Answer to Question #343910 in Statistics and Probability for qwerty

Question

The average height of a random sample of 400 people from San Pablo City is 1.75m. It is known that the heights of the population are random variables that follow a normal distribution with a variance of 0.16. Determine the interval of 99% confidence for the average heights of the population. 

Answer 1

The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} = 2.5758.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$ $$=(1.75-2.5758\times\dfrac{0.4}{\sqrt{400}}, 1.75+2.5758\times\dfrac{0.4}{\sqrt{400}})$$ $$=(1.698484, 1.801516)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $1.698484< \mu < 1.801516,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(1.698484, 1.801516).$