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 standart deviation of milk production is 2.25 kg/d.
Requried:
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) 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 $\alpha$ 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 the margin of error 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.

$M = 2.575*\frac{2.25}{\sqrt{12}} = 1.67$

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.

$M = z*\frac{2.25}{\sqrt{12}} = 1.67$

$1.25 = z*\frac{2.25}{\sqrt{12}} = 1.67$

$2.25z = 1.25\sqrt{12}$

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

$z = 1.92$

When $z = 1.92$, it has a pvalue of 0.9725.

1 - 2*(1 - 0.9725) = 0.945

So we should use a confidence level of 94.5%.