Question

Gender discrimination suit. The Journal of Business & Economic Statistics (July 2000) presented a case in which a charge of gender discrimination was filed against the U.S. Postal Service. At the time, there were 302 U.S. Postal Service employees (229 men and 73 women) who applied for promotion. Of the 72 employees who were awarded promotion, 5 were female. Make an inference about whether or not females at the U.S. Postal Service were promoted fairly.

Short answer

Females were not promoted fairly.

Step-by-step solution

Given information

From the given information, 302 U.S. Postal Service employees (229 men and 73 women) applied for promotion. There were 302 U.S. Postal Service employees (229 men and 73 women) who applied for promotion.

Finding if observing five females who were promoted was fair

x follows a hypergeometric distribution with$N=302, n=72 and r=73$

So, the expected value (mean) of x is

$\begin{array}{rcl}E\left(x\right)& =& \mu \\ & =& \frac{nr}{N}\\ & =& \frac{72\times 73}{302}\\ & =& 17.403974\\ & \approx & 17\\ & & \end{array}$

The variance of x is

$\begin{array}{rcl}V\left(x\right)& =& {\sigma }^2\\ & =& \frac{nr\left(N-r\right)\left(N-n\right)}{N^2\left(N-1\right)}\\ & =& \frac{72\times 73\times \left(302-73\right)\times \left(302-72\right)}{{302}^2\times \left(302-1\right)}\\ & =& \frac{72\times 73\times 229\times 230}{91204\times 301}\\ & =& 10.0841\\ & & \end{array}$

The variance of x is 10.0841.

The standard deviation of x is

$\begin{array}{rcl}\sigma & =& \sqrt{10.0841}\\ & =& 3.1755\\ & & \end{array}$

The standard deviation of x is 3.1755.

The interval is,$\mu \pm 3\sigma$

So,

$\begin{array}{rcl}\mu \pm 3\sigma & =& 17.403974\pm \left(3\times 3.1755\right)\\ & =& 17.403974\pm 9.5265\\ & =& \left(17.403974-9.5265,17.403974+9.5265\right)\\ & =& \left(7.8775,26.9305\right)\\ & & \end{array}$

There is no discrimination in promoting females.

Therefore,observed that only five females werepromoted.

So, females were not promoted fairly.