Question

g In 2017 the average credit score for mortgage loans purchased by Fannie Mae was 745. Recently a sample of 20 mortgages were randomly selected and it was found that the average credit score was 750 with a sample standard deviation of 25. Assume the data was normally distributed. Compute a 95% confidence interval for the average credit score.

Answer 1

Answer:

The 95% confidence interval for the average credit score is between 697.675 and 802.325

Step-by-step explanation:

We are in posession of the sample standard deviation, so the t-distribution is used 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 = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.093

The margin of error is:

M = T*s = 2.093*25 = 52.325

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 750 - 52.325 = 697.675

The upper end of the interval is the sample mean added to M. So it is 750 - 52.325 = 802.325

The 95% confidence interval for the average credit score is between 697.675 and 802.325