Respuesta :
Answer:
The 99% confidence interval for the mean consumption of meat among people over age 27 is between 1.4 pounds and 1.6 pounds.
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.
[tex]M = 2.575*\frac{1.2}{\sqrt{1179}} = 0.1[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 1.5 - 0.1 = 1.4 pounds
The upper end of the interval is the sample mean added to M. So it is 1.5 + 0.1 = 1.6 pounds
The 99% confidence interval for the mean consumption of meat among people over age 27 is between 1.4 pounds and 1.6 pounds.
