The housing market has recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 45 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss was $9411 with a standard deviation of $1505.
a. Find a 99% confidence interval for the mean loss in value per home.

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 ). 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

the reason why your question keeps getting deleted because this is not a school related question : )

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