Answer:
The 99% confidence interval for the mean loss in value per home is between $5359 and $13463
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the student's t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 45 - 1 = 44
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995([tex]t_{995}[/tex]). So we have T = 2.6923
The margin of error is:
M = T*s = 1505*2.6923 = 4052.
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 9411 - 4052 = $5359
The upper end of the interval is the sample mean added to M. So it is 9411 + 4052 = $13463
The 99% confidence interval for the mean loss in value per home is between $5359 and $13463
