Answer to Question #127576 in Statistics and Probability for Jessica

Question

Dairy cows at large commercial farms often receive injections of bST, a hormone used to spur milk production. Bauman et al. 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 the 99% confidence interval for the true mean milk production. (Round the answer to two decimal places)

B. If the farms want the confidence interval to be no wider than + - 1.10 kg/d, what level of confidence would they need to use? ( Round answers to the nearest integer) (Express the answer in percent)

Answer 1

A. The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha /2}=2.576.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times{\sigma \over \sqrt{n}}, \bar{X}+z_c\times{\sigma \over \sqrt{n}})$$

$$=(28-2.576\times{2.25\over \sqrt{12}}, 28+2.576\times{2.25 \over \sqrt{12}})$$

$$\approx(26.33, 29.67)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $26.33<\mu < 29.67,$ which indicates that we are 99% confident that the true population mean $\mu$  is contained by the interval $(26.33, 29.67).$

B. Given $E=1.10.$

$$E=z_c\times{\sigma \over \sqrt{n}}$$

$$z_c=E\times{\sqrt{n} \over \sigma}$$

$$z_c=1.1\times{\sqrt{12} \over 2.25}\approx1.69356$$

The two-tailed P value equals $0.0903$.

$$1-0.0903=0.9097\approx0.91$$

So we use 91 % confidence interval.