Respuesta :
Answer:
99% of the sample means will occur between 5,650.20 pounds and 5,849.80.
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 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{260}{\sqrt{45}} = 99.80[/tex]
The lower end of the interval is the mean subtracted by M. So it is 5,750 - 99.80 = 5,650.20
The upper end of the interval is the mean added to M. So it is 99.80 + 5,750 = 5,849.80
99% of the sample means will occur between 5,650.20 pounds and 5,849.80.
