A.) Cal Culator obtained the following results for replicate determinations of c

Question

A.)

Cal Culator obtained the following results for replicate determinations of calcium in limestone: 14.35%, 14.41%, 14.40%, 14.32%, and 14.37%. Calculate the confidence interval at the 95% confidence level.

B.)

A new procedure for determining trace amounts of zinc in vegetables was evaluated by using it to determine the zinc content of a NIST standard sample, with the following results: 0.083, 0.088, 0.087, and 0.086 ppm Zn. The certified value of the standard sample is 0.082 ppm Zn. Is there a significant difference between the mean of the results and the certified value?

Explanation / Answer

Solution

A) Mean of the data = 14.37 %

No. of data (n) = 5

Deviation = square root [ (-0.02)^2 + (0.04)^2 + (0.03)^2 + (-0.05)^2 + 0 )/5]

= 0.03286

Confidence Interval = Mean +- ( deviation/squareroot(n)) * Z

Z for 95% confidence interval = 1.96

= 14.37% +- ( 0.0326*1.96/square root 5 )

= 14.37% +-(0.0285) %

Upper end = 14.3985%

Lower end = 14.3414%

B) Mean = 0.083 + 0.088 + 0.087 + 0.086 / 4

= 0.086

Certified value = 0.082

Difference = 0.086-0.082 = 0.004

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