Dairy cows at large commercial farms often receive
injections of bst (bovine somatotropin), a hormone used to
spur milk production. bauman et al. (journal of dairy science,
1989) reported that 12 cows given bst produced an average of
28.0 kg/d of milk. assume that the standard deviation of milk
production is 2.25 kg/d.
(a) find a 99% confidence interval for the true mean milk
production.
(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?

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%.

you have orange juice on wednesday. (idk why this is the answer)

step-by-step explanation:

yes the data was not prsented

step-by-step explanation: