Question

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 3.3 mm, how many of these components should she consider to be 95% sure of knowing the mean will be within ± 0.3mm?
- 328

- 5062

- 465

- 33

- 22

Answer 1

Answer:

n=465 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=3.3$ represent the population standard deviation

Confidence =0.95 or 95%

Me= 0.3, represent the margion of error

n represent the sample size (variable of interest)

The margin of error is given by this formula:

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

And on this case we have that ME =0.3 mm and we are interested in order to find the value of n, if we solve n from equation (4) we got:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got $z_{\alpha/2}=1.960$, replacing into formula (b) we got:

$n=(\frac{1.960(3.3)}{0.3})^2 =464.8336 \approx 465$

So the answer for this case would be n=465 rounded up to the nearest integer