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 p…

Question

asked Jan 27, 2021 214k views

1 Answer

Mike Kantor · Answer 1 · 2021-01-31T12:19:07+0000

Answer:

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.