Respuesta :
Answer:
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.
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].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
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.
[tex]M = 2.575\frac{1.7}{\sqrt{36}} = 0.73[/tex]
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.
