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7 June, 13:02

A university newspaper is conducting a survey to determine what fraction of students support a $200 per year increase in fees to pay for a new football stadium. How big of a sample is required to ensure the margin of error is smaller than 0.04 using 95% confidence level?

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  1. 7 June, 16:18
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    The formula to calculate sample size is Z * √ ((P (1-P) / n)

    Z is the Z-score for the confidence level, in this problem the confidence level is given as 95%, so Z = 1.96

    P is the population, which is unknown, but we want it to be the largest value possible, so in the equation above, P is 0.5.

    n is the sample size.

    Now we have: 1.96 * √ ((0.5 * 0.5) / n)

    They want this sample size to be less than 0.04

    So now we have:

    1.96 * √ ((0.5 * 0.5) / n) < 0.04

    Multiply both sides by 1/1.96:

    √ ((0.5 * 0.5) / n) < 0.04 * 1/1.96

    Simplify:

    0.5 / √n < 4/196

    Take the reciprocal of both sides:

    √n / 0.5 < 196/4

    Simplify:

    √n/0.5 < 49

    Multiply both sides by 0.5:

    √n = 49*0.5

    √n = 24.5

    to solve for n, raise both sides to the power of 2:

    n = 24.5^2

    n = 600.25

    The sample needs to be about 600 people.
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