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

i need help with how to solve this step by step

Answer 2

thats a waste of own bucks

Answer 3

owl

Answer 4

any ideas?

Answer 5

I'm sorry I don't know your answer.
Please wait, I ask to another tutor

Answer 6

@CGGURUMANJUNATH can you help here please?

Answer 7

thank you

Answer 8

Hi there :)

Answer 9

Are you familiar with how to use Z tables

Answer 10

you need to first identity, what Z score gives you a 99% confidence interval

Answer 11

2.567?

Answer 12

sounds about right, ill have to check myself too but, now whats left for you is to find the true standard deviation

Answer 13

dan can you help me on a question after this plz

Answer 14

I'm getting stuck at 73 +- 28.8

Answer 15

the ratio 73/219 is the sample percentage that have atleast 1000 dollars of debt
73/219 = 33.33%

Answer 16

that mean is the same for the actual distribution as well

Answer 17

you will be using the formula
33.33 +- (2.57*sigma)
you need to find the sigma which is the true standard deiviation

Answer 18

also if you notice the choices given to you the middle values are all 33.33 in that range

Answer 19

(25.0 + 41.6)/2 =33.33

(27.5 + 39.1)/2=33.33

(30.1 + 36.5)/2 = 33.33

(31.0, + 35.6)/2 = 33.33

Answer 20

~clapping~

Answer 21

B?

Answer 22

uhh wait im thinking how to get the true STD from a mean and sample size

Answer 23

:s

Answer 24

so would the answer be B?

Answer 25

Show me some of your work, maybe ill see if it makes sense at least

Answer 26

the answer is A

Answer 27

Ok in order to find the sample standard deviation we need to realize that we are using proportions here. Since the answer gives the percentage as a whole number for computational sake we will do the same.

p=73/219 =33%
q=(1-p)=1-33=67%

We know that our sample standard dev is \( \sqrt{ \frac{p*q}{n}}\)
So S=\( \sqrt{ \frac{33*67}{216}}=3.17\)

Therefore our confidence interval is
\( 33.33 \pm 2.58*3.17 \to (25.15,41.5)\)