Unit 1: Descriptive Statistics & Relationships

Course - Inferential Statistics
Component - LLT Assignment
Unit - Unit 1

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Question 1

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Question Statement

A company wants to evaluate the performance of two teams (Team X and Team Y). The performance score is measured on a scale from 1 to 100, and productivity data for 6 employees from each team is provided below.

Data:

• Team X: 88, 92, 85, 79, 90, 95
• Team Y: 82, 75, 78, 80, 70, 85
  (Note : consider it as population data)

Questions:
i. Calculate the mean, median, and mode (if any) for both teams to understand the central tendency of performance scores.
ii. Compute the range, variance, standard deviation , coefficient of variation for both teams to assess the dispersion of performance scores.
iii. Based on your analysis, which team shows higher consistency in employee performance? Provide a statistical justification for your conclusion.
iv. Find skewness and kurtosis for given data.

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Solution 1

(i) Measures of Central Tendency

Given population size $N = 6$ for both teams.

1. Team X Calculations

• Raw Data: $\{88, 92, 85, 79, 90, 95\}$
• Sorted Data: $\{79, 85, 88, 90, 92, 95\}$
• Mean ($\mu_X$): $\mu_X = \frac{\sum X}{N} = \frac{79 + 85 + 88 + 90 + 92 + 95}{6} = \frac{529}{6} = \mathbf{88.17}$
• Median: Average of the 3rd and 4th terms: $\text{Median}_X = \frac{88 + 90}{2} = \mathbf{89.00}$
• Mode:
  • By Inspection: Each value appears exactly once $\implies$ No unique mode.
  • By Karl Pearson's Empirical Formula: $\text{Mode}_X = 3(\text{Median}) - 2(\text{Mean}) = 3(89.0) - 2(88.17) = 267 - 176.34 = \mathbf{90.66}$

2. Team Y Calculations

• Raw Data: $\{82, 75, 78, 80, 70, 85\}$
• Sorted Data: $\{70, 75, 78, 80, 82, 85\}$
• Mean ($\mu_Y$): $\mu_Y = \frac{\sum Y}{N} = \frac{70 + 75 + 78 + 80 + 82 + 85}{6} = \frac{470}{6} = \mathbf{78.33}$
• Median: Average of the 3rd and 4th terms: $\text{Median}_Y = \frac{78 + 80}{2} = \mathbf{79.00}$
• Mode:
  • By Inspection: Each value appears exactly once $\implies$ No unique mode.
  • By Karl Pearson's Empirical Formula: $\text{Mode}_Y = 3(\text{Median}) - 2(\text{Mean}) = 3(79.0) - 2(78.33) = 237 - 156.66 = \mathbf{80.34}$

Central Tendency Summary Table

Team	Mean ($\mu$)	Median	Mode (Empirical)
Team X	88.17	89.00	90.66
Team Y	78.33	79.00	80.34

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(ii) Measures of Dispersion (Population Data: $N = 6$)

1. Team X Dispersion Table

• $\text{Range}_X = \text{Maximum} - \text{Minimum} = 95 - 79 = \mathbf{16}$

Team X ($X$)	Deviation $(X - \mu_X)$	Squared Deviation $(X - \mu_X)^2$
88	$88 - 88.17 = -0.17$	$0.03$
92	$92 - 88.17 = 3.83$	$14.67$
85	$85 - 88.17 = -3.17$	$10.05$
79	$79 - 88.17 = -9.17$	$84.09$
90	$90 - 88.17 = 1.83$	$3.35$
95	$95 - 88.17 = 6.83$	$46.65$
Total ($\sum$)	$0.00$	$158.84$

• Population Variance ($\sigma_X^2$): $\sigma_X^2 = \frac{\sum (X - \mu_X)^2}{N} = \frac{158.84}{6} = \mathbf{26.47}$
• Population Standard Deviation ($\sigma_X$): $\sigma_X = \sqrt{\sigma_X^2} = \sqrt{26.47} = \mathbf{5.15}$ (Note: If evaluated using sample degrees of freedom $s^2 = \frac{158.84}{6-1} = 31.77 \implies s = 5.64$)
• Coefficient of Variation ($CV_X$): $CV_X = \left( \frac{\sigma_X}{\mu_X} \right) \times 100\% = \left( \frac{5.15}{88.17} \right) \times 100\% = \mathbf{5.84\%}$

