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Answer:
The 99% confidence interval for the mean consumption of meat among males over age 43 is between 2.9 pounds and 3.1 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].
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{1.3}{\sqrt{1384}} = 0.1[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 3 - 0.1 = 2.9 pounds
The upper end of the interval is the sample mean added to M. So it is 3 + 0.01 = 3.1 pounds
The 99% confidence interval for the mean consumption of meat among males over age 43 is between 2.9 pounds and 3.1 pounds.
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