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?

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%.

step-by-step explanation:

20% of 120 is 24. if you subtract 24 from 120 you get 96

c.

step-by-step explanation: