Question

In a simple random sample of 219 students at a college, 73 reported that they have at least $1000 of credit card debt. What is the 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt?
(Points : 3)
(25.0, 41.6)

(27.5, 39.1)

(30.1, 36.5)

(31.0, 35.6)

Answer 1

Take note of the formula for the confidence interval for population proportions.\[\displaystyle \hat{p} \pm \text{z}_{\alpha/2} * (\sqrt{\displaystyle \frac{\hat{p} * \hat{q}}{n}})\]

Answer 2

Where \[\hat p\] is the sample proportion, we apply the formula:
\[\hat p = \frac{x}{n}\]
where x is the number of successes and n is the sample size

Answer 3

The \[z_{\alpha/2}\] can be found using the z-table.

Answer 4

Let me attached an example on finding the \[z_{\alpha/2}\].

Answer 5

Suppose we have a confidence level of 90%.
Find the value of \[\alpha/2\] first.
Apply the formula: \[\frac{\alpha}{2} = \frac{1-c}{2}\]
where c is the confidence level written in decimal form.
So, for this example, c = 0.90.
\[\frac{\alpha}{2} = \frac{1-0.90}{2}= 0.05\]

Answer 6

We will locate the 0.05 on the z-table.
The 0.05 is in between 0.0495 and 0.0505.
So, we will take the average of the z-scores that correspond to 0.0495 and 0.0505.

Answer 7

We remove the negative sign.
\[z_{a/2}= \frac{1.64 + 1.65}{2} = 1.645\]