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

Answer:

The minimum sample size that can be taken is 1.

Step-by-step explanation:

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

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

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

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

$1 = 1.645*\frac{0.6}{\sqrt{n}}$

$\sqrt{n} = 1.645*0.6$

$(\sqrt{n})^{2} = (1.645*0.6)^{2}$

$n = 0.97$

Rounding up

The minimum sample size that can be taken is 1.