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2. Team Y Dispersion Table

• $\text{Range}_Y = \text{Maximum} - \text{Minimum} = 85 - 70 = \mathbf{15}$

Team Y ($Y$)	Deviation $(Y - \mu_Y)$	Squared Deviation $(Y - \mu_Y)^2$
70	$70 - 78.33 = -8.33$	$69.39$
75	$75 - 78.33 = -3.33$	$11.09$
78	$78 - 78.33 = -0.33$	$0.11$
80	$80 - 78.33 = 1.67$	$2.79$
82	$82 - 78.33 = 3.67$	$13.47$
85	$85 - 78.33 = 6.67$	$44.49$
Total ($\sum$)	$0.00$	$141.34$

• Population Variance ($\sigma_Y^2$): $\sigma_Y^2 = \frac{\sum (Y - \mu_Y)^2}{N} = \frac{141.34}{6} = \mathbf{23.56}$
• Population Standard Deviation ($\sigma_Y$): $\sigma_Y = \sqrt{\sigma_Y^2} = \sqrt{23.56} = \mathbf{4.85}$ (Note: If evaluated using sample degrees of freedom $s^2 = \frac{141.34}{6-1} = 28.27 \implies s = 5.32$)
• Coefficient of Variation ($CV_Y$): $CV_Y = \left( \frac{\sigma_Y}{\mu_Y} \right) \times 100\% = \left( \frac{4.85}{78.33} \right) \times 100\% = \mathbf{6.20\%}$

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Dispersion Summary Table

Team	Range	Variance ($\sigma^2$)	Standard Deviation ($\sigma$)	Coefficient of Variation (CV)
Team X	16	26.47	5.15	5.84%
Team Y	15	23.56	4.85	6.20%

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(iii) Consistency Analysis & Statistical Justification

• Conclusion: Team X shows higher consistency in employee performance.
• Statistical Justification Table:

Metrics	Team X	Team Y	Determination of Consistency
Mean (Average)	88.17	78.33	Team X performs higher on average.
Standard Deviation ($\sigma$)	5.15	4.85	Measures absolute dispersion only; sensitive to mean baseline.
Coefficient of Variation (CV)	5.84%	6.20%	Lower CV indicates higher consistency.

• Analytical Reasoning:
  While Team Y has a slightly lower absolute standard deviation ($4.85$ vs $5.15$), the two teams have vastly different average production scores ($88.17$ vs $78.33$). Standard deviation alone cannot measure consistency across groups with unequal baselines. The Coefficient of Variation ($CV = \frac{\sigma}{\mu} \times 100\%$) measures relative dispersion. Because $CV_X = 5.84\% < CV_Y = 6.20\%$, Team X exhibits less variability relative to its mean performance level, proving statistically superior consistency.

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(iv) Skewness and Kurtosis

1. Team X Moment Computation Table ($d = X - 88.17$)

$X$	$d = (X - 88.17)$	$d^2$	$d^3$	$d^4$
88	-0.17	0.03	-0.05	0.001
92	3.83	14.67	56.18	215.18
85	-3.17	10.05	-31.85	100.97
79	-9.17	84.09	-771.09	7070.93
90	1.83	3.35	6.13	11.22
95	6.83	46.65	318.61	2176.08
$\sum$	0.00	158.84	-420.02	9574.38

• Central Moments:
  • $m_2 = \frac{\sum d^2}{N} = \frac{158.84}{6} = \mathbf{26.47}$
  • $m_3 = \frac{\sum d^3}{N} = \frac{-420.02}{6} = \mathbf{-70.00}$
  • $m_4 = \frac{\sum d^4}{N} = \frac{9574.38}{6} = \mathbf{1595.73}$
• Skewness:
  • Karl Pearson's Skewness: $S_k = \frac{3(\text{Mean} - \text{Median})}{\sigma} = \frac{3(88.17 - 89.00)}{5.15} = \frac{-2.49}{5.15} = \mathbf{-0.48}$
  • Moment Coefficient of Skewness ($\gamma_1$): $\gamma_1 = \frac{m_3}{(m_2)^{3/2}} = \frac{-70.00}{(26.47)^{1.5}} = \frac{-70.00}{136.20} = \mathbf{-0.51}$
• Kurtosis:
  • $\beta_2 = \frac{m_4}{(m_2)^2} = \frac{1595.73}{(26.47)^2} = \frac{1595.73}{700.66} = \mathbf{2.28}$
  • Excess Kurtosis ($\gamma_2$): $\gamma_2 = \beta_2 - 3 = 2.28 - 3 = \mathbf{-0.72}$

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2. Team Y Moment Computation Table ($d = Y - 78.33$)

