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anonymous

  • one year ago

I WILL FAN AND MEDAL 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. moazzam07
    • one year ago
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    margin of error suggests a confidence interval, but no width has been suggested

  2. moazzam07
    • one year ago
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    if we wanted to be 99% confident that the true population proportion was within a given interval, we would use z=2.576 and use the standard error as a multiplier

  3. moazzam07
    • one year ago
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    the standard error is just a modification of the standard deviation. in this case: and i dont really understand why, the binomial standard deviation is sqrt(npq) with a mean of np (the mean i get) z=x−npnpq−−−√ z=x−npnnpq√n z=xn−pnpqn2−−−√ z=xn−ppqn−−√ zpqn−−−√=xn−p pmidpoint±zpqn−−−√margin of error=±xnend points

  4. moazzam07
    • one year ago
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    sorry dont know how to insert equations

  5. moazzam07
    • one year ago
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    my PC wont do it

  6. moazzam07
    • one year ago
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    Hint: use the central limit theorem, and use a maximum value for the standard deviation for a bernouli random variable.

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