In jrminter/statshelpR: Convenience functions for statistical analysis

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Introduction

This example was developed from an example by Prof. Mine Çetinkaya-Rundel of Duke University presented in this YouTube video.

The problem uses data from a September 2012 Gallup poll of the unadjusted unemployment rate of the population. This survey reported the unemployment rate of the population as 7.9%.

A simple problem

Prof. Çetinkaya-Rundel first poses the following problem:

Among a random sample of three Americans, what is the probability that only one is unemployed?

We need to compute the probabilities for 3 mutually exclusive scenarios

We begin with the following probabilities: $P(U) = 0.079$ and $P(E) = 1 - P(U) = 0.921$ where $P(U)$ is the probability that the person is unemployed and $P(E)$ is the probability that the person is employed. We now compute the probability for three scenarios:

1. $\frac{0.079}{U} \times \frac{0.921}{E} \times \frac{0.921}{E} = 0.067$
2. $\frac{0.921}{E} \times \frac{0.079}{U} \times \frac{0.921}{E} = 0.067$
3. $\frac{0.921}{E} \times \frac{0.921}{E} \times \frac{0.079}{E} = 0.067$

Note that the order does not change the result. We may then sum the probability from the three permutations to get the probability of $P = 3*0.067 = 0.201$ or a 20.1% probability of selecting three Americans where only one is unemployed.

A slightly more complicated problem

Now consider a slightly more complicated problem:

Among a random sample of 8 Americans, what is the probability that only two are unemployed?

This has more permutations...

1. $\frac{0.079}{U} \times \frac{0.079}{U} \times \frac{0.921}{E} \times \frac{0.921}{E} \times \frac{0.921}{E}\times \frac{0.921}{E}\times \frac{0.921}{E}\times \frac{0.921}{E} = 0.079^2 \times 0.921^6 = 0.0038$

2. $\frac{0.079}{U} \times \frac{0.931}{E} \times \frac{0.079}{U} \times \frac{0.921}{E} \times \frac{0.921}{E}\times \frac{0.921}{E}\times \frac{0.921}{E}\times \frac{0.921}{E} = 0.079^2 \times 0.921^6 = 0.0038$

3. $\ldots$

This would take a while by hand. Happily, R has the choose() function that is useful calculating the number of scenarios to organize a given number of successes (k) in a given number of trials (n).

$\dbinom{n}{k} = \frac{n!}{k!(n-k)!}$

Recall that our task was to arrange 2 unemployed people in 8 total people, so $n=8$ and $k=2$. So,

$\dbinom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2! \times 6!} = \frac{8 \times 7 \times 6!}{2 \times 1 \times 6!} = 28$

We can also use the R function choose()

ans <- choose(8, 2)
ans

$P\dbinom{2 \ unemployed}{among \ 8} = number \ of \ scenarios \times probability \ of \ one \ scenario$

or

$P\dbinom{2 \ unemployed}{among \ 8} = 28 \times (0.079^2 \times 0.921^6) = 28 \times 0.0038 = 0.1064$

there is a 10.64% chance that we could find 2 unemployed people among a random sample of 8 people.

It will be interesting to repeat the analysis from data from the Bureau of Labor and Statistics when we reach this point in the class each year.

jrminter/statshelpR documentation built on May 2, 2020, 12:08 a.m.