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Thread: Confidence interval

  1. #1
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    Confidence interval

    The health ministry of a country would like to acquire an estimate of proportion p of the local adults that are diabetic. A survey conducted shows that 82 our of 120 local adults are diabetic. Determine the number of local adults that needs to be taken at random so that the difference in the estimated value of p and actual value of p differs by not more than 0.05 with a confidence level of 99%.
    Here is how I do it:
    z_0.005√((41/60*19/60)/n) < 0.05 and solve it so I get n>574.36, is that correct?
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  2. #2
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    Re: Confidence interval

    Yes that is correct
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