Derive Sample Size Equation used for Proportions

[@royals2013]

What about for two proportions? and how do you get on top of that (z_alpha/2 + z_beta)^2

[@sudarshanachakra3.022]

exactly ....i was thinking the same .... if you have any link please send

[@FreespiritFrancon]

solved my problem! Many thanks!

[@firefrostpeacemaker]

Um... how did the z-score get squared? In the margin of error equation, the z-score was in a square root. So, when you squared both sides, in the beginning of your derivation, how did you miraculously come to gain z-score squared?

[@firefrostpeacemaker]

Never mind. The reason is because the z-score wasn't in a square root to begin with, in the confidence interval. It was a minor mistake.

[@shaoboliu]

@@firefrostpeacemaker yes, you are right.

[@apersnn9508]

If you can answer this I would be truly grateful, if my paper questioning a population knowledge of sth and I need to use the proportion formula to get my Sample size , if the paper I am using the "p" from displaying knowledge as a score ,how can I calculate the proportion to compensate with in the formula mentioned ?

[@mohammedomar4012]

Please, explain me. How i use the method of semi average to find trend value.???

[@statisticsfun]

Mohammed Omar Thanks for your question. I am not sure what you mean by a semi average. Can you give me an example or two of what you are trying to calculate?

[@mohammedomar4012]

Thanks for your attention.. I have an example: Use the semi average method to find trend value for the following data per gross incom of Pakistan for the year. Year 2001 to 2010. Income 318, 326, 337, 340, 359, 365, 372, 381, 402, 410.

[@mohammedomar4012]

It is much better. Please if you explain me through video.. Thanks :)

[@darraghkelly1079]

There is a mistake in the formula, please remove this video. The the z-score should not be in the square root.

[@mohammedomar4012]

I'll be waiting for ur answer...$ir:)