Question

When consumers apply for credit, their credit is rated using FICO (Fair, Isaac, and Company) scores. Credit ratings are given below for a sample of applicants for car loans.

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

a. Use the sample data to construct a 99% confidence interval for the mean FICO score of all applicants for credit.
b. If one bank requires a credit rating of at least 620 for a car loan, does it appear that almost all applicants will have suitable credit ratings? Why or why not?

Answer 1

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)`

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}`

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is
`\bar X= 660.313`

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by
`\alpha=1-0.99=0.01` and
`\alpha/2 =0.005`. The degrees of freedom are given by:

`df=n-1=16-1=15`

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for
`t_(\alpha/2)=-2.95`

"=T.INV(1-0.005,15)" for
`t_(1-\alpha/2)=2.95`

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))`

And if we find the limits we got:

`660.313- 2.95(95.898)/(√(16))=589.588`

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.