Suppose you have a random sample of 2179 credit scores from a population of mortgage applicants with a sample mean of 620 and known population standard deviation of 70, and would like to calculate the 99% confidence interval of the population mean credit score. Which of the following would be the correct construction?
A) 620 ± 1.65 $\left( \frac { 70 } { \sqrt { 2179 } } \right)$
B) 2179 ± 1.96 $\left( \frac { 70 } { \sqrt { 620 } } \right)$
C) 620 ± 2.58 $\left( \frac { 70 } { \sqrt { 2179 } } \right)$
D) 620 ± 2.58 $\frac { 70 } { 2179 }$

Question

Quizplus · Accepted Answer

We use the formula for a confidence interval for the population mean when the population standard deviation is known:

$$\text{CI} = \bar{x} ± z \frac{\sigma}{\sqrt{n}}$$

where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, and $z$ is the critical value of the standard normal distribution for the desired level of confidence.

Since we want a 99% confidence interval, we use a critical value of $z = 2.58$ (which can be found using a table or calculator). Plugging in the numbers, we get:

$$\text{CI} = 620 ± 2.58 \left(\frac{70}{\sqrt{2179}}\right)$$

Therefore, the correct construction is option C.

Suppose You Have a Random Sample of 2179 Credit Scores (702179)\left( \frac { 70 } { \sqrt { 2179 } } \right)(2179​70​)

Suppose You Have a Random Sample of 2179 Credit Scores $\left( \frac { 70 } { \sqrt { 2179 } } \right)$