Question #335117

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.

1
Expert's answer
2022-05-05T09:43:01-0400

The critical value for α=0.05\alpha = 0.05 and df=n1=149df = n-1 = 149 degrees of freedom is tc=z1α/2;n1=1.976013.t_c = z_{1-\alpha/2; n-1} = 1.976013.

The corresponding confidence interval is computed as shown below:


CI=(xˉtc×sn,xˉ+tc×sn)CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}} ,\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})

=(3.71.976013×1.2150,=(3.7-1.976013\times\dfrac{1.2}{\sqrt{150}},

3.7+1.976013×1.2150)3.7+1.976013\times\dfrac{1.2}{\sqrt{150}})

=(3.5064,3.8936)=(3.5064,3.8936)

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


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