Question

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

Answer 1

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

Answer 2

what is our formula for standard error?

Answer 3

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

Answer 4

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?

Answer 5

not really i dont know how to do standard deviation

Answer 6

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
\[s^2 = \frac xn \frac{n-x}{n}\]

Answer 7

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

Answer 8

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

Answer 9

after that sqrt it

Answer 10

is it going to be a decimal?