Question

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottles to within LaTeX: \pm± 0.01 liter with 95% confidence and also assumes that the standard deviation is 0.05 liter, what sample size is needed?

Answer 1

Answer:

n=97

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=0.01)$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The next step would be find the value of $\z_{\alpha/2}$, $\alpha=1-0.95=.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

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

And we have everything to replace into the formula:

$n=(\frac{1.96(0.05)}{0.01})^2 =96.04$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=0.01 Liters$ is n=97.