Answer to Question #264174 in Statistics and Probability for Myra

Question

A statistics practitioner took a random sample of 50 observations from a population with a standard deviation of 25 and sample mean of 100. The 95% confidence interval of population mean is

Answer 1

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:

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

$$=(100-1.96\times\dfrac{25}{\sqrt{50}}, 100+1.96\times\dfrac{25}{\sqrt{50}})$$

$$=(93.07, 106.93)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $93.07 < \mu < 106.93,$ which indicates that we are 95 %confident that the true population mean $\mu$ is contained by the interval $(93.07, 106.93).$