Question

To assess the accuracy of a laboratory scale, a standard weight known to weigh 1 gram is repeatedly weighed a total of n times How large should n be so that a 95% confidence interval for µ has a margin of error of ± 0.0001?

Answer 1

`n=((1.960(1))/(0.0001))^2 =384160000`

So the answer for this case would be n=384160000 rounded up to the nearest integer

Explanation:

We know the following info:

`ME = 0.0001` represent the margin of error desired

`\sigma= 1` we assume that the population deviation is this value

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =0.0001 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. If we use the normal standard distribution or excel we got:
`z_(\alpha/2)=1.960`, replacing into formula (b) we got:

`n=((1.960(1))/(0.0001))^2 =384160000`

So the answer for this case would be n=384160000 rounded up to the nearest integer