MATH SOLVE

3 months ago

Q:
# A doctor is interested in people who have experienced an unexplained episode of vitamin D intoxication. She took a SRS of size 12 from the people who have experienced an unexplained episode of vitamin D intoxication and measured the calcium level in mmol/l in each individual. The sample mean is x = 2.87 mmol/l. Assume that the distribution of the calcium level of people with an unexplained episode of vitamin D intoxication is normally distributed with standard deviation of σ = 0.40.(a) Calculate a 99% confidence interval for μ = the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication. Choose the answer closest to your answer.(2.573, 3.167)(2.784, 2.956) (0.294, 5.446)(2.643, 3.096)We can't answer this question because we don't know the shape of the distribution of the sample mean.(b) Which of the following is the best interpretation of the interval calculated above?We are 99% confident that the average sample mean calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.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. We are 99% confident that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.1% of the time the confidence interval doesnt contain x.None of the above.

Accepted Solution

A:

Answer:(a) (2.573, 3.167) is the 99% confidence interval for [tex]\mu[/tex] 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], then we know that [tex]\bar{X}[/tex] is normally distributed with mean [tex]\mu[/tex] and standard deviation [tex]\sigma/n[/tex]. Therefore we can use [tex](\bar{X}-\mu)/\frac{\sigma}{\sqrt{n}}[/tex] as a pivotal quantity.We have a sample size of n = 12. The sample mean is [tex]\bar{x}[/tex] = 2.87 mmol/l and the standard deviation is [tex]\sigma = 0.40[/tex]. The confidence interval is given by [tex]\bar{x}\pm z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})[/tex] where [tex]z_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the standard normal distribution. As we want the 99% confidence interval, we have that [tex]\alpha = 0.01[/tex] and the confidence interval is [tex]2.87\pm z_{0.005}(\frac{0.40}{\sqrt{12}})[/tex] where [tex]z_{0.005}[/tex] is the 0.5th quantile of the standard normal distribution, i.e., [tex]z_{0.005} = -2.5758[/tex]. Then, we have [tex]2.87\pm (-2.5758)(\frac{0.40}{\sqrt{12}})[/tex] 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.