Answer:
189 components must be sampled.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Assume that component lifetimes are normally distributed with population standard deviation of 16 hours.
This means that [tex]\sigma = 16[/tex]
How many components must be sampled so that a 99% confidence interval will have margin of error of 3 hours?
n components must be sampled.
n is found when M = 3. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 2.575\frac{16}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 2.575*16[/tex]
[tex]\sqrt{n} = \frac{2.575*16}{3}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*16}{3})^2[/tex]
[tex]n = 188.6[/tex]
Rounding up:
189 components must be sampled.
