Na simple random sample of 219 students at a college, 73 reported that they have at least $1000 of credit card debt.

which interval is the 99% confidence interval for the percent of all the students at that college who have at least $1000 in credit card debt?

(31.0 ,35.6)

( 30.1 , 36.5)

(25.0 ,41.6)

(27.5 ,39.1 )

The 99% confidence interval would be given (0.251;0.416).

(25.0 ,41.6)

Step-by-step explanation:

1) Data given and notation  

n=219 represent the random sample taken    

X=73 represent the students that reported that they have at least $1000 of credit card debt.

estimated proportion of students that reported that they have at least $1000 of credit card debt.

represent the significance level

z would represent the statistic (variable of interest)    

p= population proportion of students that reported that they have at least $1000 of credit card debt.

2) Confidence interval

The confidence interval would be given by this formula

For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

And replacing into the confidence interval formula we got:

And the 99% confidence interval would be given (0.251;0.416).

We are confident that about 25.1% to 41.6% of students have at least $1000 of credit card debt.

And for this case the most accurate option is:

(25.0 ,41.6)

i would say the third one.

step-by-step explanation: liner would mean in a line and the only one that doesn't look like it's even thinking about forming a line would be the third one

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1