Respuesta :
Answer:
a) 26.33 kg/d and 29.67 kg/d
b) 94.5%
Step-by-step explanation:
a. Find a 99% confidence interval for the true mean milk production.
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{2.25}{\sqrt{12}} = 1.67[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 28 - 1.67 = 26.33 kg/d
The upper end of the interval is the sample mean added to M. So it is 28 + 1.67 = 29.67 kg/d
The 99% confidence interval for the true mean milk production is between 26.33 kg/d and 29.67 kg/d
b. If the farms want the confidence interval to be no wider than ± 1.25 kg/d, what level of confidence would they need to use?
We need to find z initially, when M = 1.25.
[tex]M = z*\frac{2.25}{\sqrt{12}} = 1.67[/tex]
[tex]1.25 = z*\frac{2.25}{\sqrt{12}} = 1.67[/tex]
[tex]2.25z = 1.25\sqrt{12}[/tex]
[tex]z = \frac{1.25\sqrt{12}}{2.25}[/tex]
[tex]z = 1.92[/tex]
When [tex]z = 1.92[/tex], it has a pvalue of 0.9725.
1 - 2*(1 - 0.9725) = 0.945
So we should use a confidence level of 94.5%.
