早教吧作业答案频道 -->英语-->
帮帮忙啊,翻译问题!ForthePEFRdatas=38.8.Thestandarderrorofdisthus9.4,Forthe95%confidenceintervalwehave16degreesoffreedomandt=2.12.Hencethe95%confidenceintervalforthebiasis-2.1-(2.12´9.4)to-2.
题目详情
帮帮忙啊,翻译问题!
For the PEFR data s = 38.8. The standard error of d is thus 9.4, For the 95% confidence interval we have 16 degrees of freedom and t = 2.12. Hence the 95% confidence interval for the bias is - 2.1- (2.12 ´ 9.4) to - 2.1+ (2.12 ´ 9.4) , giving - 22.0 to 17.8 l/min. The standard error of the limit d - 2s is 16.3 l/min. The 95% confidence interval for the lower limit of agreement is - 79.7 - (2.12 ´16.3) to - 79.7 + (2.12 ´16.3) , giving -114.3 to
- 45.1 l/min. For the upper limit of agreement the 95% confidence interval is 40.9 to 110.1l/min. These intervals are wide, reflecting the small sample size and the great variation of the
differences. They show, however, that even on the most optimistic interpretation there can be considerable discrepancies between the two meters and that the degree of agreement is not
acceptable.
谢绝机译啊!
For the PEFR data s = 38.8. The standard error of d is thus 9.4, For the 95% confidence interval we have 16 degrees of freedom and t = 2.12. Hence the 95% confidence interval for the bias is - 2.1- (2.12 ´ 9.4) to - 2.1+ (2.12 ´ 9.4) , giving - 22.0 to 17.8 l/min. The standard error of the limit d - 2s is 16.3 l/min. The 95% confidence interval for the lower limit of agreement is - 79.7 - (2.12 ´16.3) to - 79.7 + (2.12 ´16.3) , giving -114.3 to
- 45.1 l/min. For the upper limit of agreement the 95% confidence interval is 40.9 to 110.1l/min. These intervals are wide, reflecting the small sample size and the great variation of the
differences. They show, however, that even on the most optimistic interpretation there can be considerable discrepancies between the two meters and that the degree of agreement is not
acceptable.
谢绝机译啊!
▼优质解答
答案和解析
For the PEFR data s = 38.8.The standard error of d is thus 9.4,对于PEFR（呼气流速变化率）数据s＝38.8来说,那d的标准误差因此为9.4.For the 95% confidence interval we have 16 degrees of freedom and t = 2.12.对于95％的置信区间来说,我们有16的自由度,而t＝2.12.Hence the 95% confidence interval for the bias is - 2.1- (2.12 ´ 9.4) to - 2.1+ (2.12 ´ 9.4) ,giving - 22.0 to 17.8 l/min.因此,对于偏差95％的置信区间为 - 2.1- (2.12 ´ 9.4) to - 2.1+ (2.12 ´ 9.4）,给出为- 22.0 to 17.8 l/min.The standard error of the limit d - 2s is 16.3 l/min.极限d－2s的标准误差为16.3 l/min.The 95% confidence interval for the lower limit of agreement is - 79.7 - (2.12 ´16.3) to - 79.7 + (2.12 ´16.3) ,giving -114.3 to
- 45.1 l/min.对于协议下限的95％的置信区间为 - 79.7 - (2.12 ´16.3) to - 79.7 + (2.12 ´16.3),给出为 -114.3 to
- 45.1 l/min.For the upper limit of agreement the 95% confidence interval is 40.9 to 110.1l/min.对于协议的上限,95％的置信区间为40.9 to 110.1l/min.These intervals are wide,reflecting the small sample size and the great variation of the differences.这些区间比较宽,反映了较小的样本量和差别的巨大变化.They show,however,that even on the most optimistic interpretation there can be considerable discrepancies between the two meters and that the degree of agreement is not acceptable.可是它们表明了,即使是在最乐观的解释下,两台仪表之间都可能存在相当大的 差异,而且符合程度不能让人接受
- 45.1 l/min.对于协议下限的95％的置信区间为 - 79.7 - (2.12 ´16.3) to - 79.7 + (2.12 ´16.3),给出为 -114.3 to
- 45.1 l/min.For the upper limit of agreement the 95% confidence interval is 40.9 to 110.1l/min.对于协议的上限,95％的置信区间为40.9 to 110.1l/min.These intervals are wide,reflecting the small sample size and the great variation of the differences.这些区间比较宽,反映了较小的样本量和差别的巨大变化.They show,however,that even on the most optimistic interpretation there can be considerable discrepancies between the two meters and that the degree of agreement is not acceptable.可是它们表明了,即使是在最乐观的解释下,两台仪表之间都可能存在相当大的 差异,而且符合程度不能让人接受
看了 帮帮忙啊,翻译问题!Fort...的网友还看了以下：
观察：1+2=3=2^2-1,1+2+2^2=7=2^3-1,1+2+2^2+2^3=15=2^4- 2020-03-31 …
lim x趋于0 1-cosx^2/x^2sinX^2(1-cosx^2)/x^2sinx^2=[ 2020-05-16 …
急用!1.2(x-2)-3(4x-1)=9(1-x)2.11x+64-2x=100-9x3.15- 2020-05-17 …
1+2+3+n=2分之1n(n+1),n是正整数,研究1*2+2*3+你(n+1),观察1*2=3 2020-05-20 …
1.为了求1+2+2²+2³+…+2^2008的值,可令S=1+2+2²+2³+…+2^2008, 2020-06-02 …
计算题,1：(x+1/2)^2+(x-1/2)^2+(x+1/2)(x-1/2)（1／3a-3/4 2020-06-03 …
吐血题,找下规律下面有一组数列,请找出它的规律来：第一列：1第二列：1,1第三列：2,1第四列：1 2020-06-14 …
巧算分数的和,会的同学帮忙,1.1/2*3*4+1/3*4*5+...+1/18*19*202.1 2020-06-16 …
在○里填上适当的运算符合,使等式成立（1）1/2○1/2○1/2○1/2○1/2=0（2）1/2○ 2020-07-08 …
（2000•内江）（1）观察下列等式：1(1+1×2)(1+2×2)＝12(11+1×2−11+2× 2020-11-12 …