Question

n a 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 )

Answer 1

Answer:

$0.333 - 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.251$

$0.333 + 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.416$

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.

$\hat p=\frac{73}{219}=0.333$ estimated proportion of students that reported that they have at least $1000 of credit card debt.

$\alpha=0.01$ 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

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.333 - 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.251$

$0.333 + 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.416$

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)