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anonymous one year ago A random sample of 100 undergraduate students at a university found that 78 of them had used the university library’s website to find resources for a class. What is the margin of error for the true proportion of all undergraduates who had used the library’s website to find resources for a class? 0.04 0.08 0.1 0.12

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1. amistre64

margin of error is not the same as 'the standard error'. this question does not contain all the information needed for a 'margin of error' calculation since it does not provide for a confidence level. so, it most likely is asking for the standard error instead

2. amistre64

what is our formula for standard error?

3. anonymous

i attached it cus i couldnt get the line in top of the x

4. amistre64

thats a good start, now, we simply need to define s what can you tell me of the standard deviation of a proportion? any ideas?

5. anonymous

not really i dont know how to do standard deviation

6. amistre64

also, lets go this route: include the s inside of the sqrt $\frac{s}{\sqrt{n}}\implies \sqrt{\frac{s^2}{n}}$ well, for lack of a better way to express it $s^2 = \frac xn \frac{n-x}{n}$

7. amistre64

therefore: $SE=\sqrt{\frac{x(n-x)}{n^3}}$ or letting p = x/n, and q+p = 1 $SE=\sqrt{\frac{pq}{n}}$

8. amistre64

what is 78(100-78)/100^3 ?

9. amistre64

after that sqrt it

10. anonymous

is it going to be a decimal?

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