Statistics and Probability: correlation (Karl Pearson, rank) and regression lines; conditional probability

$P(A \cap B) = P(A) \cdot P(B \mid A)$

For dependent events, the multiplication theorem gives $P(A \cap B) = P(A) \cdot P(B \mid A)$ (equivalently $P(B) \cdot P(A \mid B)$). The product $P(A) \cdot P(B)$ holds only when the events are independent.

Take assumed means $63$ for $X$ and $64$ for $Y$: $u=X-63$, $v=Y-64$.

Here $n=7$, $\sum u=0$, $\sum v=-1$, $\sum u^2=28$, $\sum v^2=27$, $\sum uv=19$.

Correlation coefficient

$$r=\frac{\sum uv-\frac{\sum u\sum v}{n}}{\sqrt{\left(\sum u^2-\frac{(\sum u)^2}{n}\right)\left(\sum v^2-\frac{(\sum v)^2}{n}\right)}} =\frac{19}{\sqrt{28\times\frac{188}{7}}}=\frac{19}{\sqrt{752}}\approx0.693$$

Regression coefficient of $X$ on $Y$

$b_{xy}=\frac{\sum uv-\frac{\sum u\sum v}{n}}{\sum v^2-\frac{(\sum v)^2}{n}}=\frac{19}{188/7}=\frac{133}{188}\approx0.707$

So $r\approx0.693$ (moderate positive correlation) and $b_{xy}\approx0.707$.

(a) $0.3$

Since $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.12}{0.4}=0.3$.

Let $X$ = supply and $Y$ = price, $n=6$.

a.

$$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^2-(\sum X)^2][n\sum Y^2-(\sum Y)^2]}} =\frac{6(27410)-348(465)}{\sqrt{(6\cdot20630-348^2)(6\cdot36475-465^2)}}$$

$r=\frac{2640}{\sqrt{2676\times2625}}=\frac{2640}{2650.4}\approx0.996$

There is a very high positive correlation.

b. To estimate supply from price, use the regression of $X$ on $Y$:

$b_{xy}=\frac{n\sum XY-\sum X\sum Y}{n\sum Y^2-(\sum Y)^2}=\frac{2640}{2625}=1.0057$

$\bar X=\frac{348}{6}=58,\qquad\bar Y=\frac{465}{6}=77.5$

$X-58=1.0057(Y-77.5)$

For $Y=150$:

$X=58+1.0057(72.5)=58+72.91\approx130.9$

The estimated supply is about $131$ units when the price is Rs 150.

(a) Increase in x, increase in y

Production and consumption of milk move together — as more milk is produced, more is consumed — so the two variables are positively correlated.

Let $x$ = rainfall (mm) and $y$ = rice production (quintals). Since rainfall is to be estimated from production, use the regression line of $x$ on $y$.

$n=5$, $\sum x=150$, $\sum y=300$, $\sum y^2=19000$, $\sum xy=9500$.

$b_{xy}=\frac{n\sum xy-\sum x\sum y}{n\sum y^2-(\sum y)^2}=\frac{5(9500)-150(300)}{5(19000)-300^2}=\frac{2500}{5000}=0.5$

$\bar x=30,\qquad\bar y=60$

$x-\bar x=b_{xy}(y-\bar y)\;\Rightarrow\;x-30=0.5(y-60)\;\Rightarrow\;x=0.5y$

For $y=90$: $x=0.5\times90=45$.

The rainfall corresponding to $90$ quintals of rice is 45 mm.

Total students $=20+30=50$.

Ways to choose 2 students: $\dbinom{50}{2}=1225$.

Ways to choose 1 boy and 1 girl: $\dbinom{20}{1}\dbinom{30}{1}=600$.

$P=\frac{600}{1225}=\frac{24}{49}$

(c) $3/4$

$P(M\mid S)=\dfrac{P(M\cap S)}{P(S)}=\dfrac{15\%}{20\%}=\dfrac{15}{20}=\dfrac34$.

