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

`n=((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

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)(s)/(√(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=((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=((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