Question

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?

Answer 1

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample. So

$M = 2.575*\frac{2.25}{\sqrt{12}} = 1.65$

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:

$1.25 = z*\frac{2.25}{\sqrt{12}}$

$0.6495z = 1.25$

$z = 1.92$, 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%.