问题如下图:
题目中给了标准差也不能用Z分布吗?标准差平方不是方差了吗?这样就可以用2.58了,为什么不对呢?
选项:
A.
B.
C.
解释:
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