Question

A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a sample mean of 15.3 oz. Given the standard deviation is 1.7 oz, obtain a 99% confidence interval for the true mean weight of the boxes. Interpret your results in the context of the problem. State the margin of error.

Answer 1

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

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 - 0.99)/(2) = 0.005`

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

That 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(\sigma)/(√(n))`

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

`M = 2.575(1.7)/(√(36)) = 0.73`

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.