Question

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

chip bags for quality control purposes. She'll sample n bags of chips to build a 90% confidence interval for the
mean with a margin of error of no more than 15 g. Preliminary data suggests that 0 = 60 g is a reasonable
estimate for the standard deviation of these weights.
Which of these is the smallest approximate sample size required to obtain the desired margin of error?

Answer 1

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

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 = (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*(\sigma)/(√(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*(\sigma)/(√(n))`

`15 = 1.645*(60)/(√(n))`

`15√(n) = 1.645*60`

Simplifying by 15

`√(n) = 1.645*4`

`(√(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.