$Y$	$d = (Y - 78.33)$	$d^2$	$d^3$	$d^4$
82	3.67	13.47	49.43	181.41
75	-3.33	11.09	-36.93	122.98
78	-0.33	0.11	-0.04	0.01
80	1.67	2.79	4.66	7.78
70	-8.33	69.39	-578.01	4814.80
85	6.67	44.49	296.74	1979.25
$\sum$	0.00	141.34	-264.15	7106.23

• Central Moments:
  • $m_2 = \frac{\sum d^2}{N} = \frac{141.34}{6} = \mathbf{23.56}$
  • $m_3 = \frac{\sum d^3}{N} = \frac{-264.15}{6} = \mathbf{-44.03}$
  • $m_4 = \frac{\sum d^4}{N} = \frac{7106.23}{6} = \mathbf{1184.37}$
• Skewness:
  • Karl Pearson's Skewness: $S_k = \frac{3(\text{Mean} - \text{Median})}{\sigma} = \frac{3(78.33 - 79.00)}{4.85} = \frac{-2.01}{4.85} = \mathbf{-0.41}$
  • Moment Coefficient of Skewness ($\gamma_1$): $\gamma_1 = \frac{m_3}{(m_2)^{3/2}} = \frac{-44.03}{(23.56)^{1.5}} = \frac{-44.03}{114.33} = \mathbf{-0.39}$
• Kurtosis:
  • $\beta_2 = \frac{m_4}{(m_2)^2} = \frac{1184.37}{(23.56)^2} = \frac{1184.37}{555.07} = \mathbf{2.13}$
  • Excess Kurtosis ($\gamma_2$): $\gamma_2 = \beta_2 - 3 = 2.13 - 3 = \mathbf{-0.87}$

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Shape Interpretation

• Skewness: Both datasets have a negative skewness value ($S_{k, X} = -0.48$, $S_{k, Y} = -0.41$), which reveals both distributions are left-skewed. A few lower scores pull the tail towards the lower end while the majority of values sit at the higher end.
• Kurtosis: Both teams have excess kurtosis less than zero ($\gamma_{2, X} = -0.72$, $\gamma_{2, Y} = -0.87$). This indicates platykurtic distributions with flatter peaks and lighter tails compared to a standard normal curve.

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Question 2

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Question Statement

A school psychologist is examining the effects of two factors, "Study Hours" and "Sleep Duration," on students’ "Exam Scores." The following data has been collected for 5 students over a specific academic term: (Note : consider it as population data)

Individual	Study Hours	Sleep Duration (hours)	Exam Score (%)
1	10	7	85
2	8	6	78
3	12	5	90
4	6	8	74
5	9	6	80

a) Compute the mean,median,mode, range, variance, and standard deviation for both independent variables to analyze the dispersion of study hours and sleep duration.
b) Determine which independent variable (Study Hours or Sleep Duration) has a stronger relationship with Exam Scores. Use statistical techniques to justify your analysis.
c) Find skewness and kurtosis for all last three columns.
d) Compute the correlation and coefficient between dependent and independent variables.

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Solution 2

(a) Central Tendency and Dispersion of Independent Variables ($N = 5$)

1. Calculation for Study Hours ($X_1$)

• Dataset: $\{10, 8, 12, 6, 9\}$
• Sorted: $\{6, 8, 9, 10, 12\}$
• Mean ($\mu_{X_1}$): $\mu_{X_1} = \frac{10 + 8 + 12 + 6 + 9}{5} = \frac{45}{5} = \mathbf{9.0\text{ hours}}$
• Median: The middle (3rd) value is $9.0\text{ hours}$.
• Mode:
  • By Inspection: All values appear once $\implies$ No unique mode.
  • By Empirical Formula: $\text{Mode} = 3(9.0) - 2(9.0) = \mathbf{9.0\text{ hours}}$.
• Range: $\text{Maximum} - \text{Minimum} = 12 - 6 = \mathbf{6\text{ hours}}$.
• Population Variance ($\sigma_{X_1}^2$): $\sum (X_1 - \mu)^2 = (10 - 9)^2 + (8 - 9)^2 + (12 - 9)^2 + (6 - 9)^2 + (9 - 9)^2 = 1 + 1 + 9 + 9 + 0 = \mathbf{20.00}$ $\sigma_{X_1}^2 = \frac{20.00}{5} = \mathbf{4.00}$
• Population Standard Deviation ($\sigma_{X_1}$): $\sigma_{X_1} = \sqrt{4.00} = \mathbf{2.00\text{ hours}}$

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2. Calculation for Sleep Duration ($X_2$)

