Answer to Question #152865 in Statistics and Probability for Muhammad Arfeen Khan

Question

For a given week, a random sample of 30 hourly employees selected from a very large number of employees in a manufacturing firm has a sample mean wage of x̅ = $180.00, with a sample standard deviation of

Answer 1

The provided sample mean is $\bar{X}=180$ and the population standard deviation is $\sigma=14.$ The required confidence level is $95\%,$ and the significance level is $\alpha=0.05.$ Based on the provided information, the critical $z$-value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The 95% confidence for the population mean $\mu$ is computed using the following expression

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

$$=(180-1.96\times\dfrac{14}{\sqrt{30}},180+1.96\times\dfrac{14}{\sqrt{30}})$$

$$=(174.99, 185.01)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $174.99<\mu <185.01,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(174.99, 185.01).$