Step 1: State the formula for Karl Pearson's coefficient of correlation.

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

Step 2: Substitute the given values into the formula.

$$r=\frac{10\left(220\right)-\left(40\right)\left(50\right)}{\sqrt{10\left(200\right)-(40{)}^2}\sqrt{10\left(262\right)-(50{)}^2}}$$

Step 3: Simplify the numerator and the terms inside the square roots.

Numerator: $2200-2000=200$
Denominator Term 1: $\sqrt{2000-1600}=\sqrt{400}=20$
Denominator Term 2: $\sqrt{2620-2500}=\sqrt{120}\approx 10.954$

Step 4: Calculate the final value.

$$r=\frac{200}{20\times 10.954}=\frac{10}{10.954}\approx 0.9128$$

Note: The answer key states 0.89, which would be the case if $\sum y^2=262.5$ or similar. With exactly 262, the mathematically precise answer is 0.91.

(The data table for this question is missing from the provided textbook source text. If a generic table is provided, the standard Karl Pearson formula is applied similarly to the other problems.)

Step 1: Create a calculation table.

Let $n=7$. We first find the means:
$\overline{X}=\frac{67+66+65+68+69+71+70}{7}=\frac{476}{7}=68$
$\overline{Y}=\frac{66+68+67+69+72+69+72}{7}=\frac{483}{7}=69$

Let $x=X-\overline{X}$ and $y=Y-\overline{Y}$.

Step 2: Use the deviation formula.

$$r=\frac{\sum xy}{\sqrt{\sum x^2}\sqrt{\sum y^2}}$$

$$r=\frac{20}{\sqrt{28}\sqrt{32}}=\frac{20}{\sqrt{896}}\approx \frac{20}{29.933}\approx 0.668$$

Step 1: Find the mean of X and Y.
$\sum X=300\Rightarrow \overline{X}=300/10=30$
$\sum Y=250\Rightarrow \overline{Y}=250/10=25$

Step 2: Create a calculation table. Let $x=X-30$ and $y=Y-25$.

Step 3: Substitute into the correlation formula.

$$r=\frac{\sum xy}{\sqrt{\sum x^2}\sqrt{\sum y^2}}$$

$$r=\frac{123}{\sqrt{138}\sqrt{164}}=\frac{123}{\sqrt{22632}}\approx \frac{123}{150.439}\approx 0.8176$$

Step 1: Adjust the summations by subtracting incorrect values and adding correct ones.

• Correct $\sum x$: $125-8-6+6+8=125$
• Correct $\sum y$: $100-6-14+8+12=100$
• Correct $\sum x^2$: $650-(8^2+6^2)+(6^2+8^2)=650$
• Correct $\sum y^2$: $436-(6^2+{14}^2)+(8^2+{12}^2)=436-(36+196)+(64+144)=436-232+208=412$
• Correct $\sum xy$: $520-(8\times 6+6\times 14)+(6\times 8+8\times 12)=520-(48+84)+(48+96)=520-132+144=532$

Step 2: Use the formula for $r$.

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

$$r=\frac{25\left(532\right)-\left(125\right)\left(100\right)}{\sqrt{25\left(650\right)-(125{)}^2}\sqrt{25\left(412\right)-(100{)}^2}}$$

$$r=\frac{13300-12500}{\sqrt{16250-15625}\sqrt{10300-10000}}=\frac{800}{\sqrt{625}\sqrt{300}}$$

$$r=\frac{800}{25\times 17.32}=\frac{32}{17.32}\approx 1.847$$

Note: A correlation coefficient computationally cannot exceed 1. This suggests that the initial problem data provided in the textbook (specifically the initial Σy² = 436 value) was inherently flawed or a typo. However, proceeding with the exact mathematical substitution based on the data provided yields the derived answer.

Step 1: Simplify calculations using Step Deviation Method.

Let Assumed Mean for $X$ be $A=80$ and scale $h=20$. So, $u=\frac{X-80}{20}$.
Let Assumed Mean for $Y$ be $B=600$ and scale $k=100$. So, $v=\frac{Y-600}{100}$.

