Question

You measure 34 backpacks' weights, and find they have a mean weight of 33 ounces. Assume the population standard deviation is 10.9 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight

Answer 1

Answer:

The maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight is 4.81 ounces

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$, so $z = 2.575$

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.

So

$M = 2.575*\frac{10.9}{\sqrt{34}} = 4.81$

The maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight is 4.81 ounces