Question

The standard deviation for the lifetimes of lawn mowers is estimated to be 800 hours (for the entire population). We'd like to collect data and be 97% confident that our margin of error will not exceed 50 hours. What is the minimum sample size we would need to select, to achieve this

Answer 1

Answer:

$n=(\frac{2.17(800)}{50})^2 =1205.48 \approx 1206$

So the answer for this case would be n=1206 rounded up to the nearest integer

Step-by-step explanation:

We know the following info:

$\sigma = 800$ the standard deviation estimated7

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 =50 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 critical value for 97% of confidence interval and using a significance level of $\alpha=0.03$ and $\alpha/2= 0.015$ $z_{\alpha/2}=2.17$, replacing into formula (5) we got:

$n=(\frac{2.17(800)}{50})^2 =1205.48 \approx 1206$

So the answer for this case would be n=1206 rounded up to the nearest integer