Step 2: Apply the coefficient formula.

$$r=\frac{n\sum uv-(\sum u)(\sum v)}{\sqrt{n\sum u^2-(\sum u{)}^2}\sqrt{n\sum v^2-(\sum v{)}^2}}$$

$$r=\frac{7\left(30\right)-\left(0\right)\left(1\right)}{\sqrt{7\left(28\right)-0}\sqrt{7\left(35\right)-1^2}}=\frac{210}{\sqrt{196}\sqrt{245-1}}$$

$$r=\frac{210}{14\times \sqrt{244}}=\frac{210}{14\times 15.620}=\frac{210}{218.68}\approx 0.96$$

Step 1: State the formula for Spearman's rank correlation.

$$r=1-\frac{6\sum d^2}{n(n^2-1)}$$

Step 2: Substitute the known values.
Given $r=0.8$ and $\sum d^2=33$.

$$0.8=1-\frac{6\times 33}{n(n^2-1)}$$

Step 3: Solve for $n$.

$$\frac{198}{n(n^2-1)}=1-0.8=0.2$$

$$n(n^2-1)=\frac{198}{0.2}=990$$

Expanding the left side as consecutive integers: $(n-1)\times n\times (n+1)=990$

We look for three consecutive integers whose product is 990.
$9\times 10\times 11=990$. Thus, $n=10$.

Step 1: Assign ranks to the data.
We assign ranks from 1 to 10 in descending order (highest score gets rank 1).

Step 2: Calculate Spearman's Rank Correlation.

$$r=1-\frac{6\sum d^2}{n(n^2-1)}$$

Here $n=10$.

$$r=1-\frac{6\times 30}{10(100-1)}=1-\frac{180}{990}$$

$$r=1-\frac{2}{11}=\frac{9}{11}\approx 0.818$$

Step 1: Calculate the squared differences for each pair of judges.

Step 2: Calculate Spearman's rank correlation for each pair.
Using $n=10$, so the denominator $n(n^2-1)=990$.

Between Judge 1 and Judge 2:
$r_{12}=1-\frac{6\times 48}{990}=1-\frac{288}{990}=1-0.291=0.709$

Between Judge 2 and Judge 3:
$r_{23}=1-\frac{6\times 26}{990}=1-\frac{156}{990}=1-0.158=0.842$

Between Judge 1 and Judge 3:
$r_{13}=1-\frac{6\times 88}{990}=1-\frac{528}{990}=1-0.533=0.467$

Step 1: Assign ranks to the values.
Note that there are repeating values, so we assign average ranks to tied items.

• In X: 13 occurs twice (ranks 5 and 6) → Average rank = 5.5. Number 6 occurs twice (ranks 8 and 9) → Average rank = 8.5.
• In Y: 16 occurs three times (ranks 7, 8, 9) → Average rank = 8.

Step 2: Apply the correction factor (CF) for tied ranks.
$$CF=\sum \frac{m(m^2-1)}{12}$$ Ties in X: $m_1=2$ (for 13), $m_2=2$ (for 6). Ties in Y: $m_3=3$ (for 16).
$$CF=\frac{2(4-1)}{12}+\frac{2(4-1)}{12}+\frac{3(9-1)}{12}=\frac{6}{12}+\frac{6}{12}+\frac{24}{12}=0.5+0.5+2=3$$

Step 3: Calculate the Spearman's rank correlation.
$$r=1-\frac{6(\sum d^2+CF)}{n(n^2-1)}$$ $$r=1-\frac{6(41+3)}{10(100-1)}=1-\frac{6\times 44}{990}=1-\frac{264}{990}=1-0.2667$$

Step 1: Assign ranks to the values.
We assign ranks from 1 to 8 in descending order.

Step 2: Calculate Spearman's Rank Correlation.

$$r=1-\frac{6\sum d^2}{n(n^2-1)}$$

Here $n=8$.

$$r=1-\frac{6\times 6}{8(64-1)}=1-\frac{36}{504}$$

$$r=1-\frac{1}{14}=\frac{13}{14}\approx 0.928$$

Note: A highly strong positive rank correlation exists between ad spending and sales.

Step 1: Check the mathematical relationship.
We can observe a direct linear relationship: $Y=\frac{2}{3}X-2$. Since the relationship is perfectly linear and the slope is positive, the coefficient of correlation should be exactly 1. We will verify this computationally.

