The housing market has recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 38 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss was $9379 with a standard deviation of $3000. Suppose a 95% confidence interval to estimate the average loss in home value is found.

a) Suppose the standard deviation of the losses had been $9000 instead of $3000.
b) What would the larger standard deviation do to the width of the confidence interval (assuming the same level of confidence)?

Given that the housing market has recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 38 bids from potential buyers to estimate the average loss in home value.

s = sample std deviation = 3000

Sample mean = 9379

Sample size n = 38

df = 37

Std error of sample mean = [tex]\frac{s}{\sqrt{n} } \\=486.66[/tex]

confidence interval 95% = Mean ± t critical * std error

=Mean ±1.687*486.66 = Mean ±821.003

=(8557.997, 10200.003)

a) If std deviation changes to 9000 instead of 3000, margin of error becomes 3 times

Hence 2463.008

b) The more the std deviation the more the width of confidence interval.