STT 201 Business Statistics Practice Sets & Model Questions — TU BBM Semester 3

PAPER KHOJ

Practice Examination — Set 1

For BBM · Semester 3

Based on the prescribed TU Faculty of Management syllabus and examination structure

Independently prepared by Paper Khoj. This is not an official Tribhuvan University or Dean's Office question paper.

Course: STT 201 · Business Statistics

Level: Bachelor of Business Management (BBM) · Semester 3

Full Marks: 60

Time: 3 hrs.

Candidates are required to give their answers in their own words as far as practicable. The figures in the margin indicate full marks.

Group 'A'

Brief Answer Questions. Attempt ALL questions.

[5 × 2 = 10]

[2]
Define Coefficient of Variation (CV) and state its practical utility in comparing business performance.

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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.
[2]
State the conditions under which the Poisson Distribution serves as a limiting form of the Binomial Distribution.
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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.
[2]
Differentiate between Null Hypothesis ( $H_0$ ) and Alternative Hypothesis ( $H_1$ ).
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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., $H_1: \mu \neq \mu_0$ ).
[2]
If the two regression coefficients are $b_{yx} = 0.80$ and $b_{xy} = 0.45$ , compute the Correlation Coefficient ( $r$ ).

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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).
[2]
What is the Additive Model versus Multiplicative Model of a Time Series?
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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).

Group 'B'

Descriptive Answer Questions. Attempt any THREE questions.

[3 × 10 = 30]

[10]

The weekly wages of factory workers in two industrial manufacturing units in Biratnagar are summarized below:

Wage Group (Rs.)	Unit Alpha (No. of Workers)	Unit Beta (No. of Workers)
4,000 – 5,000	12	15
5,000 – 6,000	18	25
6,000 – 7,000	35	30
7,000 – 8,000	25	20
8,000 – 9,000	10	10

a) Which manufacturing unit pays a higher total weekly wage bill? b) Which unit has greater uniformity (consistency) in wage distribution?

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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}

Wage Range	Mid-point ( $m$ )	$d$	$f_A$	$f_A d$	$f_A d^2$	$f_B$	$f_B d$	$f_B d^2$
4,000 – 5,000	4,500	-2	12	-24	48	15	-30	60
5,000 – 6,000	5,500	-1	18	-18	18	25	-25	25
6,000 – 7,000	6,500	0	35	0	0	30	0	0
7,000 – 8,000	7,500	1	25	25	25	20	20	20
8,000 – 9,000	8,500	2	10	20	40	10	20	40
Total			$N_A = 100$	$\sum f_A d = 3$	$\sum f_A d^2 = 131$	$N_B = 100$	$\sum f_B d = -15$	$\sum f_B d^2 = 145$

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 ( $\text{Rs. } 653,000 > \text{Rs. } 635,000$ ).

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.

[10]

The following data represent Advertising Expenditure ( $X$ in Lakh Rs.) and Sales Revenue ( $Y$ in Crore Rs.) of an FMCG brand over 6 consecutive quarters:

Quarter	1	2	3	4	5	6
Advertising ( $X$ )	10	12	14	16	18	20
Sales ( $Y$ )	25	28	34	38	42	49

a) Find the Linear Regression Equation of Sales ( $Y$ ) on Advertising ( $X$ ). b) Estimate the expected sales revenue if the advertising expenditure is increased to Rs. 25 Lakhs. c) Compute the Coefficient of Determination ( $R^2$ ) and interpret its meaning.

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Solution: Linear Regression Analysis

Let $n = 6$ .

$X$	$Y$	$x = X - 15$	$y = Y - 36$	$x^2$	$y^2$	$xy$
10	25	-5	-11	25	121	55
12	28	-3	-8	9	64	24
14	34	-1	-2	1	4	2
16	38	1	2	1	4	2
18	42	3	6	9	36	18
20	49	5	13	25	169	65
$\sum X = 90$	$\sum Y = 216$	$\sum x = 0$	$\sum y = 0$	$\sum x^2 = 70$	$\sum y^2 = 398$	$\sum xy = 166$

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

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.}}

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.

[10]

State the assumptions of the One-Way Analysis of Variance (ANOVA). A commercial bank tests the average transaction processing times (in seconds) across three different branch counter formats with the following sample observations:

Counter A: 12, 14, 16, 18
Counter B: 10, 11, 13, 14
Counter C: 15, 17, 18, 22

At the 5% significance level, test whether there is a significant difference in mean processing times across the three counter formats. (Critical $F_{0.05, 2, 9} = 4.26$ ).

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Solution: One-Way ANOVA Hypothesis Test

1. Assumptions of ANOVA

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

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.

3. Computations

Counter A ( $X_1$ )	Counter B ( $X_2$ )	Counter C ( $X_3$ )
12	10	15
14	11	17
16	13	18
18	14	22
$T_1 = 60$	$T_2 = 48$	$T_3 = 72$
$n_1 = 4$	$n_2 = 4$	$n_3 = 4$
$\sum X_1^2 = 920$	$\sum X_2^2 = 586$	$\sum X_3^2 = 1,322$

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

4. ANOVA Summary Table

Source of Variation	Sum of Squares ( $SS$ )	Degrees of Freedom ( $df$ )	Mean Square ( $MS$ )	Calculated $F$	Critical $F_{0.05, 2, 9}$
Between Groups	72	$k - 1 = 2$	$\frac{72}{2} = 36$	$\frac{36}{6.22} = \mathbf{5.79}$	4.26
Within Groups (Error)	56	$N - k = 9$	$\frac{56}{9} = 6.22$
Total	128	11

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.

