Answer to Question #346760 in Statistics and Probability for Bethsheba Kiap

Question

A random sample is drawn from a normally distributed population of known standard deviation 5. Construct a 99.8% confidence interval for the population mean based on the information given (not all of the information given need be used).

a. n = 16, 𝑥̅= 98, 𝑠 = 5.6

b. n = 9, 𝑥̅= 98, 𝑠 = 5.6

Answer 1

1. The critical value for $\alpha = 0.002$ $z_c = z_{1-\alpha/2} = 3.0902$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$ $$=(98-3.0902\times\dfrac{5}{\sqrt{16}},$$ $$98+3.0902\times\dfrac{5}{\sqrt{16}})$$ $$=(94.137,101.863)$$

Therefore, based on the data provided, the 95 confidence interval for the population mean is $94.137 < \mu <101.863,$ which indicates that we are 99.8% confident that the true population mean $\mu$ is contained by the interval $(94.137,101.863).$

2. The critical value for $\alpha = 0.002$ $z_c = z_{1-\alpha/2} = 3.0902$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$ $$=(98-3.0902\times\dfrac{5}{\sqrt{9}},$$ $$98+3.0902\times\dfrac{5}{\sqrt{9}})$$ $$=(92.850,103.150)$$

Therefore, based on the data provided, the 95 confidence interval for the population mean is $92.850 < \mu <103.150,$ which indicates that we are 99.8% confident that the true population mean $\mu$ is contained by the interval $(92.85,103.150).$