Business Analytics · Quiz 1 · Jan 2026

**1. Interpret right skew:** A longer right tail corresponds to positive skewness under the standard moment-skewness convention. A positive coefficient of variation measures relative spread, not tail direction. **Answer: B — Skewness is positive.** A: **Incorrect:** Positive relative spread does not establish skew direction. B: **Correct:** Positive skewness is the standard indicator of a right-skewed distribution. C: **Incorrect:** Negative skewness indicates a longer left tail. D: **Incorrect:** A stated positive skewness identifies direction without needing a histogram.

**1. Count fitted regression parameters:** The simple linear model estimates an intercept and one price slope, totaling two coefficients. **2. Subtract from observations:** Residual df=20−2=18. **Answer: 18.**

**1. Interpret market size:** For the fitted linear curve \(D=5482-175p\), market size is demand at zero price. **2. Substitute:** \(D(0)=5482\). **Answer: 5482.**

**1. Set demand to zero:** The satiating or choke price satisfies \(0=5482-175p\). **2. Solve:** \(p=5482/175\approx31.325714\), which rounds to 31.33. **Answer: 31.33.**

**1. Evaluate demand and slope:** At p=20, \(D=5482-175(20)=1982\), and \(D\prime(p)=-175\). **2. Calculate elasticity magnitude:** \[\varepsilon=-D\prime(p)\frac pD=175\frac{20}{1982}\approx1.765893\to1.77.\] **Answer: 1.77.** **Convention note:** The course reports the nonnegative magnitude. The signed price elasticity is −1.77.

**1. Compute the two demands:** \(D_0=5482-175(20)=1982\), \(D_1=5482-175(25)=1107\). Thus demand falls by 875 while price rises by 5. **2. Use initial values as percentage bases:** \[\varepsilon=\frac{875/1982}{5/20}=\frac{3500}{1982}\approx1.765893\to1.77.\] **Answer: 1.77.** **Convention note:** This matches the key’s initial-base percentage-change convention. Midpoint arc elasticity would be [(875/1544.5)/(5/22.5)]≈2.55; the question does not specify which convention.

**1. Build the observed table:** In Foundation,Diploma,Degree,Discontinued order, male counts are (55,45,20,53), female counts (85,25,45,25). Row totals are 173 and 180; column totals are 140,70,65,78; N=353. **2. Calculate independence expectations:** \(E_{ij}=R_iC_j/353\). The male row is approximately (68.61,34.31,31.86,38.23), and the female row (71.39,35.69,33.14,39.77). **3. Sum Pearson contributions:** \[\chi^2=\sum_{i,j}\frac{(O_{ij}-E_{ij})^2}{E_{ij}}\approx31.683172\to31.68.\] Keep expected frequencies unrounded until the final statistic. **Answer: 31.68.**

**1. Identify table dimensions:** Gender has two categories and completion status has four categories. **2. Calculate independence-test df:** \((r-1)(c-1)=(2-1)(4-1)=3\), entered as 3.00. **Answer: 3.00.**

**1. State the test:** H0 is independence. At α=0.20, reject H0 for p≤0.20; equivalently reject when the statistic exceeds the upper-tail critical value. **2. Apply each decision rule:** p=0.25 leads to nonrejection, whereas p=0.15 leads to rejection. A statistic below the critical value gives nonrejection; one above it gives rejection. **Answer: A, B, E, F — A, B, E and F.** **Convention note:** Nonrejection does not prove independence. A and E use “conclude independent” as the course’s shorthand for insufficient evidence against independence. A: **Correct:** Intended correct: 0.25>0.20 means fail to reject independence. B: **Correct:** 0.15<0.20 means reject independence in favor of association. C: **Incorrect:** It reverses the decision for p=0.25. D: **Incorrect:** It reverses the decision for p=0.15. E: **Correct:** Intended correct: a statistic below the upper critical value does not reject independence. F: **Correct:** A statistic above the upper critical value rejects independence. G: **Incorrect:** It reverses the below-critical-value rule. H: **Incorrect:** It reverses the above-critical-value rule.

**1. Partition by bulb type:** Use production proportions 0.30,0.35,0.35 and survival probabilities 0.80,0.60,0.70. **2. Apply total probability:** \(P(L\ge100)=0.30(0.80)+0.35(0.60)+0.35(0.70)=0.24+0.21+0.245=0.695\), rounding to 0.70. **Answer: 0.70.**

**1. Compute the joint probability:** \(P(C,L\ge100)=0.35(0.70)=0.245\). Total survival probability is \(0.30(0.80)+0.35(0.60)+0.35(0.70)=0.695\). **2. Use Bayes’ rule:** \(P(C\mid L\ge100)=0.245/0.695\approx0.352518\), giving 0.35. **Answer: 0.35.**

**1. Calculate expected warranty costs by type:** Type A contributes \(200000(0.30)(0.20)(8)=96000\); B contributes \(200000(0.35)(0.40)(5)=140000\); C contributes \(200000(0.35)(0.30)(6)=126000\). **2. Sum:** Expected total cost is \(96000+140000+126000=362000.00\) rupees. **Answer: 362000.00.**

**1. Recover zero-defect observations:** The listed counts sum to 120+80+150+50=400, while the sample has 500 toys and none has more than four defects. Thus 100 have zero defects. **2. Estimate the Poisson mean:** \[\hat\lambda=\frac{0(100)+1(120)+2(80)+3(150)+4(50)}{500}=\frac{930}{500}=1.86.\] **3. Compute the expected four-defect count:** \(E_4=500e^{-1.86}1.86^4/4!\approx38.817075\), rounding to 38.82. **Answer: 38.82.**

**1. Recover zero-defect observations:** The listed counts sum to 120+80+150+50=400, while the sample has 500 toys and none has more than four defects. Thus 100 have zero defects. **2. Estimate the Poisson mean:** \[\hat\lambda=\frac{0(100)+1(120)+2(80)+3(150)+4(50)}{500}=\frac{930}{500}=1.86.\] **3. Include the unobserved upper tail:** A Poisson distribution still assigns positive probability to ≥5 defects. Use six categories 0,1,2,3,4,≥5. Since one parameter λ is estimated, \(df=6-1-1=4\). **Answer: 4.** **Convention note:** Zero observed toys in the last category does not justify dropping its positive expected frequency.

**1. Use the valid goodness-of-fit statistic:** Fitting λ=1.86 to the 500 toys and including categories 0,1,2,3,4,≥5 gives df=4 and χ²≈109.41. **2. Compare critical values:** For df=4, the 5% upper critical value is 9.49 and the 2.5% value is 11.14. The statistic exceeds both by a wide margin; its p-value is below 0.005. **3. Select rejection statements:** Reject H0 at significance 5%. The printed 5%-confidence statement also says reject; whether read literally as α=95% or as the intended 5%-significance wording, rejection remains the outcome. **Answer: C, E — C and E.** **Convention note:** The preceding scoring-key statistic is inconsistent, but even its much smaller value near 82 would lead to the same selections. A: **Incorrect:** The statistic is far above the tabulated 0.5% critical value, so p is not between 2.5% and 5%. B: **Incorrect:** Nonrejection contradicts the large statistic. C: **Correct:** Rejection holds under either reading of its unusual confidence-level wording. D: **Incorrect:** 109.41 exceeds the 5% critical value 9.49. E: **Correct:** It correctly rejects at 5% significance. F: **Incorrect:** 12.83 is the 2.5% critical value for df=5, not df=4; the correct value is 11.14. G: **Incorrect:** There are applicable rejection statements.