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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
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.
When , it has a pvalue of 0.9725.
1 - 2*(1 - 0.9725) = 0.945
So we should use a confidence level of 94.5%.
