anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • 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
amistre64
  • amistre64
what is our formula for standard error?
anonymous
  • anonymous
i attached it cus i couldnt get the line in top of the x

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amistre64
  • 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?
anonymous
  • anonymous
not really i dont know how to do standard deviation
amistre64
  • 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 \[s^2 = \frac xn \frac{n-x}{n}\]
amistre64
  • amistre64
therefore: \[SE=\sqrt{\frac{x(n-x)}{n^3}}\] or letting p = x/n, and q+p = 1 \[SE=\sqrt{\frac{pq}{n}}\]
amistre64
  • amistre64
what is 78(100-78)/100^3 ?
amistre64
  • amistre64
after that sqrt it
anonymous
  • anonymous
is it going to be a decimal?

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