Question

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)?

Answer 1

Answer:

Step-by-step explanation:

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 = $\frac{s}{\sqrt{n} } \\=486.66$

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.