Intro Statstics Question

Question

1.The Governor of California proposes to give all state employees a flat raise of $250 a month. What would this do to the average monthly salary of state employees?

A) Nothing

B)Increase by $250 times the number of employees

C)Increase by $250

D)Increase by the square root of $250

2. What would a flat raise of $250 a month do to the Standard deviation?

A) Nothing

B)Increase by $250 times the number of employees

C)Increase by $250

D)Increase by the square root of $250

So for 1 I believe it should be B, but what will happen to the SD?? Can someone explain 2 to me please it will be greatly appreciated!

Answer 1

First and foremost, the answer to 1 should be C. I shall explain via an example. Let there be 3 employees, with salaries 100, 200, and 300, the average being 200. Upon a raise, the salaries will be increased by 250, leading to salaries of 350,450,550, the average being 450. This can be extended to n employees, with salaries $p_1, p_2, ... , p_n$. The average will be $\frac{\sum_{i=1}^n p_i}{n}$. Upon a salary increase, the average will be $\frac{\sum_{i=1}^n (p_i+250)}{n}$ = $\frac{\sum_{i=1}^n (p_i) + 250n}{n}$ = $\frac{\sum_{i=1}^n p_i}{n}$ + 250, an increase of 250.

There will be no change to the standard deviation. Population SD is defined as $\sqrt{\frac{\sum_{i=1}^n (p_i - \mu_p)^2}{n}}$, where $\mu_p =\frac{\sum_{i=1}^n (p_i)}{n}$. When the salary increase is applied,every $p_i$ value will increase by 250. Similarly, $\mu_p$ will increase by 250. The increases cancel each other out.

Additional question: What if the increase is 5%?

There will be an increase to the standard deviation. Population SD is defined as $\sqrt{\frac{\sum_{i=1}^n (p_i - \mu_p)^2}{n}}$, where $\mu_p =\frac{\sum_{i=1}^n (p_i)}{n}$. When the salary increase is applied,every $p_i$ value will be multiplied by 1.05. Similarly, $\mu_p$ will be multiplied by 1.05. Final Population SD = $\sqrt{\frac{\sum_{i=1}^n (1.05p_i - 1.05\mu_p)^2}{n}}$ = $\sqrt{\frac{(1.05)^2\sum_{i=1}^n (p_i - \mu_p)^2}{n}}$ = ${1.05\sqrt\frac{\sum_{i=1}^n (p_i - \mu_p)^2}{n}}$ = 1.05* Initial Population SD

Hope this helps.

Answer 2

The answers to your questions are: $1) C$ and $2) A$. To see these are true, set $a = 250 \implies \mu_{\text{new}} = \dfrac{\sum (x+a)}{N}=\dfrac{\sum x}{N}+a=\mu+a$. And $\sigma_{\text{new}} = \sqrt{\dfrac{\sum ((x+a)-(\mu +a))^2}{N}}= \sigma$

Answer 3

The basic properties of the mean (expected value) and variance: $$E(aX+b)=aE(X)+b; Var(aX+b)=a^2Var(X).$$ Proof: $$E(aX+b)=\frac{\sum (aX+b)}{N}=\frac{a\sum X+bN}{N}=a\frac{\sum X}{N}+b=aE(X)+b.$$ $$\begin{align} Var(aX+b)=&\frac{\sum((aX+b)-E(aX+b))^2}{N}=\\ &\frac{\sum((aX+b)^2-2(aX+b)E(aX+b)+(E(aX+b))^2)}{N}= \\ &\frac{\sum ((a^2X^2+2abX+b^2)-(2aX+2b)(aE(X)+b)+(aE(X)+b)^2)}{N}=\\ &\frac{\sum ((\color{red}{a^2X^2}+2abX+b^2)-(\color{red}{2a^2XE(X)}+2abX+2abE(X)+2b^2)+(\color{red}{a^2E^2(X)}+2abE(X)+b^2))}{N}= &\frac{\sum (a^2(X-E(X))^2)}{N}=\\ &\frac{a^2\sum (X-E(X))^2}{N}=\\ &a^2Var(X).\end{align}$$