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解释：
老师请问到底怎么区分什么时候直接用标准差？什么时候用标准误？有时候在描述组合管理中的标准差就可以不除以根号n，为什么在这里的同样也是样本，还需要转换成标准误？可以详细列出什么时候如何判断吗？谢谢
NO.PZ2015120604000118 问题如下 li Bur is a high school. A recent survey of 25 stunt of the high school incates ththe metime they spengoing to school is 40 minutes . This sample's stanrviation is 8 minutes. The stribution of the population is supposeto normal. The 99% confinintervfor the metime thall li Bur stunts spengoing to the school is: A.30.72 to 34.55 B.38.52 to 54.48 C.35.52 to 44.48 C is correct.Because the simple size is less th30, so the confinintervfor the population whose varianis unknow is : x - ± t α/2 (s/ n ) . To calculate criticvalue: t 0.005 an = 24 is 2.797.So, the confinintervis 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48本题由于样本数量小于30，且总体方差未知。可知均值X bar=40，对应的X bar的标准差（即标准误）=8/√25=1.6。 而criticvalue需要查表求得，如果是正态分布，这个值就是2.58。但这道题对应的是t分布，需要查的是t表（总体方差未知用t）。本题α=1%，则α/2=0.005。t分布需要考虑自由度，=n-1=24。通过对应单尾概率0.005和自由度24查表可得t criticvalue=2.797.代入公式40±2.797×1.6即可得到答案 如何看出这道题总体方差未知呢？
NO.PZ2015120604000118问题如下 li Bur is a high school. A recent survey of 25 stunt of the high school incates ththe metime they spengoing to school is 40 minutes . This sample's stanrviation is 8 minutes. The stribution of the population is supposeto normal. The 99% confinintervfor the metime thall li Bur stunts spengoing to the school is:A.30.72 to 34.55B.38.52 to 54.48C.35.52 to 44.48 C is correct.Because the simple size is less th30, so the confinintervfor the population whose varianis unknow is : x - ± t α/2 (s/ n ) . To calculate criticvalue: t 0.005 an = 24 is 2.797.So, the confinintervis 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48本题由于样本数量小于30，且总体方差未知。可知均值X bar=40，对应的X bar的标准差（即标准误）=8/√25=1.6。 而criticvalue需要查表求得，如果是正态分布，这个值就是2.58。但这道题对应的是t分布，需要查的是t表（总体方差未知用t）。本题α=1%，则α/2=0.005。t分布需要考虑自由度，=n-1=24。通过对应单尾概率0.005和自由度24查表可得t criticvalue=2.797.代入公式40±2.797×1.6即可得到答案百分之99对应的不是2.58吗，为啥答案是2.7几
NO.PZ2015120604000118 问题如下 li Bur is a high school. A recent survey of 25 stunt of the high school incates ththe metime they spengoing to school is 40 minutes . This sample's stanrviation is 8 minutes. The stribution of the population is supposeto normal. The 99% confinintervfor the metime thall li Bur stunts spengoing to the school is: A.30.72 to 34.55 B.38.52 to 54.48 C.35.52 to 44.48 C is correct.Because the simple size is less th30, so the confinintervfor the population whose varianis unknow is : x - ± t α/2 (s/ n ) . To calculate criticvalue: t 0.005 an = 24 is 2.797.So, the confinintervis 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48本题由于样本数量小于30，且总体方差未知。可知均值X bar=40，对应的X bar的标准差（即标准误）=8/√25=1.6。 而criticvalue需要查表求得，如果是正态分布，这个值就是2.58。但这道题对应的是t分布，需要查的是t表（总体方差未知用t）。本题α=1%，则α/2=0.005。t分布需要考虑自由度，=n-1=24。通过对应单尾概率0.005和自由度24查表可得t criticvalue=2.797.代入公式40±2.797×1.6即可得到答案
NO.PZ2015120604000118 问题如下 li Bur is a high school. A recent survey of 25 stunt of the high school incates ththe metime they spengoing to school is 40 minutes . This sample's stanrviation is 8 minutes. The stribution of the population is supposeto normal. The 99% confinintervfor the metime thall li Bur stunts spengoing to the school is: A.30.72 to 34.55 B.38.52 to 54.48 C.35.52 to 44.48 C is correct.Because the simple size is less th30, so the confinintervfor the population whose varianis unknow is : x - ± t α/2 (s/ n ) . To calculate criticvalue: t 0.005 an = 24 is 2.797.So, the confinintervis 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48本题由于样本数量小于30，且总体方差未知。可知均值X bar=40，对应的X bar的标准差（即标准误）=8/√25=1.6。 而criticvalue需要查表求得，如果是正态分布，这个值就是2.58。但这道题对应的是t分布，需要查的是t表（总体方差未知用t）。本题α=1%，则α/2=0.005。t分布需要考虑自由度，=n-1=24。通过对应单尾概率0.005和自由度24查表可得t criticvalue=2.797.代入公式40±2.797×1.6即可得到答案 25这个数字对于这道题有什么样的影响？是不是小于30，置信区间计算就不能用讲义里的公式？另外为什么这道题是的是t分部谢谢老师
NO.PZ2015120604000118问题如下li Bur is a high school. A recent survey of 25 stunt of the high school incates ththe metime they spengoing to school is 40 minutes . This sample's stanrviation is 8 minutes. The stribution of the population is supposeto normal. The 99% confinintervfor the metime thall li Bur stunts spengoing to the school is:A.30.72 to 34.55B.38.52 to 54.48C.35.52 to 44.48 C is correct.Because the simple size is less th30, so the confinintervfor the population whose varianis unknow is : x - ± t α/2 (s/ n ) . To calculate criticvalue: t 0.005 an = 24 is 2.797.So, the confinintervis 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48本题由于样本数量小于30，且总体方差未知。可知均值X bar=40，对应的X bar的标准差（即标准误）=8/√25=1.6。 而criticvalue需要查表求得，如果是正态分布，这个值就是2.58。但这道题对应的是t分布，需要查的是t表（总体方差未知用t）。本题α=1%，则α/2=0.005。t分布需要考虑自由度，=n-1=24。通过对应单尾概率0.005和自由度24查表可得t criticvalue=2.797.代入公式40±2.797×1.6即可得到答案老师好，请问下，是单尾、为什么查表不是按1%查
