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

`n=((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

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)(s)/(√(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=((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=((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