Sanderson M. Smith
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Hypothesis Testing (proportions)
[Copy of note sent to students in STATISTICS (non AP) class]
OK, I'd really like to see you "graduate" from this course with a solid understanding of hypothesis testing, so I'll provide a few examples during the next few days to give you something to read (and hopefully understand). This might help you with your projects.
What is between the ======= lines is what I would call a good statistical report. (That is, someone who knows statistics would understand what is being said.)
In a state election, Measure B needs 2/3 of the vote to pass. A qualified polling organization finds that 64% of a random sample of 1,256 registered voters will vote for Measure B. Does this suggest that the Measure will fail?
As a statistician, I'll test the null hypothesis
Ho: m = 2/3
(That 2/3 of the registered voters will vote for Measure B. Basically, this includes 2/3 or more.)
against the alternate hypothesis
Ha: m < 2/3
(Basically, that Measure B will fail.)
This is a 1-tail situation.
If Ho is true, then, considering all possible sample proportions for samples of size 1256, the mean of the sample proportions is 2/3 and the standard deviation of the sample proportions is
SQRT[2/3)(1/3)/1256] = 0.0133, or about 1.33%.
In the obtained sample, p(hat) = 64%.
The z statistic calculated from the sample is z = (0.635-2/3)/.0133 = -2.38.
At the 1% level of significance (1-tail), the critical z values are z < -2.33.
The statistic z = -2.38 is in the critical region. The null hypothesis is rejected at the 1% level. In other words, there is strong statistical evidence that Measure B will not pass.
OK, now we could take other approaches to this problem.
Here are just a few related statements that "jive" with the report above:
If the null hypothesis Ho is true (that is, if the true population proportion is 2/3), then the probability that a sample of size 1256 will have p(hat) = 0.64 is normalcdf(-1E99,0.635,2/3,0.0133) = 0.008634, or about 0.86% (which less than 1%). If Ho is true, it is highly unlikely that one would obtain a sample with p(hat) = 48%.
We could also use the z-score calculated above. Note that normalcdf(-1E99,-2.38,0,1) = 0.008656, which pretty much "jives" with the previously-calculated probability.
In a nutshell, our analysis leads us to believe that Measure B will be defeated, and there is approximately a 1% chance that this conclusion is incorrect.
OK, now suppose that the polling was done about two weeks before the election. The results obtained are based on what would probably happen at the time of the polling. If you want Measure B to pass, the statistical analysis suggests that it would be a good idea to get out and make some noise about what you see to be good points in Measure B. Perhaps you can get change the minds of some voters who don't support it at the present time.
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