Quantitative Instrumental Analysis 2) (25 pts) A series of eight control experim

Question

Quantitative Instrumental Analysis
2) (25 pts) A series of eight control experiments were done and the measured values are 7.43689, 7.44001, 7.43876, 7.43897,7.44002, 7.44010, 7.43992, and 7.43989. Similarly, a series of seven samples were measured and their values are 7.41025, 7.40987, 741046, 7.40998, 7.39988, 7.41005, and 7.41034 Using student's t, calculate the 99.9% confidence intervals of the controls and the samples? At 99.9% confidence, are the samples different from, the controls? a) b) Student& values confidence level(%) -929 8 9 10 15 20 25 6.87 5.96 5.41 5.04 4.78 4.59 4.07 3.85 3.73

Explanation / Answer

here first we have to calculate mean and standard deviation of control and samples seperately.

Here 99.9% confidence interval for control = xcontrol +- tdf,0.05 scontrol /sqrt(n)

= 7.43932 +- t7,0.05 * (0.001108/sqrt(8)

= 7.43932 +- 0.00212

Here 99.9% confidence interval for samples = xsample +- tdf,0.05 ssample/sqrt(n)

= 7.40869 +- t6,0.05 * (0.00389/sqrt(8)

= 7.43932 +- 0.00876

= (7.39993, 7.41745)

(b) So here we can see that both confidence interval doesn't overlap with each other. SO, that means samples are different from controls.

Control Samples 7.43689 7.41025 7.44001 7.40987 7.43876 7.41046 7.43897 7.40998 7.44002 7.39988 7.4401 7.41005 7.43992 7.41034 7.43989 Mean 7.43932 7.40869 Std. Dev. 0.001108 0.00389 Sample size 8 7

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