Let price be $x$ and supply be $y$.

Here, $n=6$, $\sum x=697$, $\sum y=245$, $\sum x^2=81869$, $\sum y^2=10975$, $\sum xy=29040$.

a. Coefficient of correlation

Karl Pearson’s coefficient of correlation is

$r=\frac{n\sum xy-\sum x\sum y}{\sqrt{\left[n\sum x^2-(\sum x)^2\right]\left[n\sum y^2-(\sum y)^2\right]}}$

Substituting the values,

$r=\frac{6(29040)-697(245)}{\sqrt{\left[6(81869)-697^2\right]\left[6(10975)-245^2\right]}}$

$r=\frac{174240-170765}{\sqrt{(491214-485809)(65850-60025)}}$

$r=\frac{3475}{\sqrt{5405\times 5825}}$

$r=\frac{3475}{5611.07}=0.6193$

$\boxed{r=0.6193}$

So, there is a moderate positive correlation between price and supply.

b. Estimate supply when price is Rs 150 per kg

Since supply is to be estimated from price, we use the regression line of $y$ on $x$:

$y=a+bx$

where

$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$

$b=\frac{3475}{5405}=0.6429$

Also,

$a=\bar y-b\bar x$

$\bar x=\frac{697}{6}=116.1667,\qquad \bar y=\frac{245}{6}=40.8333$

$a=40.8333-0.6429(116.1667)$

$a=-33.8529$

Hence the regression equation is

$y=-33.8529+0.6429x$

For price $x=150$,

$y=-33.8529+0.6429(150)$

$y=-33.8529+96.4385$

$y=62.5856$

$\boxed{\text{Estimated supply}\approx 62.59\text{ kg}}$

So, when the price is Rs 150 per kg, the estimated supply is about 62.59 kg.

(c) $\dfrac1{13}$

The values in the question are missing; the intended data are $P(A)=\tfrac14$ and $P(B\mid A)=\tfrac4{13}$. For dependent events the multiplication rule gives

$P(A\cap B)=P(A)\,P(B\mid A)=\frac14\times\frac4{13}=\frac1{13}$

$\bar x=\dfrac{810}{5}=162$, $\bar y=\dfrac{310}{5}=62$. Let $dx=x-162$, $dy=y-62$.

Correlation coefficient

$r=\frac{\sum dx\,dy}{\sqrt{\sum dx^2\sum dy^2}}=\frac{6}{\sqrt{140}}\approx0.507$

Regression coefficient of $Y$ on $X$

$b_{yx}=\frac{\sum dx\,dy}{\sum dx^2}=\frac6{14}=\frac37\approx0.429$

So $r\approx0.507$ (moderate positive correlation) and $b_{yx}\approx0.429$ kg per cm.

Karl Pearson's coefficient of correlation is

$r=\frac{\sum xy}{n\,\sigma_x\sigma_y}$

where $x=X-\bar X$, $y=Y-\bar Y$, $n$ is the number of pairs, and $\sigma_x,\sigma_y$ are the standard deviations of $X$ and $Y$. Equivalently,

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{n\sum X^2-(\sum X)^2}\,\sqrt{n\sum Y^2-(\sum Y)^2}}$

The formula for rank correlation (Spearman's rank correlation coefficient) is:

$R = 1 - \frac{6\sum D^2}{n^3 - n}$

where $D$ is the difference between the ranks of corresponding values and $n$ is the number of pairs of observations.

Given:
$n = 5,\ \sum x = 20,\ \sum y = 30,\ \sum xy = 150,\ \sum x^2 = 90,\ \sum y^2 = 210$

Karl Pearson’s correlation coefficient: $r=\frac{n\sum xy-\sum x\sum y}{\sqrt{\left[n\sum x^2-(\sum x)^2\right]\left[n\sum y^2-(\sum y)^2\right]}}$

Substituting the values: $r=\frac{5(150)-20(30)}{\sqrt{\left[5(90)-20^2\right]\left[5(210)-30^2\right]}}$ $r=\frac{750-600}{\sqrt{(450-400)(1050-900)}}$ $r=\frac{150}{\sqrt{50\times150}}=\frac{150}{50\sqrt3}=\sqrt3\approx1.732$

Since the correlation coefficient must lie between $-1$ and $1$, the given data are inconsistent.
(If $\sum xy$ were $130$, then $r=0.577$.)

