Question

2. We take a random sample of 250 Maryland police officers and find that their average annual salary is $42,500, with a standard deviation of $6,000. Construct a 98% confidence interval around your point estimate. Interpret your results.

Answer 1

The critical value for $\alpha = 0.02$ and $df = n-1 =249$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.341417$

The corresponding confidence interval is computed as shown below:

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

$$=(42500-2.341417\times\dfrac{6000}{\sqrt{250}},$$

$$42500+2.341417\times\dfrac{6000}{\sqrt{250}})$$

$$=(41611.495, 43388.505)$$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $41611.495 < \mu < 43388.505,$ which indicates that we are 98% confident that the true population mean $\mu$ is contained by the interval $(41611.495, 43388.505).$