Question

Stats: In a large population of high school students, 20% have experienced math anxiety. You take a random sample of 10 students from this population. The standard deviation of the number of students in the sample who have experienced math anxiety is?? :)

Answer 1

Breee: I recognize this problem. I recommended that you try to find the proper formula for sample standard deviation when working with proportions. What did you find through your Internet search for that topic?

Answer 2

My point here, Bree, is that developing the ability to find needed info is essential, the higher you go in education.

I will share my own Internet search for "proportion standard deviation."

https://www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&es_th=1&ie=UTF-8&q=proportion%20sample%20standard%20deviation&oq=proportion%20sample%20standard%20deviation&aqs=chrome..69i57j0l3.9336j0j7

Answer 3

sqrt((p(1-p)/n) is what I found but I'm not sure what to do...

Answer 4

You're getting there.

Please refer to the following:

http://sites.stat.psu.edu/~dhunter/100/spring2005/lecture/lecture28.pdf

There are 4 boxes with info in them at the top of this page. Look at the box in the 2nd column and the second row.

This is the formula you need for "sample standard deviation" for a sample proportion.

Answer 5

In your particular problem, you may use p=0.20 and n=10 (since you have 10 samples).

Answer 6

If you can evaluate that formula correctly, you're done.

Answer 7

copy down the formula from

http://sites.stat.psu.edu/~dhunter/100/spring2005/lecture/lecture28.pdf

onto paper and then subst. p=0.2 and n=10. share your work, if possible.
Your result??

Answer 8

let \(X\) be the number of students out of the 10 sampled that have math anxiety

Then \(X\sim\)Bin(10,.2)

the s.d. of \(X\) is given by \(\sqrt{np(1-p)}\)

Answer 9

Zarkon: I know there are criteria for approximating the sample standard deviation in this manner, but have forgotten them. Are they worth bringing up in this discussion? Thank you for the correction to my statement earlier.

Answer 10

you are not looking for the standard deviation of the proportion

\[\frac{X}{n}\]

which would be \[\sqrt{\frac{p(1-p)}{n}}\]

you are looking for the s.d. of \(X\)

Answer 11

Sorry! My computer crashed :o

Answer 12

So it would be sqrt((.20(1-.20))/10)?

Answer 13

no...use the formula

\[\sqrt{np(1-p)}\]

Answer 14

Oh okay one sec

Answer 15

http://www.dummies.com/how-to/content/how-to-find-the-mean-variance-and-standard-deviati.html

Answer 16

sqrt(10*.2(1-.2)) = 1.265?

Answer 17

yes

Answer 18

Both equations gave me the same answer, what's the difference between the two??

Answer 19

they don't...you must have put it in your calculator incorrectly

Answer 20

Thanks again, Zarkon.
Bree: Sorry for having given you the wrong formula.

Answer 21

Okey dokey, thanks for the help! :)

Answer 22

It's all good :)

Answer 23

the previous formula gives .1265

the latter is 1.265

Answer 24

Ohhh oops, my bad :)

Answer 25

Thanks again! :D