to 19.13%.C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]= 什么时候会考这个区间，一般问这个区间的问题是怎么问的这个区间的标准差指的是总体标准差吗
NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]= SignificanLevel都是要除以2吗？
NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13% 请题干以及该题涉及的原理所对应知识点
NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13% 这里的Stanrerror 不应该是25%/ √100 吗？解析里的式子是什么意思
NO.PZ2015120604000130问题如下We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis :A.- 10.88% to 19.13%.B.10.88% to 19.13%.C.10.88% to 20.57%.B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13%老师，既然题目说是正态分布，而且是选择题。想问下能否利用正态分布的特性，代入解题？正太分布中轴是均值，题干给出的置信区间应该也是轴对称。直接带入，分位点-均值19.13-15=4.13均值-分位点到均值的距离15-4.13=10.87然后选出最符合的的