Binomial Test
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Transcript of Binomial Test
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SM439: Nonparametric Statistics Prof. Jager
Binomial Test
Example: (Rossman and Chance, 2008) The ancient Greeksextensively used the golden ratio in art and literature
they believed that a width-to-length ratio of 1+
52
= 0.618
was aesthetically pleasing. Some have conjectured that
American Indians also used this ratio (Hand et al., 1993).
The following data are width-to-length ratios for a random
sample of 20 beaded rectangles used by the Shoshoni
Indians to decorate their leather goods.
0.693 0.662 0.690 0.606 0.570
0.749 0.672 0.628 0.609 0.844
0.654 0.615 0.668 0.601 0.576
0.670 0.606 0.611 0.553 0.933
Use this data to determine whether there is evidence that
the Shoshoni Indians also employed the golden ratio in
their art.
What are we trying to do here?
Old Way: Parametric methods
New Way: Nonparametric methods BINOMIAL TEST!
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SM439: Nonparametric Statistics Prof. Jager
Old Way: One-sample t-test
Let =
We want to test the following hypotheses:
Our test statistic is:
In our sample:
If H0 is true, then:
So ourp
-value is:
Conclusion:
Assumptions:
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SM439: Nonparametric Statistics Prof. Jager
New Way: Binomial Test
Test uses the median instead of the mean.
For a continuous distribution, the median (.5) satisfies:
P r(X > .5) =P r(X < .5) =.5
For the golden rule example, our hypotheses are then:
Our tests statistic is: B =
IfH0 is true, then the distribution ofB is:
Our p-value is:
Conclusion:
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SM439: Nonparametric Statistics Prof. Jager
Normal Approximation for Binomial Test
When the sample is large enough (n 30), we can use the
normal approximation to the binomial distribution to
perform this test:
Note: Since we have computers, it is just as easy to
calculate the p-value exactly, rather than approximately!
For the approximation, our test statistic is:
For our golden rule example:
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SM439: Nonparametric Statistics Prof. Jager
Using R
The sign.test() function from the BSDA library will
perform a binomial test for us:
> library(BSDA)
> goldenratio = read.table(file.choose(), header=T)
> names(goldenratio)
[1] "ratios"
> attach(goldenratio)
> sign.test(ratios, md = .618, alternative="two.sided")
$rval
One-sample Sign-Test
data: ratios
s = 11, p-value = 0.8238
alternative hypothesis: true median is not equal to 0.618
95 percent confidence interval:
0.6063494 0.6717671
sample estimates:
median of x
0.641
$Confidence.Intervals
Conf.Level L.E.pt U.E.pt
Lower Achieved CI 0.8847 0.6090 0.6700
Interpolated CI 0.9500 0.6063 0.6718
Upper Achieved CI 0.9586 0.6060 0.6720
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