Question

The average weight of 25 oranges selected from a normally distributed population is 400g with a standard deviation of 20g. Find the confidence interval using 95% confidence level.

Answer 1

The critical value for $\alpha = 0.05$ and $df = n-1 = 24$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.063899.$

The corresponding confidence interval is computed as shown below:

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

$$=(400-2.063899\times\dfrac{20}{\sqrt{25}},$$

$$400+2.063899\times\dfrac{20}{\sqrt{25}})$$

$$=(391.7444,408.2556)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $391.7444<\mu<408.2556,$ which indicates that we are 95% confident that the true population mean $μ$ is contained by the interval $(391.7444,408.2556).$