Question

a volumetric calcium analysis on triplicate samples of the blood serum of a patient believed to be suffering from a hyper parathyroid candition produced the following data: meq Ca/L=3.15 ,3.25 ,3.26. what is the 95% confidence limit for the mean assuming a) no prior information about precision of analysis is known b) s--> sigma =0.056 meq Ca/L

Answer 1

3.15,3.25, 3.26

Mean = (∑xi)/N= 9.66/3 = 3.22

Standard deviation (s) =√([∑(xi–x)2]/N-1)=√(0 0074 / 3) =√0 00247 = 0 0497 = 0 05

=(0.0074 / 3) =0.00247 = 0.0497 = 0.05

95% CI (μ) =x ± t*s/√N–= 3.22 ± 4.30*0.05 /√3

= 3.22 ± 0.215 / 1.732= 3.22 ± 0.12

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Mean = (∑xi)/N = 9.66/3 = 3.22

sÆσ= 0.05695%

CIμ) =x±z*σ/√N–

95% CI (μ) = x±z*σ/√N

= 3.22 ± 1.96*0.056 /√3

= 3.22 ± 0.1098/1.732 = 3.22 ± 0.06

On the basis of new CI, should 3.15 be rejected?

Qexp= (xq-xn)/w= 0.10 / 0.11 = 0.91

Qcrit= 0.97

Qexp<Q crit