I = $125 r = 6% t = 1
1 answer:
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Answer:
a) The 99% confidence interval for the true mean milk production is (26.35 kg/d, 29.65 kg/d).
b) They would need to use a confidence level of 94.52%.
Step-by-step explanation:
a)
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 M as such
In which is the standard deviation of the population and n is the size of the sample. So
The lower end of the interval is the mean subtracted by M. So it is 28-1.65 = 26.35 kg/d
The upper end of the interval is the mean added to M. So it is 28 + 1.65 = 29.65 kg/d.
The 99% confidence interval for the true mean milk production is (26.35 kg/d, 29.65 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?
Now we want to find the value of z when M = 1.25. So:
, that has a pvalue of 0.9726.
So the confidence level should be 1 - 2*(1-0.9726) = 0.9452
They would need to use a confidence level of 94.52%.
