Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Discover in-depth and trustworthy answers from our extensive network of knowledgeable professionals.
Sagot :
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions are important to us at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
