Analyzing the Ratio of Sums of Squared Deviations \(P\)

The question asks us to determine the value of \(P\) based on the given formula:

\(P=\dfrac{\sum^n_{i=1}(x_i-A)^2}{\sum^n_{i=1}(x_i-\bar x)^2}\)

We are given that \(A \ne \bar x\), where \(\bar x\) is the arithmetic mean of the data points \(x_i\).

Understanding Sum of Squared Deviations

The term \(\sum^n_{i=1}(x_i-c)^2\) represents the sum of the squared differences between each data point \(x_i\) and a constant value \(c\). This sum is a measure of the dispersion or spread of the data around the value \(c\).

Key Property of the Arithmetic Mean

A fundamental property in statistics is that the sum of squared deviations is minimized when the constant \(c\) is equal to the arithmetic mean \(\bar x\). In other words, for any set of data points \(x_1, x_2, \ldots, x_n\), the value of \(\sum^n_{i=1}(x_i-c)^2\) is smallest when \(c = \bar x\).

Mathematically, this means:

\(\sum^n_{i=1}(x_i-c)^2 \ge \sum^n_{i=1}(x_i-\bar x)^2\)

The equality holds only when \(c = \bar x\). If \(c \ne \bar x\), the inequality is strict:

\(\sum^n_{i=1}(x_i-c)^2 > \sum^n_{i=1}(x_i-\bar x)^2 \quad \text{if } c \ne \bar x\)

Evaluating P Given \(A \ne \bar x\)

In the given formula for \(P\), the constant \(A\) is used in the numerator, while the arithmetic mean \(\bar x\) is used in the denominator. The numerator is \(\sum^n_{i=1}(x_i-A)^2\) and the denominator is \(\sum^n_{i=1}(x_i-\bar x)^2\).

Since we are given that \(A \ne \bar x\), according to the property of the mean minimizing the sum of squared deviations, the numerator must be strictly greater than the denominator:

\(\sum^n_{i=1}(x_i-A)^2 > \sum^n_{i=1}(x_i-\bar x)^2\)

Since both sums represent sums of squares, they are non-negative. The sum of squared deviations from the mean is zero only if all data points are identical, but even in that case, the argument still holds: if \(x_i = k\) for all \(i\), then \(\bar x = k\), and \(\sum(x_i-\bar x)^2 = 0\). However, the problem statement implies \(A \ne \bar x\), suggesting a non-trivial case where the denominator is generally positive (unless all \(x_i\) are the same and \(A\) is different, in which case the ratio is undefined or needs careful consideration, but typically data with some variation is assumed). Assuming the denominator is positive (i.e., not all \(x_i\) are identical), we can divide by it.

Dividing the inequality by the denominator \(\sum^n_{i=1}(x_i-\bar x)^2\) (which is positive if there is any variance in the data, or zero if all \(x_i\) are the same):

\(\dfrac{\sum^n_{i=1}(x_i-A)^2}{\sum^n_{i=1}(x_i-\bar x)^2} > 1\)

This shows that \(P > 1\).

Conclusion

Because the sum of squared deviations from any value \(A\) other than the mean \(\bar x\) is always greater than the sum of squared deviations from the mean \(\bar x\), the ratio \(P\) will be greater than 1.

Revision Table: Key Statistical Concepts

Additional Information: Why Mean Minimizes Sum of Squares

We can formally show that \(\sum^n_{i=1}(x_i-c)^2\) is minimized when \(c=\bar x\). Let \(S(c) = \sum^n_{i=1}(x_i-c)^2\). To find the minimum, we can take the derivative with respect to \(c\) and set it to zero:

\(\dfrac{dS}{dc} = \dfrac{d}{dc} \sum^n_{i=1}(x_i-c)^2\)

We can move the derivative inside the sum:

\(\dfrac{dS}{dc} = \sum^n_{i=1} \dfrac{d}{dc} (x_i-c)^2\)

Using the chain rule, \(\dfrac{d}{dc} (x_i-c)^2 = 2(x_i-c) \cdot (-1) = -2(x_i-c)\). So,

\(\dfrac{dS}{dc} = \sum^n_{i=1} -2(x_i-c) = -2 \sum^n_{i=1} (x_i-c)\)

Set the derivative to zero to find critical points:

\(-2 \sum^n_{i=1} (x_i-c) = 0\)

\(\sum^n_{i=1} (x_i-c) = 0\)

Distribute the sum:

\(\sum^n_{i=1} x_i - \sum^n_{i=1} c = 0\)

\(\sum^n_{i=1} x_i - nc = 0\)

\(\sum^n_{i=1} x_i = nc\)

Solving for \(c\):

\(c = \dfrac{\sum^n_{i=1} x_i}{n}\)

This value of \(c\) is the arithmetic mean, \(\bar x\).

To confirm this is a minimum, we can check the second derivative:

\(\dfrac{d^2S}{dc^2} = \dfrac{d}{dc} (-2 \sum^n_{i=1} (x_i-c)) = -2 \sum^n_{i=1} \dfrac{d}{dc} (x_i-c) = -2 \sum^n_{i=1} (-1) = -2(-n) = 2n\)

Since \(n\) is the number of data points, \(n > 0\), so \(2n > 0\). A positive second derivative at the critical point indicates a minimum. This confirms that the sum of squared deviations is minimized at the arithmetic mean.