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Hypothesis Testing

Posted May 22, 11:40 pm

The new Leaving Cert Syllabus (Project Maths) has Hypothesis testing on it as a prominent part of the course, well, at least enough so that two of my students I help in grinds this week said they find it difficult. And it is difficult concept.

I was noticing that some books and notes use the Margin of Error formula $ME = \frac{1}{\sqrt{n}}$ and some places use the Standard Error $SE = \sqrt{\frac{\left(p\right)\left(1-p\right)}{n}}$ when calculating the Confidence Interval (so that one could test the Hypothesis) in the case of Binomial Trials. Is there a difference? Well there is.

I much prefer using the SE formula although it is acceptable that some places at some times use ME. The reason I prefer teaching to use SE is that ME is just a special case of SE. If you are wondering when you can use ME instead of SE (since ME is a simpler formula), then keep reading.

In the SE case with a 5% test and $p$ as approximately 0.5 the Confidence Interval of $\pm 1.96 SE$ would equal $\pm 1.96 \sqrt{\frac{\left(0.5\right)\left(1-0.5\right)}{\sqrt{n}}} = \frac{0.98}{\sqrt{n}} \approx \frac{1}{\sqrt{n}} = ME$. Basically whenever you have $p$ near a half and are doing a 5% Hypothesis Test then the approximation is acceptable (and widely used).

Although if you are doing something with pen and paper I do feel SE is what should be used, I often think of ME as a quick and easy way to do a pretty good Hypothesis Test in my head. For example, if I see a claim on a poster that one Cola is prefer over enough 75% of the time (and in the footnote this said it is based off of a survey of 50 people) then I can judge that claim rather quickly. In my head I know the square-root of 50 is about 7 and one over 7 is about 0.14 or in other words 14%. 75% $\pm$ 14% is still above 50% so I feel claiming that this Cola is more often preferred is reasonable.



Author
Stephen Easley-Walsh


 

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