Question

A random sample is drawn from a population of known standard deviation 11.3.

Construct a 90% confidence interval for the population mean based on the information

given (not all of the information given need be used).

a. n = 36, 𝑥̅= 105.2, 𝑠 = 11.2

b. n = 100, 𝑥̅= 105.2, 𝑠 = 11.2

Answer 1

a.

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

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

$$=(105.2-1.6449\times\dfrac{11.3}{\sqrt{36}}, 105.2+1.6449\times\dfrac{11.3}{\sqrt{36}})$$

$$=(102.1021, 108.2979)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $102.1021 < \mu < 108.2979,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(102.1021, 108.2979).$

b.

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

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

$$=(105.2-1.6449\times\dfrac{11.3}{\sqrt{100}}, 105.2+1.6449\times\dfrac{11.3}{\sqrt{100}})$$

$$=(103.3413, 107.0587)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $103.3413 < \mu < 107.0587,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(103.3413, 107.0587).$