Using a new method we just developed, a student in my group recently recorded th

Question

Using a new method we just developed, a student in my group recently recorded the following data for the external
standard determination of an unknown solution of IL-6 (a pro-inflammatory cytokine) in human cerebrospinal fluid.
Using the data provided, answer the questions below:

a. After obtaining the provided external standard calibration curve, four replicate samples of unknown IL-6
concentration were measured (± standard deviation) to be 756 ±12 units. Determine the concentration of
IL-6 in the sample.

b. Determine the 95% confidence interval associated with this measurement.

Concentration Response Of IL-6 (pg/mL) | (Apm) Standard # IL-6 calibration curve 4.2 12.5 66.4 396.6 1712.5 2000 1800 1600 1400 E 1200 0 R2 0.9988 5 25 125 3 1000 800 0 600 400 200 0 -200 SUMMARY OUTPUT Regression Statistics MultipleR 0.999384178 R Square Adjusted RS 0.998358313 Standard Er 29.66637154 Observation 0.998768734 50 100 150 IL-6 concentration (pg/mL) Coefficients Standard Error t Stat P-value Lower 95% Upper 95% ower 95.09 Upper 95.0% 8.4595259815.8464975 0.533842 0.630474 -41.971101 58.890153 -41.971158.8901534 X Variable 1 13.70129724 0.277743848 49.3307 1.83E-05 12.817392 14.585202 12.81739 14.58520213 Intercept

Explanation / Answer

a)
y^ = 8.45952598 + 13.70129724 *x
when y^ = 756
x = (756 - 8.4595)/13.7013
=54.559
= 54.6

b)
12 = 8.46 + 13.70 *sd (X)
sd(X) = (12 - 8.46)/13.70
= 0.258

95 % confidence interval
(mean - z* sd/sqrt(n) , mean + z * sd/sqrt(n))

=(54.6 - 1.96* 0.258 / sqrt(4) , 54.6+1.96* 0.25/8 /sqrt(4))

= (54.35 , 54.85)

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