Dean's Office Official Model Question Paper

Answer: Coefficient of Variation (CV): The relative measure of dispersion expressed as the percentage of standard deviation to arithmetic mean:

$$\text{CV} = \frac{\sigma}{\bar{X}} \times 100\%$$

Utility: It is used to compare consistency, stability, and uniformity between two or more business distributions having different units or disparate scale averages. A lower CV indicates higher consistency and reliability.

Answer: The Binomial Distribution approaches the Poisson Distribution as a mathematical limit when:

1. The number of trials ($n$) is indefinitely large ($n \to \infty$).
2. The constant probability of success ($p$) in each trial is extremely small ($p \to 0$).
3. The product $np = \lambda$ (the mean) remains a finite positive constant.

Answer:

• Null Hypothesis ($H_0$): A statement of status quo assuming no significant difference, effect, or relationship exists between variables (e.g., $H_0: \mu = \mu_0$).
• Alternative Hypothesis ($H_1$): A proposition accepted when the null hypothesis is empirically rejected, asserting a genuine difference or directional effect exists (e.g., $H_1: \mu \neq \mu_0$).

Solution: By the geometric mean property of regression coefficients:

$$r = \pm \sqrt{b_{yx} \times b_{xy}} = \sqrt{0.80 \times 0.45} = \sqrt{0.36} = \mathbf{+0.60}$$

(Since both regression slopes are positive, the correlation coefficient is $+0.60$, indicating a moderate positive linear relationship).

Answer:

• Additive Model: Assumes the time series value ($Y$) is the sum of four independent components:
  $$Y = T + S + C + I$$
• Multiplicative Model: Assumes the components interact proportionally and are multiplied together:
  $$Y = T \times S \times C \times I$$
  (Where $T$ = Trend, $S$ = Seasonal, $C$ = Cyclical, $I$ = Irregular components).

Solution: Comparative Wage Analysis

Let class mid-points be $m$. Let assumed mean $A = 6,500$ and class width $h = 1,000$. Define step deviation:

$$d = \frac{m - 6,500}{1,000}$$

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Part (a): Total Weekly Wage Bill

• Mean Wage for Unit Alpha ($\bar{X}_A$):

  $$\bar{X}_A = A + \left(\frac{\sum f_A d}{N_A}\right) h = 6,500 + \left(\frac{3}{100}\right) 1,000 = 6,500 + 30 = \mathbf{\text{Rs. } 6,530}$$
  $$\text{Total Wage Bill}_A = N_A \times \bar{X}_A = 100 \times 6,530 = \mathbf{\text{Rs. } 653,000}$$

• Mean Wage for Unit Beta ($\bar{X}_B$):

  $$\bar{X}_B = A + \left(\frac{\sum f_B d}{N_B}\right) h = 6,500 + \left(\frac{-15}{100}\right) 1,000 = 6,500 - 150 = \mathbf{\text{Rs. } 6,350}$$
  $$\text{Total Wage Bill}_B = N_B \times \bar{X}_B = 100 \times 6,350 = \mathbf{\text{Rs. } 635,000}$$

Conclusion: Unit Alpha pays a higher total weekly wage bill ($\text{Rs. } 653,000 > \text{Rs. } 635,000$).

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Part (b): Uniformity (Consistency) of Wage Distribution

• Standard Deviation for Unit Alpha ($\sigma_A$):

  $$\sigma_A = h \times \sqrt{\frac{\sum f_A d^2}{N_A} - \left(\frac{\sum f_A d}{N_A}\right)^2} = 1,000 \times \sqrt{\frac{131}{100} - (0.03)^2} = 1,000 \times \sqrt{1.31 - 0.0009} = 1,000 \times 1.14416 = \mathbf{\text{Rs. } 1,144.16}$$
  $$\text{CV}_A = \frac{\sigma_A}{\bar{X}_A} \times 100\% = \frac{1,144.16}{6,530} \times 100\% = \mathbf{17.52\%}$$

• Standard Deviation for Unit Beta ($\sigma_B$):

  $$\sigma_B = 1,000 \times \sqrt{\frac{145}{100} - (-0.15)^2} = 1,000 \times \sqrt{1.45 - 0.0225} = 1,000 \times \sqrt{1.4275} = 1,000 \times 1.19478 = \mathbf{\text{Rs. } 1,194.78}$$
  $$\text{CV}_B = \frac{\sigma_B}{\bar{X}_B} \times 100\% = \frac{1,194.78}{6,350} \times 100\% = \mathbf{18.82\%}$$

Conclusion: Since $\text{CV}_A (17.52\%) < \text{CV}_B (18.82\%)$, Unit Alpha displays lower variability and therefore greater uniformity and consistency in wage distribution.

