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BINOMIAL VS. NORMAL..A NOTE TO STUDENTS

Hello AP STATers....

This is not required reading, but it does relate to a question asked by Emily today in class. If you are really serious about understanding the amazing academic discipline of statistics, I urge you to see if what is below makes sense. (If it does, you are in good shape.)

Suppose we have a population that is 60% YES. (The 60% is a parameter.)

We take an SRS of size 200, and find that 57% of the sample are YES.

HOW LIKELY IS IT THAT WE WOULD GET A SAMPLE LIKE THIS?

This is definitely binomial. Let x be the count of YES in a sample of size 200. OK, the sample we took had a count of 200(.57) = 114. The probability of getting 114 or fewer is binomcdf(200,.6,114) = .2131, or about 21%. In other words, such a sample is not unlikely. Note that binomcdf does not require input of a standard deviation. This approach to answering the question would be perfectly acceptable on the AP.

Now, 200 is not a large sample size. Our calculator could handle the computation required in this case. However, as Anne pointed out, the calculator cannot compute binomial probabilities if the sample is too large.

We could do this using a normal approximation to the binomial. Note that Np = 200(.6) and N(1-p) = 200(.4) are definitely greater than 10, so a normal approximation is reasonable here. Working with proportions, we have

mean[p(hat)] = .6

st. dev.[p(hat)] = ÷[.6)(.4)/200] = .03464

Remember that the normal distribution is a continuous curve, and the number .57 "extends" from .565 to .575 on a continuous curve. Probability [p(hat) is at most .57] = normalcdf(-1E99,.575,.6,.03464) = 0..2352. As is true above, we would conclude that such a sample is not unlikely. This approach to answering the question would be perfectly acceptable on the AP.

In this example, both the binomial and normal approaches represent an attempt to fit a mathematical model to a real-life situation. The models are not giving us exact theoretical probabilities for the real-life situation, but rather just giving us some indication as to what we can expect. As a sample size increases, the binomial distribution approaches the normal distribution, and we can use it in place of the binomial. (Doing this has many advantages.) Note that in this problem both models suggest to us that the chance of getting the sample obtained is a bit more than 20%. In other words, such a sample is not unexpected. (Note that saying it is not unexpected is not equivalent to saying that it is expected.)

Note that you can get answers of approximately 21% or 24%, depending on how you approach the problem. Either answer is acceptable and reasonable, as long as you make it clear how it was obtained.

For those who have read this far, I hope this provides you with some additional statistical insight. I would really like to see all of you graduate with a full appreciation of the power of statistical thinking.

MATH POWER TO ALL

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