For the regression line of $y$ on $x$,

$$b_{yx}=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$$

Given:

$$n=5,\quad \sum x=20,\quad \sum y=30,\quad \sum xy=150,\quad \sum x^2=90$$ $$b_{yx}=\frac{5(150)-(20)(30)}{5(90)-(20)^2}$$ $$b_{yx}=\frac{750-600}{450-400}=\frac{150}{50}=3$$

Also,

$$\bar x=\frac{20}{5}=4,\qquad \bar y=\frac{30}{5}=6$$

The regression line is

$$y-\bar y=b_{yx}(x-\bar x)$$ $$y-6=3(x-4)$$ $$y-6=3x-12$$ $$\boxed{y=3x-6}$$

So, the regression line of $y$ on $x$ is $\boxed{y=3x-6}$.

(a) 1

The number of heads $X\sim B(n=4,\ p=\tfrac12)$. Mean $=np=2$, variance $=npq=4\cdot\tfrac12\cdot\tfrac12=1$. Difference $=2-1=1$.

Given $\bar X=20,\ \bar Y=100,\ \sigma_x=3,\ \sigma_y=12,\ r=0.8$.

(a) Regression coefficients: $b_{yx}=r\frac{\sigma_y}{\sigma_x}=0.8\times\frac{12}{3}=3.2,\qquad b_{xy}=r\frac{\sigma_x}{\sigma_y}=0.8\times\frac{3}{12}=0.2.$

Regression of $Y$ on $X$: $Y-\bar Y=b_{yx}(X-\bar X)$ $Y-100=3.2(X-20)\ \Rightarrow\ Y=3.2X+36.$

Regression of $X$ on $Y$: $X-\bar X=b_{xy}(Y-\bar Y)$ $X-20=0.2(Y-100)\ \Rightarrow\ X=0.2Y.$

(b) To estimate the expenditure $X$ for a given sales figure $Y=200$, use the regression of $X$ on $Y$: $X=0.2\times200=40.$ Advertisement expenditure of Rs 40 lakhs is needed.

(d) less than $P(B/A)$

$P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{2}{5}}{\frac{3}{4}}=\frac{8}{15}\approx 0.533$

Also, $P(B/A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{2}{5}}{\frac{1}{2}}=\frac{4}{5}=0.8$

Since $\frac{8}{15}<\frac{4}{5}$, we get $P(A/B)<P(B/A)$.

Let $X$ = price (Rs) and $Y$ = demand (units).

$$\bar X=\frac{78}{6}=13,\qquad \bar Y=\frac{246}{6}=41$$

Pearson’s coefficient of correlation

$$r=\frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}}$$ $$r=\frac{-6}{\sqrt{24\times 50}} =\frac{-6}{\sqrt{1200}} =\frac{-6}{20\sqrt3} =-\frac{\sqrt3}{10}$$ $$\boxed{r=-0.1732}$$

So there is a weak negative correlation between price and demand.

Regression coefficient of $X$ on $Y$

$$b_{xy}=\frac{\sum xy}{\sum y^2}$$ $$b_{xy}=\frac{-6}{50}$$ $$\boxed{b_{xy}=-0.12}$$

Hence, the regression coefficient of $X$ on $Y$ is $\boxed{-0.12}$.

If both regression coefficients are required, then

$$b_{yx}=\frac{\sum xy}{\sum x^2}=\frac{-6}{24}=-0.25$$ $$\boxed{b_{yx}=-0.25}$$