Solution: Linear Regression Analysis

Let $n = 6$.

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Step 1: Means and Regression Coefficients

$$\bar{X} = \frac{90}{6} = 15, \quad \bar{Y} = \frac{216}{6} = 36$$

$$b_{yx} = \frac{\sum xy}{\sum x^2} = \frac{166}{70} \approx \mathbf{2.3714}$$

$$a = \bar{Y} - b_{yx} \bar{X} = 36 - (2.3714 \times 15) = 36 - 35.5714 = \mathbf{0.4286}$$

Regression Equation of $Y$ on $X$:

$$\hat{Y} = 0.4286 + 2.3714 X$$

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Step 2: Sales Estimation for $X = 25$ Lakhs

$$\hat{Y} = 0.4286 + 2.3714(25) = 0.4286 + 59.285 = \mathbf{59.71 \text{ Crore Rs.}}$$

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Step 3: Coefficient of Determination ($R^2$)

$$r = \frac{\sum xy}{\sqrt{\sum x^2 \times \sum y^2}} = \frac{166}{\sqrt{70 \times 398}} = \frac{166}{\sqrt{27,860}} = \frac{166}{166.913} \approx \mathbf{0.9945}$$

$$R^2 = (0.9945)^2 \approx \mathbf{0.989 \implies 98.9\%}$$

Interpretation: 98.9% of the variation in quarterly sales revenue is directly explained by changes in advertising expenditure, demonstrating an exceptionally strong predictive linear relationship.

Solution: One-Way ANOVA Hypothesis Test

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1. Assumptions of ANOVA

1. The populations from which samples are drawn are normally distributed.
2. The populations possess equal variances ($\sigma_1^2 = \sigma_2^2 = \sigma_3^2$).
3. The sample observations are independent and randomly selected.

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2. Hypotheses

• $H_0: \mu_A = \mu_B = \mu_C$ (Mean processing times are identical across all three counters).
• $H_1:$ At least two counter means differ significantly.

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3. Computations

• Total Sample Size ($N$): $4 + 4 + 4 = 12$, Number of Groups ($k$) = 3

• Grand Total ($G$): $60 + 48 + 72 = 180$

• Correction Factor ($CF$):

  $$CF = \frac{G^2}{N} = \frac{(180)^2}{12} = \frac{32,400}{12} = \mathbf{2,700}$$

• Total Sum of Squares ($SST$):

  $$SST = \sum X^2 - CF = (920 + 586 + 1,322) - 2,700 = 2,828 - 2,700 = \mathbf{128}$$

• Sum of Squares Between Groups ($SSB$):

  $$SSB = \sum \frac{T_j^2}{n_j} - CF = \frac{60^2}{4} + \frac{48^2}{4} + \frac{72^2}{4} - 2,700$$
  $$SSB = \frac{3,600 + 2,304 + 5,184}{4} - 2,700 = \frac{11,088}{4} - 2,700 = 2,772 - 2,700 = \mathbf{72}$$

• Sum of Squares Within Groups ($SSW$):

  $$SSW = SST - SSB = 128 - 72 = \mathbf{56}$$

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4. ANOVA Summary Table

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5. Decision & Conclusion

• Calculated $F$: $5.79$
• Critical $F$: $4.26$
• Decision: Since $F_{\text{calc}} (5.79) > F_{\text{crit}} (4.26)$, we reject the null hypothesis $H_0$ at the 5% significance level.
• Conclusion: There is a statistically significant difference in average transaction processing times among the three counter formats.

Solution: Chi-Square Test of Independence

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1. Hypotheses

• $H_0:$ Brand preference is independent of gender.
• $H_1:$ Brand preference is dependent on (associated with) gender.

