Question

The data below show the wages earned by six workers at three different firms (A, B, and C). The firm's name is shown in column 1 and the workers' wages in column 2. The third column shows the number of times that each firm has been audited for financial fraud while the worker was employed at the firm. The fourth column is a binary variable indicating whether or not the employee works at the accounting department. In the following questions you may check your work using a calculator but must show how you set up the expression you are solving (i.e. show your work). 1. firm wage audits accountant 14 19 15 12 16

Answer 1

New data when the wages and audit is larger 3 times is

firm wage audits A 33 9 A 42 6 B 57 0 B 45 0 C 36 3 C 48 6

Using the formula, $\text{mean of wages}=\mu={\sum_{i=1}^{6} x_i\over 6}=43.5$ and  

Also $\text{Variance in wages}=\sigma^2={\sum_{i=1}^{6}(\mu- x_i)\over 6}= 62.25$

The earlier mean was 14.5, and new mean is 43.5, clearly

3 × 14.5-43.5

and earlier varaince is 6.916667 and new varaince is 62.25, clearly

32 × 6.916667= 62.25

b) The new covaraince is given by

$\text{Cov}=\sum_{i=1}^{6}(x-\mu)(y-\mu_1)$

Where y denotes audit, and $\mu_1$ its mean.

Using the new values we have

new Cov-_16.5

Earlier covariance is -1.833333333, and $3^2\times -1.8333333= -16.5$

c) Let $X=x_1,x_2,\cdots, x_n$ and  $Y=y_1,y_2,\cdots, y_n$ then

$aX=ax_1, ax_2,\cdots, ax_n~~~,~~~aY=ay_1,ay_2,\cdots,a y_n$

Then mean of X is $\mu$ and mean of Y is $\mu_1$ . Also mean of aX is $a\mu$ and mean of aY is $a\mu_1$

$\text{Cov}(aX,aY)=\sum_{i=1}^{n}(ax_i-a\mu)(ay_i-a\mu_1) \\\\ ~~~~~~~~~~ ~~~~~~~~~~~~=\sum_{i=1}^{n}a^2(x_i-\mu)(y_i-\mu_1)\\\\ ~~~~~~~~~~ ~~~~~~~~~~~~=a^2\sum_{i=1}^{n}(x_i-\mu)(y_i-\mu_1)\\\\ ~~~~~~~~~~ ~~~~~~~~~~~~=a^2\text{Cov}(X,Y)$