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?

Answer 1

what is your idea in creating the interval?

Answer 2

do i need to find the margin of error?

Answer 3

margin of error is part of it sure, but in essense:

we are trying to find 2, x values, associated with the z scores that canter 99% of the information about the mean. with some adjustments this is just the z formula
\[z=\frac{x-\bar x}{\sigma}\]
solving for x we get
\[\bar x\pm z\sigma=\pm x\]
the adjustment is in sigma, we use the standard error instead of the standard devaition

Answer 4

SE = sqrt(pq/n) for a proportion setup, and xbar = p

Answer 5

\[p\pm z\sqrt{\frac{pq}{n}}\]
this can be stated also as
\[\frac {\bar x}{n}\pm z\sqrt{\frac{\bar x~(n-\bar x)}{n^3}}\]

Answer 6

is it 30.1 and 36.5?

Answer 7

I'm really confused on this

Answer 8

weve started with something basic, the z formula, and adapted it to our needs. now all we do is input the information into it.

\[\frac {73}{219}\pm z\sqrt{\frac{73~(219-73)}{219^3}}\]
and out z value for 99% is what ... 2.567? 2.576?

\[\frac {73}{219}\pm 2.576\sqrt{\frac{73~(219-73)}{219^3}}\]
multiplying thru by 219 gives us proportions of the sample size needed

\[73\pm 2.576\sqrt{\frac{73~(219-73)}{219}}\]

Answer 9

73 +- 28.8

Answer 10

and no it cant be 30.1 to 36.5 simply because we are centering around 73 ...

73 is nowhere in between 30 and 36