-2.35 - (-1.27) as a fraction in simplest form

(a) (2.573, 3.167) is the 99% confidence interval for $\mu$ the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication.

(b) There is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

Step-by-step explanation:

If we have a random sample of size n from a normal distribution with mean $\mu$ and standard deviation $\sigma$ , then we know that $\bar{X}$ is normally distributed with mean $\mu$ and standard deviation $\sigma/n$ . Therefore we can use $(\bar{X}-\mu)/\frac{\sigma}{\sqrt{n}}$ as a pivotal quantity.

We have a sample size of n = 12. The sample mean is $\bar{x}$ = 2.87 mmol/l and the standard deviation is $\sigma = 0.40$ . The confidence interval is given by $\bar{x}\pm z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})$ where $z_{\alpha/2}$ is the $\alpha/2$ th quantile of the standard normal distribution. As we want the 99% confidence interval, we have that $\alpha = 0.01$ and the confidence interval is $2.87\pm z_{0.005}(\frac{0.40}{\sqrt{12}})$ where $z_{0.005}$ is the 0.5th quantile of the standard normal distribution, i.e., $z_{0.005} = -2.5758$ . Then, we have $2.87\pm (-2.5758)(\frac{0.40}{\sqrt{12}})$ and the 99% confidence interval is given by (2.573, 3.167)

(b) We found a 99% confidence interval for the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication. Therefore, there is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

8 0

4 years ago

Graph y =-7x-1 il do anything

Goshia [24]

Answer:

here you go!