分析化学-chap7-notes-误差分析

本文探讨了误差分析中的置信区间(CI)和置信水平(CL),介绍了如何根据样本大小计算CI,并区分系统误差与随机误差。通过z和t分布进行假设检验,用于判断实验结果与已知值的差异是否显著。此外,还提到了F检验和Q、Grubbs测试在评估数据质量中的应用。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

confidence interval(CI) 置信度(区间)
certain probability,true mean,the range of values
the probability is called confidence level (CL)

 CI for  μ = x ˉ ± z σ N \text { CI for } \mu=\bar{x} \pm \frac{z \sigma}{\sqrt{N}}  CI for μ=xˉ±N zσ

we might say that it is 99% probable that the true population mean…
Confidence level: the probability that true mean lies in the given range
Significant level: the probability that a result is outside the C L \mathrm{CL} CL
Confidence interval for the mean: the range of values within which the population mean is expected to lie with a certain probability

To estimate the true mean from
large sample: z, σ \sigma σ
small sample: t,s

  • Single measurement:
     CI for  μ = x ± z σ \text { CI for } \mu=x \pm z \sigma  CI for μ=x±zσ

  • Mean of N measurements (large sample):
     CI for  μ = x ˉ ± z σ N \text { CI for } \mu=\bar{x} \pm \frac{z \sigma}{\sqrt{N}}  CI for μ=xˉ±N zσ

  • Mean of N measurements (small sample):

 CI for  μ = x ˉ ± t s N \text { CI for } \mu=\bar{x} \pm \frac{t s}{\sqrt{N}}  CI for μ=xˉ±N ts

z 的值
 Confidence Levels for Various   Values of  z  Confidence Level, %  z 50 0.67 68 1.00 80 1.28 90 1.64 95 1.96 95.4 2.00 99 2.58 99.7 3.00 99.9 3.29 \begin{array}{cc} \hline \text { Confidence Levels for Various } \\ \text { Values of } z \\ \hline \text { Confidence Level, \% } & \boldsymbol{z} \\ \hline 50 & 0.67 \\ 68 & 1.00 \\ 80 & 1.28 \\ 90 & 1.64 \\ 95 & 1.96 \\ 95.4 & 2.00 \\ 99 & 2.58 \\ 99.7 & 3.00 \\ 99.9 & 3.29 \\ \hline \end{array}  Confidence Levels for Various  Values of z Confidence Level, % 506880909595.49999.799.9z0.671.001.281.641.962.002.583.003.29

Finding the Confidence Interval When σ Is Unknown:
 CI for  μ = x ˉ ± t s N \text { CI for } \mu=\bar{x} \pm \frac{t s}{\sqrt{N}}  CI for μ=xˉ±N ts

t = x ˉ − μ s / N t=\frac{\bar{x}-\mu}{s / \sqrt{N}} t=s/N xˉμ

评价-精密度:
Systematic Error or Random Error?

  1. Comparing an Experimental Mean with a Known Value (z/t Test)
    z = x ˉ − μ 0 σ / N z=\frac{\bar{x}-\mu_{0}}{\sigma / \sqrt{N}} z=σ/N xˉμ0

t = x ˉ − μ 0 s / N t=\frac{\bar{x}-\mu_{0}}{s / \sqrt{N}} t=s/N xˉμ0

If z > z crit  z>z_{\text {crit }} z>zcrit  or t > t crit t>t_{\text {crit}} t>tcrit, systematic error exists and the analysis method needs improvements, otherwise retain it!

