Question

A sample of 20 observations is selected from normal population distribution with

population standard deviation 9 and sample mean 51. Determine the highest

value of population mean with 95% confidence interval (z-1.96). Round the

answer to the two decimal places.

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:

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

$$=(51-1.96\times\dfrac{9}{\sqrt{20}}, 51+1.96\times\dfrac{9}{\sqrt{20}})$$

$$=(47.06, 54.94)$$

The highest value of population mean with 95% confidence interval is $54.94.$