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Thanks Jason ** Question answered **

Posted By: Jason Lee
Date: Thursday, 29 January 2015, at 8:52 p.m.

In Response To: Thanks Jason ** Question answered ** (John McDonald)

If I am doing this correctly, then p +/- Square root [(1/n)*p*(1-p)] for a 95% CI

You're close. 95% CI requires a z-score of 2, so your CI should be p ± 2 sqrt(...)

Technically, the 2 is more like 1.96, but everybody just uses 2.

JLee

Messages In This Thread

• Statistical C.I. Question
  John McDonald -- Thursday, 29 January 2015, at 1:58 p.m.
  • Statistical C.I. Question
    sita -- Thursday, 29 January 2015, at 2:40 p.m.
    • Statistical C.I. Question
      Daniel Murphy -- Thursday, 29 January 2015, at 3:21 p.m.
  • Statistical C.I. Question
    Jason Lee -- Thursday, 29 January 2015, at 2:57 p.m.
    • Thanks Jason ** Question answered **
      John McDonald -- Thursday, 29 January 2015, at 3:41 p.m.
      • Thanks Jason ** Question answered **
        Jason Lee -- Thursday, 29 January 2015, at 8:52 p.m.
        • Thanks Jason ** Question answered **
          John McDonald -- Friday, 30 January 2015, at 2:05 p.m.