Question

The following information will be used for the following two questions:
The analytical laboratory of a pharmaceutical company is asked to evaluate the claim that the concentration of an active ingredient in a product is 0.80%. The mean of four repeated analyses of the specimen x-bar is 0.82%. The standard deviation is known to be sigma equals 0.01 percent The true concentration of the active ingredient is the mean of the population of repeated analyses and the population has a normal distribution.
Using the information above, give a 99% confidence interval for µ.

Answer 1

Choose one answer.
A. (0.8071%, 0.8328%)
B. (0.7871%, 0.8129%)
C. (0.79%, 0.81%)
D. (0.7942%, 0.8458%)
E. (0.75%, 0.80%)

Answer 2

the concentration of an active ingredient in a product is 0.80%.

The mean of four (sample size of 4)
repeated analyses of the specimen x-bar is (with mean of) 0.82%.

The standard deviation is known to be sigma equals 0.01 percent

sample - claim
-------------
claim stats/sqrt(n)

Answer 3

what is out zscore for a 99% confidence interval?

Answer 4

our z score related to some margin of error
\[\Large z_{99\%}=\frac{.0082-.0080}{.0001/\sqrt{4}}\]
\[\Large z_{99\%}\frac{.0001}{\sqrt{4}}=.0082-.0080\]
\[\Large z_{99\%}\frac{.0001}{\sqrt{4}}=Error~margin\]
CI = .0082 +- error margin seems fair

Answer 5

So which confidence interval is it? The margin of error you gave doesn't match any of them.

Answer 6

@amistre64

Answer 7

what is the z score for a 99% CI?

Answer 8

and also, since BCE dont contain .82% they cant possibly be it, that leaves us with A or D

Answer 9

Is it A?

Answer 10

i dont think D has an average of .82, seems a little skewed to the right

Answer 11

without any calcs, i would say A is the best shot .... but the calculations would have to verify that

Answer 12

Ok, thanks