 Values of  t  for Various Levels of Probability   Degrees of   Freedom  80 % 90 % 95 % 99 % 99.9 % 1 3.08 6.31 12.7 63.7 637 2 1.89 2.92 4.30 9.92 31.6 3 1.64 2.35 3.18 5.84 12.9 4 1.53 2.13 2.78 4.60 8.61 5 1.48 2.02 2.57 4.03 6.87 6 1.44 1.94 2.45 3.71 5.96 7 1.42 1.90 2.36 3.50 5.41 8 1.40 1.86 2.31 3.36 5.04 9 1.38 1.83 2.26 3.25 4.78 10 1.37 1.81 2.23 3.17 4.59 15 1.34 1.75 2.13 2.95 4.07 20 1.32 1.73 2.09 2.84 3.85 40 1.30 1.68 2.02 2.70 3.55 60 1.30 1.67 2.00 2.62 3.46 ∞ 1.28 1.64 1.96 2.58 3.29 \begin{aligned} &\text { Values of } t \text { for Various Levels of Probability }\\ &\begin{array}{cccccc} \begin{array}{ccccc} \text { Degrees of } \\ \text { Freedom } \end{array} & \mathbf{8 0 \%} & \mathbf{9 0 \%} & \mathbf{9 5 \%} & \mathbf{9 9 \%} & \mathbf{9 9 . 9 \%} \\ \hline 1 & 3.08 & 6.31 & 12.7 & 63.7 & 637 \\ 2 & 1.89 & 2.92 & 4.30 & 9.92 & 31.6 \\ 3 & 1.64 & 2.35 & 3.18 & 5.84 & 12.9 \\ 4 & 1.53 & 2.13 & 2.78 & 4.60 & 8.61 \\ 5 & 1.48 & 2.02 & 2.57 & 4.03 & 6.87 \\ 6 & 1.44 & 1.94 & 2.45 & 3.71 & 5.96 \\ 7 & 1.42 & 1.90 & 2.36 & 3.50 & 5.41 \\ 8 & 1.40 & 1.86 & 2.31 & 3.36 & 5.04 \\ 9 & 1.38 & 1.83 & 2.26 & 3.25 & 4.78 \\ 10 & 1.37 & 1.81 & 2.23 & 3.17 & 4.59 \\ 15 & 1.34 & 1.75 & 2.13 & 2.95 & 4.07 \\ 20 & 1.32 & 1.73 & 2.09 & 2.84 & 3.85 \\ 40 & 1.30 & 1.68 & 2.02 & 2.70 & 3.55 \\ 60 & 1.30 & 1.67 & 2.00 & 2.62 & 3.46 \\ \infty & 1.28 & 1.64 & 1.96 & 2.58 & 3.29 \end{array} \end{aligned}  Values of t for Various Levels of Probability  Degrees of  Freedom 123456789101520406080%3.081.891.641.531.481.441.421.401.381.371.341.321.301.301.2890%6.312.922.352.132.021.941.901.861.831.811.751.731.681.671.6495%12.74.303.182.782.572.452.362.312.262.232.132.092.022.001.9699%63.79.925.844.604.033.713.503.363.253.172.952.842.702.622.5899.9%63731.612.98.616.875.965.415.044.784.594.073.853.553.463.29

CI & t

t up, then CI up.面积嘛

t & f

f up, 说明曲线变高变瘦,越来越多的值接近一个值,因此 t 在置信区间不变时,变小

comparing with two experimental Means
先用F检验法,证明两组数据无显著性差异。
再是t检验法:合并标准偏差,计算t值

s p o o l e d = s 1 2 ( n 1 − 1 ) + s 1 2 ( n 2 − 1 ) ( n 1 − 1 ) + ( n 2 − 1 ) s_{pooled}=\sqrt{\frac{s_{1}^{2}\left(n_{1}-1\right)+s_{1}^{2}\left(n_{2}-1\right)}{\left(n_{1}-1\right)+\left(n_{2}-1\right)}} spooled=(n11)+(n21)s12(n11)+s12(n21)

t = x ˉ 1 − x ˉ 2 s pooled  N 1 + N 2 N 1 N 2 t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s_{\text {pooled }} \sqrt{\frac{N_{1}+N_{2}}{N_{1} N_{2}}}} t=spooled N1N2N1+N2 xˉ1xˉ2

比较t和 t c r i t i c a l t_{critical} tcritical t比较大时,说明系统误差还是比较大

F test

F = s b i g 2 s s m a l l 2 F=\frac{s_{big}^{2}}{s_{small}^{2}} F=ssmall2sbig2

F如果比较大,说明俩方法的精密度都不对。

对于一个反常数据该不该留
Q test
Q = x n − x n − 1 x n − x 1 = x 2 − x 1 x n − x 1 Q=\frac{x_{n}-x_{n-1}}{x_n-x_1}=\frac{x_{2}-x_{1}}{x_n-x_1} Q=xnx1xnxn1=xnx1x2x1
有个表Q的

Grubbs test
G = x n − x m e a n s = x m e a n − x 1 s G=\frac{x_{n}-x_{mean}}{s}=\frac{x_{mean}-x_{1}}{s} G=sxnxmean=sxmeanx1

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值