Question

A large loan company specializes in making automobile loans for used cars. The board of directors wants to estimate the average amount loaned for cars during the past year. Since the high-tech industry has cooled off, we know the data will be slightly skewed. The company decides to take a random sample of 225 customer files for this period. The mean amount loaned for this sample is $8,200 with a standard deviation of $750. To find a 90% confidence interval for the mean amount loaned we would need a z-interval.

Answer 1

Answer:

z-interval: (-1.64, 1.64)

90% Confidence interval:  (7954,8446)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = $8,200

Sample size, n = 225

Alpha, α = 0.10

Sample standard deviation, s = $750

90% Confidence interval:

$\mu \pm z_{critical}\dfrac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64$

z-interval: (-1.64, 1.64)

$8200 \pm 1.64(\dfrac{750}{\sqrt{25}} ) = 8200 \pm 246 = (7954,8446)$

(7954,8446) is the required confidence interval.