Answer to Question #335117 in Statistics and Probability for Kei

Question

corporation that owns apartment complexes wishes to estimate the average length of time residents remain in the same apartment before moving out. A sample of 150 rental contracts gave a mean length of occupancy of 3.7 years with standard deviation 1.2 years. Construct a 95% confidence interval for the mean length of occupancy of apartments owned by this corporation.

Answer 1

The critical value for $\alpha = 0.05$ and $df = n-1 = 149$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 1.976013.$

The corresponding confidence interval is computed as shown below:

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

$$=(3.7-1.976013\times\dfrac{1.2}{\sqrt{150}},$$

$$3.7+1.976013\times\dfrac{1.2}{\sqrt{150}})$$

$$=(3.5064,3.8936)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $3.5064 < \mu < 3.8936,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(3.5064, 3.8936).$