Question

The population standard deviation for the weights of boxes of breakfast cereal is 0.6 ounces. If we want to be 90% confident that the sample mean is within 1 ounce of the true population mean, what is the minimum sample size that can be taken

Answer 1

The minimum sample size that can be taken is 1.

Explanation:

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))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

In this problem

We have to find n, when
`M = 1, \sigma = 0.6`. So

`M = z*(\sigma)/(√(n))`

`1 = 1.645*(0.6)/(√(n))`

`√(n) = 1.645*0.6`

`(√(n))^(2) = (1.645*0.6)^(2)`

`n = 0.97`

Rounding up

The minimum sample size that can be taken is 1.