Business Analytics · Quiz 1 · May 2025

**1. Match visual form to purpose:** Pie slices encode category shares of one total. A pie chart is therefore suited to showing proportions, especially for a small number of mutually exclusive categories. **Answer: C — When proportions in the data need to be shown.** A: **Incorrect:** Outliers are clearer in a dot plot or box plot. B: **Incorrect:** Trends are usually clearer in a line chart. C: **Correct:** Slice areas communicate parts of a whole. D: **Incorrect:** The proportions choice is applicable.

**1. Read the two aligned bar panels:** Each model has one average weight and one average mileage. Model D has the largest weight, 5.430 tons, and smallest mileage, 3.60 km/l; model E has the smallest weight, 1.622 tons, and largest mileage, 16.78 km/l. **2. Identify the most direct summary:** The display compares model-level averages and makes the minimum and maximum across models easy to identify. **Answer: D — D — model-wise comparison and extrema of average mileage and weight.** **Convention note:** The figure does not show minimum/maximum values within each model. Its bars show one average per model; “minimum and maximum” refers to comparing those averages across models. A: **Incorrect:** It mentions only mileage and misses the weight panel. B: **Incorrect:** The inverse pattern is visible, but the graph primarily compares category-level average bars. C: **Incorrect:** It mentions only weights and misses the mileage panel. D: **Correct:** It captures comparisons and extrema of both plotted averages across car models. E: **Incorrect:** A scatter plot would show the numerical weight–mileage relationship more directly. F: **Incorrect:** It treats weight as an ordering axis rather than recognizing the categorical model comparison. G: **Incorrect:** It similarly describes a relationship plot rather than these aligned average bars.

**1. Interpret the moment entries:** Under the intended convention, the second central moment is 4 and the third central moment is 2. **2. Standardize the third moment:** \(\gamma_1=\mu_3/\mu_2^{3/2}=2/4^{3/2}=2/8=0.25>0\), indicating right skew. **Answer: A — Right skewed.** **Convention note:** The table does not label moments as central. Treating all as raw moments is impossible because E[X²]=4 would be less than E[X]²=144. The mean 12, median 6 and central variance 4 are also mutually inconsistent. A is the intended conclusion from the positive third central moment, not validation of the entire summary table. A: **Correct:** A positive standardized third central moment indicates right skew. B: **Incorrect:** The stated third central moment is nonzero. C: **Incorrect:** Left skew requires a negative third central moment. D: **Incorrect:** The second and third central moments already permit the skewness calculation.

**1. Win the semifinal:** Team 1 must beat Team 2, with probability 1/(1+2)=1/3. **2. Average over the final opponent:** Team 3 advances with probability 3/7 and Team 4 with probability 4/7. Conditional on reaching the final, Team 1 wins with probability \((3/7)(1/4)+(4/7)(1/5)=31/140\). **3. Multiply the stages:** \(P(\text{champion})=(1/3)(31/140)=31/420\approx0.073810\to0.07\). **Answer: 0.07.** **Convention note:** The question says “conditional probability” without stating a conditioning event. The key uses the overall championship probability. Conditional on reaching the final, the probability would be 31/140≈0.22.

**1. Divide total probability equally:** The problem stipulates seven equal-area buckets, so each has probability 1/7≈0.142857. **2. Round:** Enter 0.14. **Answer: 0.14.** **Convention note:** Equal probability is stipulated, not a consequence of the listed age intervals. Their unequal widths alone do not establish equal area under a normal density.

**1. Count the observations:** The seven age-bucket frequencies are 25,50,30,50,20,25,50, totaling n=250. **2. Apply the stipulated equal-area null:** Seven equal-probability buckets each have probability 1/7 and expected frequency 250/7≈35.714286. **3. Read the requested bucket:** The 25–30 age group is one of the seven buckets, hence its expected count is 250/7≈35.71. **Answer: 35.71.**

**1. Count the observations:** The seven age-bucket frequencies are 25,50,30,50,20,25,50, totaling n=250. **2. Apply the stipulated equal-area null:** Seven equal-probability buckets each have probability 1/7 and expected frequency 250/7≈35.714286. **3. Sum deviations:** \[\chi^2=\sum_{i=1}^{7}\frac{(O_i-250/7)^2}{250/7}=31.40.\] Using an exact expected count avoids rounding artifacts. **Answer: 31.40.** **Convention note:** This evaluates the explicitly stipulated equal-area null. It does not establish that the printed age cutoffs are actual normal quantiles.

**1. Use the normal-fit goodness-of-fit convention:** There are seven buckets. A fitted normal distribution estimates two parameters, its mean and standard deviation. **2. Subtract constraints and estimated parameters:** \(df=7-1-2=4\). **Answer: 4.** **Convention note:** The key assumes both normal parameters are estimated. If the seven equal cell probabilities are fully specified independently of data and no parameters are fitted, the df is 7−1=6 instead. The passage does not describe parameter estimation explicitly.

**1. Express rejection as a p-value condition:** Rejection at α=0.10 implies p≤0.10. **2. Increase the significance threshold:** The same p-value is also ≤0.50, so H0 is rejected at α=0.50. **Answer: B — Null Hypothesis is rejected.** A: **Incorrect:** The original rejection already bounds the p-value sufficiently. B: **Correct:** Any p-value triggering rejection at 0.10 also triggers it at 0.50. C: **Incorrect:** A larger significance level expands rather than shrinks the rejection region.

**1. Add the bucket frequencies:** \(25+50+30+50+20+25+50=250\). **2. Interpret the total:** Each sale contributes to one bucket, so the analysis sample size is 250, not the number of buckets. **Answer: 250.**

**1. Read the observed counts:** For FC,BL,LG, Type-1/Type-2 counts are (288,124),(622,204),(212,128). Product totals are 412,826,340; store totals 1122,456; grand total 1578. **2. Calculate the independence expectation:** \[E_{BL,2}=\frac{826(456)}{1578}\approx238.692015\to238.69.\] **Answer: 238.69.**

**1. Read the observed counts:** For FC,BL,LG, Type-1/Type-2 counts are (288,124),(622,204),(212,128). Product totals are 412,826,340; store totals 1122,456; grand total 1578. **2. Compute expected pairs:** Under independence they are approximately (292.94,119.06),(587.31,238.69),(241.75,98.25). **3. Sum Pearson terms:** \(\chi^2=\sum(O-E)^2/E\approx20.048532\to20.05\). **Answer: 20.05.**

**1. Form the independence null:** H0 says the distribution of product categories is independent of store-display type. The alternative is association between these two classifications. **2. Distinguish independence from uniformity:** Independence allows unequal product totals and unequal store-type totals; it does not require equal frequencies across cells. **Answer: A — A — product category and store type are independent.** A: **Correct:** It states independence of product category and retail-store type. B: **Incorrect:** This is the alternative, not the null. C: **Incorrect:** Independence does not require uniform cell probabilities. D: **Incorrect:** Equal product shares are not the hypothesis being tested. E: **Incorrect:** Equal store-type shares are not the hypothesis being tested. F: **Incorrect:** The first option supplies the independence null.