Step 2: Find means and deviations.
$\overline{X}=\frac{147}{7}=21$
$\overline{Y}=\frac{84}{7}=12$

Step 3: Calculate the correlation.

$$r=\frac{\sum xy}{\sqrt{\sum x^2}\sqrt{\sum y^2}}$$

$$r=\frac{168}{\sqrt{252}\sqrt{112}}=\frac{168}{\sqrt{28224}}=\frac{168}{168}=1$$

Step 1: Calculate the mean values.
$\sum X=705\Rightarrow \overline{X}=70.5$
$\sum Y=670\Rightarrow \overline{Y}=67$

Step 2: Create a calculation table. We can use assumed mean $A=70$ and $B=67$.
Let $x=X-70$ and $y=Y-67$.

Step 3: Calculate using Assumed Mean formula.

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

$$r=\frac{10\left(122\right)-\left(5\right)\left(0\right)}{\sqrt{10\left(133\right)-5^2}\sqrt{10\left(130\right)-0^2}}$$

$$r=\frac{1220}{\sqrt{1330-25}\sqrt{1300}}=\frac{1220}{\sqrt{1305}\sqrt{1300}}$$

$$r=\frac{1220}{\sqrt{1696500}}=\frac{1220}{1302.49}\approx 0.9366$$

Step 1: Assign ranks to the values.
We assign ranks from 1 to 8 in descending order.

Step 2: Calculate Spearman's Rank Correlation.

$$r=1-\frac{6\sum d^2}{n(n^2-1)}$$

$$r=1-\frac{6\times 6}{8(64-1)}=1-\frac{36}{504}=1-\frac{1}{14}\approx 0.928$$

Step 1: Assign ranks to the values.
Handle ties by averaging the ranks that the identical numbers would have occupied.

• In X: 73 appears twice (ranks 4, 5) → Average rank = 4.5. The number 70 appears twice (ranks 6, 7) → Average rank = 6.5.
• In Y: 30 appears three times (ranks 1, 2, 3) → Average rank = 2.

Step 2: Apply the correction factor (CF) for tied ranks.
$$CF=\sum \frac{m(m^2-1)}{12}$$ Ties in X: $m_1=2$ (for 73), $m_2=2$ (for 70). Ties in Y: $m_3=3$ (for 30).
$$CF=\frac{2(4-1)}{12}+\frac{2(4-1)}{12}+\frac{3(9-1)}{12}=\frac{6}{12}+\frac{6}{12}+\frac{24}{12}=0.5+0.5+2=3$$

Step 3: Calculate the Spearman's rank correlation.
$$r=1-\frac{6(\sum d^2+CF)}{n(n^2-1)}$$ $$r=1-\frac{6(56+3)}{8(64-1)}=1-\frac{6\times 59}{504}=1-\frac{354}{504}=1-0.702$$

(1) Top Rankings:
Top rating in Test Matches = 128 (Australia).
Top rating in T-20 Matches = 272 (India).

(2) Calculate Spearman's Rank Correlation:
Assign ranks in descending order for both series.

$$r=1-\frac{6\sum d^2}{n(n^2-1)}$$

$$r=1-\frac{6\times 26}{9(81-1)}=1-\frac{156}{720}=1-0.2167=0.7833$$

(3) Interpretation:
A high positive correlation ($r\approx 0.78$) signifies that countries performing well in Test Cricket also tend to perform well in T-20.

(1) Tabular Form:

(2) Coefficient of Correlation Calculation:

For Karl Pearson's Method, we analyze the numerical deviations. It provides a precise scalar metric reflecting the exact linear dependencies between actual continuous data points.

For Spearman's Rank Method, we analyze the ranked order of the observations. This evaluates monotonic relationships without factoring in the magnitude of distances between actual data points.

Yes, the two coefficients are marginally different because Karl Pearson's evaluates linear magnitude, whereas Spearman's evaluates purely monotonic ordinality. Karl Pearson's method is more accurate because it utilizes the exact values rather than losing informational detail through ranking abstraction.

(3) Interpretation of Variance:

Yes, the comparison gives deep insight into significance. The squared correlation coefficient ($r^2$), known as the Coefficient of Determination, precisely quantifies the explained variation.
If $r^2$ approaches $1$, the explained variation vastly outweighs the unexplained variation, concluding the correlation is very strong and highly significant.