[10]

What is the Chi-Square ( $\chi^2$ ) Test of Independence? A market research firm surveyed 200 consumers to determine whether product preference (Brand A vs. Brand B) is independent of gender:

Gender	Brand A	Brand B	Total
Male	60	40	100
Female	30	70	100
Total	90	110	200

Test at the 5% significance level whether brand preference is associated with gender. (Critical $\chi^2_{0.05, 1} = 3.841$ ).

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Solution: Chi-Square Test of Independence

1. Hypotheses

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

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:

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

3. Computation of $\chi^2$ Statistic

Cell ( $i, j$ )	Observed ( $O$ )	Expected ( $E$ )	$(O - E)$	$(O - E)^2$	$\frac{(O - E)^2}{E}$
Male, Brand A	60	45	+15	225	$\frac{225}{45} = 5.000$
Male, Brand B	40	55	-15	225	$\frac{225}{55} = 4.091$
Female, Brand A	30	45	-15	225	$\frac{225}{45} = 5.000$
Female, Brand B	70	55	+15	225	$\frac{225}{55} = 4.091$
Total	200	200	0		$\chi^2_{\text{calc}} = \mathbf{18.182}$

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.

Group 'C'

Comprehensive Answer / Case Analysis Question.

[1 × 20 = 20]

[20]
Read the business scenario and answer all questions:

Scenario: Quality Assurance at Himalaya Beverages Ltd. Himalaya Beverages Ltd. packages mineral water in 1-Litre bottles. The bottling machine is calibrated to deliver a population mean ( $\mu$ ) of 1,000 ml with a known standard deviation ( $\sigma$ ) of 20 ml. The quality control inspector draws a random sample of 64 bottles during the morning shift and discovers a sample mean ( $\bar{X}$ ) of 994 ml.

Later, the inspector records the net sales revenue ( $Y$ in Million Rs.) and distribution outlets ( $X$ in hundreds) across 5 regional sales territories:
- $\sum X = 25, \quad \sum Y = 50, \quad \sum X^2 = 165, \quad \sum Y^2 = 550, \quad \sum XY = 295, \quad n = 5$
Required: (a) At the 1% significance level, test whether the bottling machine is under-filling bottles. (Critical $z_{0.01}$ for one-tailed test = -2.33). (6 Marks) (b) Compute a 95% Confidence Interval for the true mean volume of mineral water delivered by the bottling line. (4 Marks) (c) Calculate Karl Pearson’s Correlation Coefficient ( $r$ ) between distribution outlets and sales revenue and interpret the result. (5 Marks) (d) Fit the regression equation of Sales ( $Y$ ) on Outlets ( $X$ ) and forecast sales for a new territory with 800 distribution outlets ( $X = 8$ ). (5 Marks)
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Solution: Comprehensive Quality & Econometric Analysis

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.

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.

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.

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.

Business Statistics

Practice sets and model questions

Practice Set 1

Practice Examination — Set 1

Brief Answer Questions. Attempt ALL questions.

Descriptive Answer Questions. Attempt any THREE questions.

Solution: Comparative Wage Analysis

Part (a): Total Weekly Wage Bill

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

Solution: Linear Regression Analysis

Step 1: Means and Regression Coefficients

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

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

Solution: One-Way ANOVA Hypothesis Test

1. Assumptions of ANOVA

2. Hypotheses

3. Computations

4. ANOVA Summary Table

5. Decision & Conclusion

Solution: Chi-Square Test of Independence

1. Hypotheses

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

3. Computation of $\chi^2$ Statistic

4. Decision and Conclusion

Comprehensive Answer / Case Analysis Question.

Solution: Comprehensive Quality & Econometric Analysis

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

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

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

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

Practice sets and model questions

Practice Set 1

Solution: Comparative Wage Analysis

Part (a): Total Weekly Wage Bill

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

Solution: Linear Regression Analysis

Step 1: Means and Regression Coefficients

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

Step 3: Coefficient of Determination (R2R^2R2)

Solution: One-Way ANOVA Hypothesis Test

1. Assumptions of ANOVA

2. Hypotheses

3. Computations

4. ANOVA Summary Table

5. Decision & Conclusion

Solution: Chi-Square Test of Independence

1. Hypotheses

2. Expected Frequencies (E=Row Total×Col TotalGrand TotalE = \frac{\text{Row Total} \times \text{Col Total}}{\text{Grand Total}}E=Grand TotalRow Total×Col Total​)

3. Computation of χ2\chi^2χ2 Statistic

4. Decision and Conclusion

Solution: Comprehensive Quality & Econometric Analysis

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

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

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

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

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

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

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

3. Computation of $\chi^2$ Statistic

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

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