Question

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. Answer:

Answer 1

Answer:

$n=(\frac{2.17(0.143)}{0.023})^2 =182.03 \approx 183$

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

Step-by-step explanation:

Information provided

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma = 0.143$ represent the population standard deviation

n represent the sample size  

$ME = 0.023$ the margin of error desired

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (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:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The confidence level is 97% or 0.97 and the significance would be $\alpha=1-0.97=0.03$ and $\alpha/2 = 0.015$ then the critical value would be: $z_{\alpha/2}=2.17$, replacing into formula (5) we got:

$n=(\frac{2.17(0.143)}{0.023})^2 =182.03 \approx 183$

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