Question

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation of all components is 2.2 mm, how many of these components should she consider to be 80% sure of know the mean will be within 0.3 mm?

Answer 1

Answer:

$n=(\frac{0.539(2.2)}{0.3})^2 =15.62 \approx 16$

So the answer for this case would be n=16 rounded up to the nearest integer

Step-by-step explanation:

For this case we have the following info given:

$\sigma = 2.2$ represent the population deviation

$Confidence =0.8$

$ME = 0.3$ represent the margin of error desired

The margin of error for the true mean is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (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 (a) we got:

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

The confidence level is 80%, the significance would be $\alpha=0.2$ and $\alpha/2 =0.1$ the critical value for this case is $z_{\alpha/2}=0.539$, replacing into formula (b) we got:

$n=(\frac{0.539(2.2)}{0.3})^2 =15.62 \approx 16$

So the answer for this case would be n=16 rounded up to the nearest integer