Question

A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.996 ounces with a standard deviation of 0.004 ounces. How many pieces of candy must we sample if we want to be 99 percent confident that the sample mean is within 0.001 ounces of the true mean

Answer 1

Answer:

$n=(\frac{2.58(0.004)}{0.001})^2 =106.50 \approx 107$

So the answer for this case would be n=107 rounded up to the nearest integer

Step-by-step explanation:

Information given

$\bar X= 0.996$ the sample mean

$s=0.004$ the sample deviation

$n =25$ the sample size

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 =0.001 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)

We can use as estimator of the real population deviation the sample deviation for this case $\hat \sigma = s$/ The critical value for 99% of confidence interval is given by $z_{\alpha/2}=2.58$, replacing into formula (b) we got:

$n=(\frac{2.58(0.004)}{0.001})^2 =106.50 \approx 107$

So the answer for this case would be n=107 rounded up to the nearest integer