Question

The coffee and soup machine at the local subway station is supposed to fill cups with 6 ounces of soup. Nine cups of soup are bought with results of a mean of 5.5 ounces and a standard deviation of 0.18 ounces. How large a sample of soups would we need to be 95 percent confident that the sample mean is within 0.03 ounces of the population mean

Answer 1

Answer:

$n=(\frac{1.960(0.18)}{0.03})^2 =138.30$

So the answer for this case would be n=139 rounded up to the nearest integer

Step-by-step explanation:

For this case we have the following info given:

$\bar X = 5.5$ the sample mean

$s = 0.18$ the standard deviation

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$    (a)

And on this case we have that ME =0.03 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution and if we look in the normal standard distirbution we got $z_{\alpha/2}=1.96$, replacing into formula (b) we got:

$n=(\frac{1.960(0.18)}{0.03})^2 =138.30$

So the answer for this case would be n=139 rounded up to the nearest integer