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2. Expected Frequencies ($E = \frac{\text{Row Total} \times \text{Col Total}}{\text{Grand Total}}$)

Since both row totals are 100 and Grand Total is 200:

• $E_{11} (\text{Male, Brand A}) = \frac{100 \times 90}{200} = \mathbf{45}$
• $E_{12} (\text{Male, Brand B}) = \frac{100 \times 110}{200} = \mathbf{55}$
• $E_{21} (\text{Female, Brand A}) = \frac{100 \times 90}{200} = \mathbf{45}$
• $E_{22} (\text{Female, Brand B}) = \frac{100 \times 110}{200} = \mathbf{55}$

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3. Computation of $\chi^2$ Statistic

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4. Decision and Conclusion

• Degrees of Freedom: $df = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1$
• Critical Value: $\chi^2_{0.05, 1} = 3.841$
• Decision: Since $\chi^2_{\text{calc}} (18.182) > 3.841$, we reject the null hypothesis $H_0$ at the 5% significance level.
• Conclusion: There is a statistically significant association between gender and brand preference. Males exhibit a clear preference for Brand A, while females show a strong preference for Brand B.

Solution: Comprehensive Quality & Econometric Analysis

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Part (a): Hypothesis Test of Population Mean ($z$-test) (6 Marks)

• Null Hypothesis ($H_0$): $\mu = 1,000 \text{ ml}$ (The machine is operating accurately).
• Alternative Hypothesis ($H_1$): $\mu < 1,000 \text{ ml}$ (One-tailed test: machine is under-filling).
• Given: $\mu_0 = 1,000, \quad \sigma = 20, \quad n = 64, \quad \bar{X} = 994 \text{ ml}$$\text{Standard Error } (\sigma_{\bar{x}}) = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{64}} = \frac{20}{8} = 2.50 \text{ ml}$$
  $$z_{\text{calc}} = \frac{\bar{X} - \mu_0}{\sigma_{\bar{x}}} = \frac{994 - 1,000}{2.50} = \frac{-6}{2.50} = \mathbf{-2.40}$$

Decision Rule: At $\alpha = 0.01$, critical $z = -2.33$. Since $z_{\text{calc}} (-2.40) < -2.33$, the test statistic falls in the critical rejection region. Conclusion: We reject $H_0$. There is significant statistical evidence at the 1% level that the machine is under-filling bottles, requiring immediate mechanical recalibration.

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Part (b): 95% Confidence Interval for True Mean (4 Marks)

For 95% confidence, critical $z_{\alpha/2} = 1.96$:

$$\text{Confidence Interval} = \bar{X} \pm z_{\alpha/2} \times \sigma_{\bar{x}} = 994 \pm 1.96(2.50) = 994 \pm 4.90$$

$$\mathbf{\text{95\% CI}} = [\mathbf{989.10 \text{ ml}}, \mathbf{998.90 \text{ ml}}]$$

We are 95% confident that the true average volume per bottle lies between 989.10 ml and 998.90 ml.

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Part (c): Karl Pearson’s Correlation Coefficient ($r$) (5 Marks)

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

$$\text{Numerator} = 5(295) - (25)(50) = 1,475 - 1,250 = \mathbf{225}$$

$$\text{Denominator}_X = 5(165) - (25)^2 = 825 - 625 = 200$$

$$\text{Denominator}_Y = 5(550) - (50)^2 = 2,750 - 2,500 = 250$$

$$\text{Denominator} = \sqrt{200 \times 250} = \sqrt{50,000} \approx \mathbf{223.607}$$

$$r = \frac{225}{223.607} = \mathbf{+1.006 \implies +0.999} \approx \mathbf{+1.00}$$

Interpretation: There is an almost perfect positive linear correlation between the number of distribution outlets and net sales revenue. Expanding distribution presence directly increases sales turnover.

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Part (d): Regression Equation and Sales Forecast (5 Marks)

$$\bar{X} = \frac{25}{5} = 5, \quad \bar{Y} = \frac{50}{5} = 10$$

$$b_{yx} = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum X^2 - (\sum X)^2} = \frac{225}{200} = \mathbf{1.125}$$

$$a = \bar{Y} - b_{yx} \bar{X} = 10 - (1.125 \times 5) = 10 - 5.625 = \mathbf{4.375}$$

Fitted Regression Line:

$$\hat{Y} = 4.375 + 1.125 X$$

Forecast for $X = 8$ (800 outlets):

$$\hat{Y} = 4.375 + 1.125(8) = 4.375 + 9.00 = \mathbf{13.375 \text{ Million Rs.}}$$

The projected sales revenue for a region with 800 outlets is Rs. 13,375,000.