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

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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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

So

[tex]M = 2.575*\frac{10.9}{\sqrt{34}} = 4.81[/tex]

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