Question

A sample of 91 circuits from a large normal population has a mean resistance of 2,4 ohms. If the population standard deviation is 1.5 ohms, determine a 95% confidence interval for the true mean resistance of the population. (Hint: critical value is 1.96)

Answer 1

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

The corresponding confidence interval is computed as shown below:

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

$$=(2.4-1.96\times \dfrac{1.5}{\sqrt{91}},2.4+1.96\times \dfrac{1.5}{\sqrt{91}})$$

$$=(2.0918, 2.7082)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $2.0918<\mu<2.7082,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(2.0918, 2.7082).$