**Freakonometrics - Tag - R-english**, and kindly contributed to R-bloggers)

No need to do politics. Just take a statistical course. And I do not

talk about misinterpretation of statistics, but I talk about the

mathematical foundations of statistical tests.

Consider the following

parametric test, with a one-dimensional parameter: versus , for some (fixed) . A standard way of doing such a test is to consider an rejection region . The test works as follows: consider a sample ,

- if , then we accept
- if , the we reject

For instance, consider the case of a Bernoulli sample, with probability . The standard idea is to define

The rejection region is then based on statistic ,

- if , then we accept
- if , the we reject

where threshold is taken so that the probability to make a first type error is (say 5%) using the Gaussian approximation for z. Here

Thus, the acceptation region is then the green area below, while the rejection region is the red one, for .

Consider now the exact opposite test (with the same ), versus . Here, we use the same statistics, and the test is

- if , then we accept
- if , the we reject

where now

Thus, now, the acceptation region is then the green area below, while the

rejection region is the red one.

So if we summarize what we just said,

- in the region on the left below, both test agree that
- in the region on the right below, both test agree that
- and in the region in blue, in the middle, the two tests disagree (one claims that , and the other one that )

Here is the evolution of the region as a function of (the size of the sample) when the sample frequency is 20%. With a small sample size, we can hardly say anything.

n=seq(1,100)

p=0.2

x1=p+qnorm(.95)*sqrt(p*(1-p)/n)

x2=p+qnorm(.05)*sqrt(p*(1-p)/n)

plot(n,x1,type="l",ylim=c(0,1))

polygon(c(n,rev(n)),c(x1,rev(x2)),col="light blue",border=NA)

lines(n,x1,lwd=2,col="red")

lines(n,x2,lwd=2,col="red")

y1=qbinom(.95,size=n,prob=p)/n

y2=qbinom(.05,size=n,prob=p)/n

polygon(c(n,rev(n)),c(y1,rev(y2)),col="blue",border=NA)

lines(n,y1,lwd=2,col="red")

lines(n,y2,lwd=2,col="red")

and we get

This is what we can observe if we use R statistical procedures, either the asymptotic one,

> prop.test(2,10,.5,alternative="less")

1-sample proportions test with continuity correction

data: 2 out of 10, null probability 0.5

X-squared = 2.5, df = 1, p-value = 0.05692

alternative hypothesis: true p is less than 0.5

95 percent confidence interval:

0.0000000 0.5100219

sample estimates:

p

0.2

> prop.test(2,10,.5,alternative="greater")

1-sample proportions test with continuity correction

data: 2 out of 10, null probability 0.5

X-squared = 2.5, df = 1, p-value = 0.943

alternative hypothesis: true p is greater than 0.5

95 percent confidence interval:

0.04368507 1.00000000

sample estimates:

p

0.2

or a more accurate one

> binom.test(2,10,.5,alternative="less")

Exact binomial test

data: 2 and 10

number of successes = 2, number of trials = 10, p-value = 0.05469

alternative hypothesis: true probability of success is less than 0.5

95 percent confidence interval:

0.0000000 0.5069013

sample estimates:

probability of success

0.2

> binom.test(2,10,.5,alternative="greater")

Exact binomial test

data: 2 and 10

number of successes = 2, number of trials = 10, p-value = 0.9893

alternative hypothesis: true probability of success is greater than 0.5

95 percent confidence interval:

0.03677144 1.00000000

sample estimates:

probability of success

0.2

Here, when the sample frequency is 20% and is equal to 10, we accept at the same time that theta is higher than 50% and lower than 50%.

And

obviously it is not only a theoretical problem: it has obviously some

strong implications. This morning, a good friend mentioned a post

published some months ago, online here, about discrimination, and the lack of women with academic positions in mathematics, in France. As claimed by the author of the post“A Paris VI, meilleure université française selon son président, sur 11

postes de maitres de conférences, 5 filles classées premières. Il y a

donc des filles excellentes ? A Toulouse, sur 4 postes, 2 filles

premières. Parité parfaite. Mais à côté de cela, Bordeaux, 4 postes, 0

fille première. Littoral, 3 postes, 0 fille, Nice, 5 postes, 0 fille,

Rennes, 7 postes, 0 fille…“.

Consider the latter one: in

Rennes, out of 7 people hired last year, no woman. So in some sense, it

looks obvious that there is some kind of discrimination ! Zero out of

seven ! Well, if we consider the fact that around 30% of PhD thesis in mathematics

were defended by women those years, we can also try to see is there if

no “*positive discrimination*“, i.e. test where theta is the probability to hire a woman (just to be a little bit provocative).

> prop.test(0,7,.3,alternative="less")

1-sample proportions test with continuity correction

data: 0 out of 7, null probability 0.3

X-squared = 1.7415, df = 1, p-value = 0.09347

alternative hypothesis: true p is less than 0.3

95 percent confidence interval:

0.0000000 0.3719021

sample estimates:

p

0

Warning message:

In prop.test(0, 7, 0.3, alternative = "less") :

Chi-squared approximation may be incorrect

> binom.test(0,7,.3,alternative="less")

Exact binomial test

data: 0 and 7

number of successes = 0, number of trials = 7, p-value = 0.08235

alternative hypothesis: true probability of success is less than 0.3

95 percent confidence interval:

0.0000000 0.3481637

sample estimates:

probability of success

0

With no woman hired that year, we can still pretend that there was some kind of “*positive discrimination*“. An note that we do accept – with more confidence – the assumption of “*positive discrimination*” if we look at all universities together,

> prop.test(5+2,11+4+4+3+5+7,.3,alternative="less")

1-sample proportions test with continuity correction

data: 5 + 2 out of 11 + 4 + 4 + 3 + 5 + 7, null probability 0.3

X-squared = 1.021, df = 1, p-value = 0.1561

alternative hypothesis: true p is less than 0.3

95 percent confidence interval:

0.0000000 0.3556254

sample estimates:

p

0.2058824

> binom.test(5+2,11+4+4+3+5+7,.3,alternative="less")

Exact binomial test

data: 5 + 2 and 11 + 4 + 4 + 3 + 5 + 7

number of successes = 7, number of trials = 34, p-value = 0.1558

alternative hypothesis: true probability of success is less than 0.3

95 percent confidence interval:

0.0000000 0.3521612

sample estimates:

probability of success

0.2058824

So obviously, with small sample, almost anything can be claimed !

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