Question

Jessa works at a potato chip factory and would like to estimate the mean weight in grams of the factory's potato

Answer 1

Answer:

The smallest sample size required to obtain the desired margin of error is 44.

Step-by-step explanation:

I think there was a small typing error, we have that $\sigma = 60$ is the standard deviation of these weighs.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.05 = 0.95$, so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

Which of these is the smallest approximate sample size required to obtain the desired margin of error?

This sample size is n.

n is found when $M = 15$

So

$M = z*\frac{\sigma}{\sqrt{n}}$

$15 = 1.645*\frac{60}{\sqrt{n}}$

$15\sqrt{n} = 1.645*60$

Simplifying by 15

$\sqrt{n} = 1.645*4$

$(\sqrt{n})^{2} = (1.645*4)^{2}$

$n = 43.3$

Rounding up

The smallest sample size required to obtain the desired margin of error is 44.