Answer to Question #270527 in Statistics and Probability for Subham

Question

Find a 99% confidence interval for the mean � of a normal population with variance 9

from the sample 30,42,40,34,48,50.

Answer 1

$$mean=\bar{x}=\dfrac{1}{n}\sum _ix_i$$ $$=\dfrac{1}{6}(30+42+40+34+48+50)=\dfrac{244}{6}=\dfrac{122}{3}$$

$$\approx40.6667$$

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:

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

$$=(\dfrac{122}{3}-2.5758\times\dfrac{3}{\sqrt{6}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(37.512, 43.821)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $37.512 < \mu < 43.821,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(37.512, 43.821).$