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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
    • one year ago
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    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
    • one year ago
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    what is our formula for standard error?

  3. anonymous
    • one year ago
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    i attached it cus i couldnt get the line in top of the x

  4. amistre64
    • one year ago
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    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
    • one year ago
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    not really i dont know how to do standard deviation

  6. amistre64
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    what is 78(100-78)/100^3 ?

  9. amistre64
    • one year ago
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    after that sqrt it

  10. anonymous
    • one year ago
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    is it going to be a decimal?

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