Question

A machine makes parts with a variance of 14.5 cm in length. A random sample of 50 parts has a mean length of 106.5 cm. What are the 95% and 99% confidence intervals for the length of parts?

Answer 1

$$\sigma=\sqrt{\sigma^2}=\sqrt{14.5}=3.8$$

a) The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

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

$$=(106.5-1.96\times\dfrac{3.8}{\sqrt{50}}, 106.5+1.96\times\dfrac{3.8}{\sqrt{50}})$$

$$=(105.4467, 107.5533)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $105.4467<\mu<107.5533,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(105.4467, 107.5533).$

b) 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:

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

$$=(106.5-2.5758\times\dfrac{3.8}{\sqrt{50}}, 106.5+2.5758\times\dfrac{3.8}{\sqrt{50}})$$

$$=(105.1158, 107.8842)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $105.1158<\mu<107.8842,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(105.1158, 107.8842).$