The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mercury or less. A laboratory is estimating the amount of mercury in tuna fish for a new company and needs to have a margin of error within 0.023 ppm of mercury with 97% confidence. Assume the population standard deviation is 0.143 ppm of mercury. What sample size is needed? Round up to the nearest integer, do not include any decimals.

So the answer for this case would be n=183 rounded up to the nearest integer

Step-by-step explanation:

Information provided

represent the sample mean

population mean (variable of interest)

represent the population standard deviation

n represent the sample size  

the margin of error desired

Solution to the problem

The margin of error is given by this formula:

   (a)

And on this case we have that ME =0.023 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The confidence level is 97% or 0.97 and the significance would be and then the critical value would be: , replacing into formula (5) we got:

So the answer for this case would be n=183 rounded up to the nearest integer

the answer on e2020 is

10

4/5

3/5

4/3

step-by-step explanation:

the correct answer is d

hope this