• Dataset: $\{7, 6, 5, 8, 6\}$
• Sorted: $\{5, 6, 6, 7, 8\}$
• Mean ($\mu_{X_2}$): $\mu_{X_2} = \frac{7 + 6 + 5 + 8 + 6}{5} = \frac{32}{5} = \mathbf{6.4\text{ hours}}$
• Median: The middle (3rd) value is $6.0\text{ hours}$.
• Mode: $6.0\text{ hours}$ (appears twice).
• Range: $\text{Maximum} - \text{Minimum} = 8 - 5 = \mathbf{3.0\text{ hours}}$.
• Population Variance ($\sigma_{X_2}^2$): $\sum (X_2 - \mu)^2 = (7 - 6.4)^2 + (6 - 6.4)^2 + (5 - 6.4)^2 + (8 - 6.4)^2 + (6 - 6.4)^2$ $= 0.36 + 0.16 + 1.96 + 2.56 + 0.16 = \mathbf{5.20}$ $\sigma_{X_2}^2 = \frac{5.20}{5} = \mathbf{1.04}$
• Population Standard Deviation ($\sigma_{X_2}$): $\sigma_{X_2} = \sqrt{1.04} = \mathbf{1.02\text{ hours}}$

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Independent Variables Summary Table

Variable	Mean ($\mu$)	Median	Mode	Variance ($\sigma^2$)	Std Dev ($\sigma$)	Range
Study Hours	9.0	9.0	9.0	4.00	2.00	6
Sleep Duration	6.4	6.0	6.0	1.04	1.02	3

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(b) & (d) Correlation and Relationship Strength Analysis

Given baseline data for dependent variable Exam Score ($Y$): $\{85, 78, 90, 74, 80\}$.

• Mean ($\mu_Y$): $\mu_Y = \frac{85 + 78 + 90 + 74 + 80}{5} = \frac{407}{5} = \mathbf{81.4\%}$
• Sum of Squared Deviations $\sum (Y - \mu_Y)^2$: $\sum (Y - \mu_Y)^2 = (85 - 81.4)^2 + (78 - 81.4)^2 + (90 - 81.4)^2 + (74 - 81.4)^2 + (80 - 81.4)^2$ $= 12.96 + 11.56 + 73.96 + 54.76 + 1.96 = \mathbf{155.20}$
• $\sigma_Y^2 = \frac{155.20}{5} = 31.04 \implies \sigma_Y = \sqrt{31.04} \approx 5.57\%$

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1. Relationship Between Study Hours ($X_1$) vs Exam Scores ($Y$)

Student	$(X_1 - \mu_{X_1})$	$(Y - \mu_Y)$	Product: $(X_1 - \mu_{X_1})(Y - \mu_Y)$
1	$10 - 9 = 1.0$	$85 - 81.4 = 3.6$	$1.0 \times 3.6 = 3.6$
2	$8 - 9 = -1.0$	$78 - 81.4 = -3.4$	$(-1.0) \times (-3.4) = 3.4$
3	$12 - 9 = 3.0$	$90 - 81.4 = 8.6$	$3.0 \times 8.6 = 25.8$
4	$6 - 9 = -3.0$	$74 - 81.4 = -7.4$	$(-3.0) \times (-7.4) = 22.2$
5	$9 - 9 = 0.0$	$80 - 81.4 = -1.4$	$0.0 \times (-1.4) = 0.0$
Total ($\sum$)			$55.0$

• Pearson Correlation ($r_{\text{study}}$): $r_{\text{study}} = \frac{\sum (X_1 - \mu_{X_1})(Y - \mu_Y)}{\sqrt{\sum (X_1 - \mu_{X_1})^2 \cdot \sum (Y - \mu_Y)^2}} = \frac{55.0}{\sqrt{20.0 \times 155.2}} = \frac{55.0}{\sqrt{3104}} = \mathbf{+0.9872}$
• Coefficient of Determination ($R^2$): $R^2 = (0.9872)^2 = 0.9746\quad (\mathbf{97.46\%})$

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2. Relationship Between Sleep Duration ($X_2$) vs Exam Scores ($Y$)

Student	$(X_2 - \mu_{X_2})$	$(Y - \mu_Y)$	Product: $(X_2 - \mu_{X_2})(Y - \mu_Y)$
1	$7 - 6.4 = 0.6$	$85 - 81.4 = 3.6$	$0.6 \times 3.6 = 2.16$
2	$6 - 6.4 = -0.4$	$78 - 81.4 = -3.4$	$(-0.4) \times (-3.4) = 1.36$
3	$5 - 6.4 = -1.4$	$90 - 81.4 = 8.6$	$(-1.4) \times 8.6 = -12.04$
4	$8 - 6.4 = 1.6$	$74 - 81.4 = -7.4$	$1.6 \times (-7.4) = -11.84$
5	$6 - 6.4 = -0.4$	$80 - 81.4 = -1.4$	$(-0.4) \times (-1.4) = 0.56$
Total ($\sum$)			$-19.80$

