Statistics — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for statistics.

UT-1: Identifying Skew from Quartile Positions

Question:

For a dataset, the quartiles are $Q_1 = 42$, $Q_2 = 55$And $Q_3 = 70$.

(a) Determine whether the data is positively skewed, negatively skewed, or symmetric.

(b) A student argues: “Since $Q_2 - Q_1 = 13$ and $Q_3 - Q_2 = 15$The data is positively skewed because $Q_3 - Q_2 \gt Q_2 - Q_1$.” Is this reasoning correct?

(c) If the interquartile range is $IQR = 28$State the outlier boundaries using the $1.5 \times IQR$ rule.

[Difficulty: hard. Tests interpretation of quartile positions to identify skewness and outlier detection.]

Solution:

(a) The distances from the median are:

• $Q_2 - Q_1 = 55 - 42 = 13$
• $Q_3 - Q_2 = 70 - 55 = 15$

Since $Q_3 - Q_2 \gt Q_2 - Q_1$The right tail is longer than the left tail, indicating positive skew.

(b) The student”s reasoning is correct in principle: positive skew means the right tail is longer. However, the student should note that this is a heuristic — formal skewness is measured by the moment coefficient $\frac{1}{n}\sum\left(\frac{x_i - \bar{x}}{s}\right)^3$Not just quartile differences. The quartile-based test is a quick check, not definitive proof.

(c) Lower fence: $Q_1 - 1.5 \times IQR = 42 - 42 = 0$. Upper fence: $Q_3 + 1.5 \times IQR = 70 + 42 = 112$.

Outliers are values below $0$ or above $112$.

UT-2: PMCC with Coded Data

Question:

A dataset has the following coded values. The coding is $y = \frac{x - 50}{10}$:

$\sum y = 45, \quad \sum y^2 = 285, \quad n = 9$

(a) Find $\bar{x}$ and $s_x$ (the standard deviation of $x$).

(b) A student computes $s_y = \sqrt{\frac{285}{9} - 25} = \sqrt{31.67 - 25} = \sqrt{6.67}$ and concludes $s_x = s_y$. Explain why this is wrong.

[Difficulty: hard. Tests coded data transformations and the effect on mean and standard deviation.]

Solution:

(a) $\bar{y} = \frac{45}{9} = 5$.

Since $y = \frac{x - 50}{10}$We have $x = 10y + 50$:

$\bar{x} = 10\bar{y} + 50 = 10(5) + 50 = 100$

For the standard deviation: $s_x = 10s_y$.

$s_y = \sqrt{\frac{\sum y^2}{n} - \bar{y}^2} = \sqrt{\frac{285}{9} - 25} = \sqrt{31.67 - 25} = \sqrt{6.67} \approx 2.58$

$s_x = 10 \times 2.58 = 25.8$

(b) The student’s error is concluding $s_x = s_y$. The coding $y = \frac{x-50}{10}$ scales by a factor of $\frac{1}{10}$ and shifts by $50$. Scaling by $c$ multiplies the standard deviation by $|c|$So $s_x = 10s_y$Not $s_y$. The student forgot to account for the scaling factor. Additionally, the student used $\frac{285}{9} \approx 31.67$ and then subtracted $25$ (where $25 = 5^2$), which is correct for computing $s_y$But then incorrectly applied the result to $s_x$.

Integration Tests

Tests synthesis of statistics with other topics.

IT-1: Least Squares Regression and Summation (with Algebra)

Question:

Given five data points $(x_i, y_i)$ with \sum x_i = 15$$\sum y_i = 20$$\sum x_i^2 = 55$$\sum x_iy_i = 68And $\sum y_i^2 = 90$:

(a) Find the equation of the least squares regression line of $y$ on $x$ in the form $y = a + bx$.

(b) Find PMCC (Pearson product-moment correlation coefficient).

(c) Predict the value of $y$ when $x = 5$.

[Difficulty: hard. Combines regression computation with correlation analysis.]

Solution:

(a)

$b = \frac{n\sum x_iy_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2} = \frac{5(68) - 15(20)}{5(55) - 225} = \frac{340 - 300}{275 - 225} = \frac{40}{50} = 0.8$

$a = \bar{y} - b\bar{x} = \frac{20}{5} - 0.8 \times \frac{15}{5} = 4 - 2.4 = 1.6$

Regression line: $y = 1.6 + 0.8x$.

(b)

$r = \frac{n\sum x_iy_i - \sum x_i\sum y_i}{\sqrt{\big[n\sum x_i^2 - (\sum x_i)^2\big]\big[n\sum y_i^2 - (\sum y_i)^2\big]}}$

$= \frac{340 - 300}{\sqrt{(275 - 225)(450 - 400)}} = \frac{40}{\sqrt{50 \times 50}} = \frac{40}{50} = 0.8$

(c) When $x = 5$: $y = 1.6 + 0.8(5) = 1.6 + 4 = 5.6$.

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