• Pearson Correlation ($r_{\text{sleep}}$): $r_{\text{sleep}} = \frac{\sum (X_2 - \mu_{X_2})(Y - \mu_Y)}{\sqrt{\sum (X_2 - \mu_{X_2})^2 \cdot \sum (Y - \mu_Y)^2}} = \frac{-19.80}{\sqrt{5.2 \times 155.2}} = \frac{-19.80}{\sqrt{807.04}} = \mathbf{-0.6970}$
• Coefficient of Determination ($R^2$): $R^2 = (-0.6970)^2 = 0.4858\quad (\mathbf{48.58\%})$

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Stronger Relationship Justification

Comparing the absolute values of the two correlation coefficients: $|r_{\text{study}}| = |+0.9872| = \mathbf{0.9872}$ $|r_{\text{sleep}}| = |-0.6970| = \mathbf{0.6970}$

• Conclusion: Study Hours has an almost perfect positive linear relationship with final exam performance ($r = +0.9872$). Since $|0.9872| > |0.6970|$, Study Hours serves as a significantly stronger predictor for Exam Scores than Sleep Duration. Furthermore, $97.46\%$ of score variation is explained by study hours alone, compared to only $48.58\%$ by sleep duration.

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(c) Skewness and Kurtosis for All Last Three Columns

Computation Summary Table

Metric	Study Hours ($X_1$)	Sleep Duration ($X_2$)	Exam Score ($Y$)
Deviations $(x - \mu)$	$1, -1, 3, -3, 0$	$0.6, -0.4, -1.4, 1.6, -0.4$	$3.6, -3.4, 8.6, -7.4, -1.4$
$\sum (x - \mu)^2$	$20.00$	$5.20$	$155.20$
$\sum (x - \mu)^4$	$164.00$	$10.56$	$8774.15$
$m_2 = \frac{\sum d^2}{N}$	$4.00$	$1.04$	$31.04$
$m_4 = \frac{\sum d^4}{N}$	$32.80$	$2.1152$	$1754.83$

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1. Study Hours

• Skewness: $S_k = \frac{3(\mu - \text{Median})}{\sigma} = \frac{3(9.0 - 9.0)}{2.0} = \mathbf{0.00}\quad (\text{Perfect symmetry})$
• Kurtosis ($\beta_2$): $\beta_2 = \frac{m_4}{(m_2)^2} = \frac{32.80}{(4.00)^2} = \frac{32.80}{16.00} = \mathbf{2.05}$
• Excess Kurtosis ($\gamma_2$): $\gamma_2 = \beta_2 - 3 = 2.05 - 3 = \mathbf{-0.95}\quad (\text{Platykurtic})$

2. Sleep Duration

• Skewness: $S_k = \frac{3(\mu - \text{Median})}{\sigma} = \frac{3(6.4 - 6.0)}{1.02} = \frac{1.20}{1.02} = \mathbf{+1.18}\quad (\text{Right-skewed})$ (Moment skewness: $\gamma_1 = \frac{m_3}{\sigma^3} = \frac{0.288}{1.060} = \mathbf{+0.27}$)
• Kurtosis ($\beta_2$): $\beta_2 = \frac{m_4}{(m_2)^2} = \frac{2.1152}{(1.04)^2} = \frac{2.1152}{1.0816} = \mathbf{1.96}$
• Excess Kurtosis ($\gamma_2$): $\gamma_2 = \beta_2 - 3 = 1.96 - 3 = \mathbf{-1.04}\quad (\text{Platykurtic})$

3. Exam Score Calculations

• Skewness: $S_k = \frac{3(\mu - \text{Median})}{\sigma} = \frac{3(81.4 - 80.0)}{5.57} = \frac{4.20}{5.57} = \mathbf{+0.75}\quad (\text{Moderately right-skewed})$ (Moment skewness: $\gamma_1 = \frac{47.088}{172.93} = \mathbf{+0.27}$)
• Kurtosis ($\beta_2$): $\beta_2 = \frac{m_4}{(m_2)^2} = \frac{1754.83}{(31.04)^2} = \frac{1754.83}{963.48} = \mathbf{1.82}$
• Excess Kurtosis ($\gamma_2$): $\gamma_2 = \beta_2 - 3 = 1.82 - 3 = \mathbf{-1.18}\quad (\text{Platykurtic})$

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